RD Sharma Solutions for CBSE Class 10 Mathematics chapter 15 - Statistics

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Chapter 15 - Statistics Excercise Ex. 15.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9



Solution 9

Question 10

 

 

 

 

 

 

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Chapter 15 - Statistics Excercise Ex. 15.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Chapter 15 - Statistics Excercise Ex. 15.3

Question 1



Solution 1

Question 2

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants 0 - 2 2 - 4 4 - 6 6 - 8 8 - 10 10 - 12 12 - 14
Number of houses 1 2 1 5 6 2 3



Which method did you use for finding the mean, and why?

Solution 2

Let us find class marks (xi) for each interval by using the relation.

Now we may compute xi and fixias following

Number of plants Number of houses (fi) xi fixi
0 - 2 1 1 1 x 1 = 1
2 - 4 2 3 2 x 3 = 6
4 - 6 1 5 1 x 5 = 5
6 - 8 5 7 5 x 7 = 35
8 - 10 6 9 6 x 9 = 54
10 - 12 2 11 2 x 11 = 22
12 - 14 3 13 3 x 13 = 39
Total 20   162



From the table we may observe that
 

So, mean number of plants per house is 8.1.
We have used here direct method as values of class marks (xi) and fi are small.

Question 3

Solution 3

Question 4

Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Fine the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per minute 65 - 68 68 - 71
71-74
74 - 77 77 - 80 80 - 83 83 - 86
Number of women 2 4 3 8 7 4 2
Solution 4

We may find class mark of each interval (xi) by using the relation.

Class size h of this data = 3
Now taking 75.5 as assumed mean (a) we may calculate di, ui, fiui as following.

Number of heart beats per minute Number of women fi xi di = xi -75.5 fiui
65 - 68 2 66.5 - 9 - 3 - 6
68 - 71 4 69.5 - 6 - 2 - 8
71 - 74 3 72.5 - 3 - 1 - 3
74 - 77 8 75.5 0 0 0
77 - 80 7 78.5 3 1 7
80 - 83 4 81.5 6 2 8
83 - 86 2 84.5 9 3 6
Total 30       4


Now we may observe from table that

So mean hear beats per minute for these women are 75.9 beats per minute.

Question 5

Solution 5

Question 6

Find the mean of the following frequency distribution:

Solution 6

Question 7

Find the mean of the following frequency distribution:

Solution 7

Question 8

Find the mean of the following frequency distribution:

Solution 8

Question 9

Find the mean of the following frequency distribution:

Solution 9

Question 10

Find the mean of the following frequency distribution:

Solution 10

Question 11

Find the mean of the following frequency distribution:

Solution 11

Question 12

Find the mean of the following frequency distribution:

Solution 12

Question 13

Find the mean of the following frequency distribution:

Solution 13

Question 14

Find the mean of the following frequency distribution:

Solution 14

Question 15

For the following distribution, calculate mean using all suitable methods :

Size of item 1-4 4-9 9-16 16-27
Frequency 6 12 26 20

 

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

The following distribution shows the daily pocket allowance given to the children of a multistorey building. The average pocket allowance is Rs 18.00. Find out the missing frequency.

Solution 19

Question 20

Solution 20

Question 21

In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes 50 - 52 53 - 55 56 - 58 59 - 61 62 - 64
Number of boxes 15 110 135 115 25



Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Solution 21
Number of mangoes Number of boxes
fi
50 - 52 15
53 - 55 110
56 - 58 135
59 - 61 115
62 - 64 25


We may observe that class intervals are not continuous. There is a gap of 1 between two class intervals. So we have to add to upper class limit and subtract from lower class limit of each interval.
And class mark (xi) may be obtained by using the relation

Class size (h) of this data = 3

Now taking 57 as assumed mean (a) we may calculate di, ui, fiui as follows:

 

Class interval fi xi di = xi - 57 fiui
49.5 - 52.5 15 51 -6 -2 -30
52.5 - 55.5 110 54 -3 -1 -110
55.5 - 58.5 135 57 0 0 0
58.5 - 61.5 115 60 3 1 115
61.5 - 64.5 25 63 6 2 50
Total 400       25


Now, we have:
 
Clearly mean number of mangoes kept in a packing box is 57.19.

Note: We have chosen step deviation method here as values of fi, di are big and also there is a common multiple between all di.

Question 22

The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rs) 100 - 150 150 - 200 200 - 250 250 - 300 300 - 350
Number of households 4 5 12 2 2


Find the mean daily expenditure on food by a suitable method.

Solution 22

We may calculate class mark (xi) for each interval by using the relation


Class size = 50

Now taking 225 as assumed mean (a) we may calculate di, ui, fiui as follows:

 

Daily expenditure (in Rs) fi xi di = xi - 225 fiui
100 - 150 4 125 -100 -2 -8
150 - 200 5 175 -50 -1 -5
200 - 250 12 225 0 0 0
250 - 300 2 275 50 1 2
300 - 350 2 325 100 2 4
Total         -7


Now we may observe that -

Question 23

To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

concentration of SO2 (in ppm) Frequency
0.00 - 0.04 4
0.04 - 0.08 9
0.08 - 0.12 9
0.12 - 0.16 2
0.16 - 0.20 4
0.20 - 0.24 2


Find the mean concentration of SO2 in the air.

Solution 23

Concentration of SO2 (in ppm) Frequency Class mark xi di = xi - 0.14 fiui
0.00 - 0.04 4 0.02 -0.12 -3 -12
0.04 - 0.08 9 0.06 -0.08 -2 -18
0.08 - 0.12 9 0.10 -0.04 -1 -9
0.12 - 0.16 2 0.14 0 0 0
0.16 - 0.20 4 0.18 0.04 1 4
0.20 - 0.24 2 0.22 0.08 2 4
Total 30       -31


Question 24

A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.


Number of days

0 - 6

6 - 10

10 - 14

14 - 20

20 - 28

28 - 38

38 - 40

Number of students

11

10

7

4

4

3

1
Solution 24

We may find class mark of each interval by using the relation

 

Now, taking 17 as assumed mean (a) we may calculate di and fidi as follows: 


Number of days

Number of students
fi

xi

di= xi - 17

fidi

0 - 6

11

3

-14

-154

6 -10

10

8

-9

-90

10 - 14

7

12

-5

-35

14 - 20

4

17

0

0

20 - 28

4

24

7

28

28 - 38

3

33

16

48

38 - 40

1

39

22

22

Total

40





-181



Now we may observe that
 sum for blank of straight f subscript straight i equals 40
sum for blank of straight f subscript straight i straight d subscript straight i equals negative 181
straight x subscript straight i equals 17 comma
Mean comma space top enclose straight x equals straight A plus fraction numerator sum for blank of straight f subscript straight i straight d subscript straight i over denominator sum for blank of straight f subscript straight i end fraction
equals 17 plus open parentheses fraction numerator negative 181 over denominator 40 end fraction close parentheses
equals 17 minus 4.525
equals 12.475 space days
So, mean number of days is 12.475 days, for which a student was absent.

Question 25

The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.


Literacy rate
(in %)

45 - 55

55 - 65

65 - 75

75 - 85

85 - 95

Number of cities

3

10

11

8

3
Solution 25

We may find class marks by using the relation



Class size (h) for this data = 10

Now taking 70 as assumed mean (a) we may calculate di, ui, and fiui as follows:  


Literacy rate
(in %)

Number of cities
fi

xi

di= xi - 70

ui =di/10

fiui

45 - 55

3


50

-20

-2

-6

55 - 65

10

60

-10

-1

-10

65 - 75

11

70

0

0

0

75 - 85

8

80

10

1

8


85 - 95

3

90

20

2

6

Total

35







-2



Now we may observe that

So, mean literacy rate is 69.43%.

Question 26

The following is the cumulative frequency distribution (of less than type) of 1000 persons each of age 20 years and above. Determine the mean age.

 

Age below (in years)

30

40

50

60

70

80

Number of persons

100

220

350

750

950

1000

 

Solution 26

Here, we have cumulative frequency distribution less than type. First we convert it into an ordinary frequency distribution.

Question 27

If the means of the following frequency distribution is 18, find the missing frequency.

 

Class interval:

11-13

13-15

15-17

17-19

19-21

21-23

23-25

Frequency:

3

6

9

13

f

5

4

 

Solution 27

Question 28

Find the missing frequencies in the following distribution, if the sum of the frequencies is 120 and the mean is 50.

 

Class:

0-20

20-40

40-60

60-80

80-100

Frequency:

17

f1

32

f2

19

 

Solution 28

Question 29

The daily income of a sample of 50 employees are tabulated as follows:

 

Income (in Rs.):

1-200

201-400

401-600

601-800

No. of employees:

14

15

14

7

 

Find the mean daily income of employees.

Solution 29

 

Chapter 15 - Statistics Excercise Ex. 15.4

Question 1

Solution 1

Question 2

Solution 2

The median height of the students is Rs 167.13.

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Find the following tables gives the distribution of the life time of 400 neon lamps:


Life time (in hours)

Number of lamps

1500 - 2000

14

2000 - 2500

56

2500 - 3000

60

3000 - 3500

86

3500 - 4000

74

4000 - 4500

62

4500 - 5000

48



Find the median life time of a lamp.

Solution 8

We can find cumulative frequencies with their respective class intervals as below -


Life time

Number of lamps (fi)

Cumulative frequency

1500 - 2000

14

14

2000 - 2500

56

14 + 56 = 70

2500 - 3000

60

70 + 60 = 130

3000 - 3500

86

130 + 86 = 216

3500 - 4000

74

216 + 74 = 290

4000 - 4500

62

290 + 62 = 352

4500 - 5000

48

352 + 48 = 400

Total (n)

400
 



Now we may observe that cumulative frequency just greater than is 216 belonging to class interval 3000 - 3500.
Median class = 3000 - 3500
Lower limit (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size (h) = 500


So, median life time of lamps is 3406.98 hours.

Question 9

The distribution below gives the weights of 30 students of a class. Find the median weight of the students.


Weight
(in kg)

40 - 45

45 - 50

50 - 55

55 - 60

60 - 65

65 - 70

70 - 75

Number of students

2


3

8

6

6

3

2
Solution 9

We may find cumulative frequencies with their respective class intervals as below

Weight (in kg) 40 - 45 45 - 50 50 - 55 55 - 60 60 - 65 65 - 70 70 - 75
Number of students (f) 2 3 8 6 6 3 2
c.f. 2 5 13 19 25 28 30


Cumulative frequency just greater than   is 19, belonging to class interval 55 - 60.
Median class = 55 - 60
Lower limit (l) of median class = 55
Frequency (f) of median class = 6
Cumulative frequency (cf) of median class = 13
Class size (h) = 5



So, median weight is 56.67 kg.

Question 10

Solution 10

Question 11

You are given that the median value is 46 and the total number of items is 230.
(i) Using the median formula fill up missing frequencies.
(ii) Calculate the AM of the completed distribution.
Solution 11

Question 12

Solution 12



Question 13

Solution 13



Question 14

Solution 14



Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.


Age (in years)

Number of policy holders

Below 20

2

Below 25

6

Below 30

24

Below 35

45

Below 40

78

Below 45

89

Below 50

92

Below 55

98

Below 60

100
Solution 18

Here class width is not same. There is no need to adjust the frequencies according to class intervals. Now given frequency table is of less than type represented with upper class limits. As policies were given only to persons having age 18 years onwards but less than 60 years, we can define class intervals with their respective cumulative frequency as below


Age (in years)

Number of policy holders (fi)

Cumulative frequency (cf)

18 - 20

2

2

20 - 25

6 - 2 = 4

6

25 - 30

24 - 6 = 18

24

30 - 35

45 - 24 = 21

45

35 - 40

78 - 45 = 33

78

40 - 45

89 - 78 = 11

89
45 - 50
92 - 89 = 3

92

50 - 55

98 - 92 = 6

98

55 - 60

100 - 98 = 2

100

Total (n)

 





Now from table we may observe that n = 100.

Cumulative frequency (cf) just greater than  is 78 belonging to interval 35 - 40
So, median class = 35 - 40
Lower limit (l) of median class = 35
Class size (h) = 5
Frequency (f) of median class = 33
Cumulative frequency (cf) of class preceding median class = 45


So, median age is 35.76 years.

Question 19

The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:


Length (in mm)

Number or leaves fi

118 - 126

3

127 - 135

5

136 - 144

9

145 - 153

12

154 - 162

5

163 - 171

4

172 - 180

2



Find the median length of the leaves.

Solution 19

The given data is not having continuous class intervals. We can observe that difference between two class intervals is 1. So, we have to add and subtract

  to upper class limits and lower class limits.
Now continuous class intervals with respective cumulative frequencies can be represented as below:

 


Length (in mm)

Number or leaves fi

Cumulative frequency

117.5 - 126.5

3

3

126.5 - 135.5

5

3 + 5 = 8

135.5 - 144.5

9

 


8 + 9 = 17

144.5 - 153.5

12

17 + 12 = 29

153.5 - 162.5

5

 


29 + 5 = 34

162.5 - 171.5

4

34 + 4 = 38

171.5 - 180.5

2

38 + 2 = 40



From the table we may observe that cumulative frequency just greater then
  is 29, belonging to class interval 144.5 - 153.5.
Median class = 144.5 - 153.5
Lower limit (l) of median class = 144.5
Class size (h) = 9
Frequency (f) of median class = 12
Cumulative frequency (cf) of class preceding median class = 17



So, median length of leaves is 146.75 mm.

Question 20

Solution 20

Question 21

The median of the distribution given below is 14.4. Find the values of x and y, if the total frequency is 20.

 

Class interval:

0-6

6-12

12-18

18-24

24-30

Frequency:

4

x

5

y

1

 

Solution 21

Question 22

The median of the following data is 50. Find the values of p and q, if the sum of all the frequencies is 90.

 

Marks:

20-30

30-40

40-50

50-60

60-70

70-80

80-90

Frequency:

p

15

25

20

q

8

10

 

Solution 22

Chapter 15 - Statistics Excercise Ex. 15.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

The following table shows the ages of the patients admitted in a hospital during a year:


Age (in years)

5 - 15

15 - 25

25 - 35

35 - 45

45 - 55

55 - 65

Number of patients

6

11

21

23

14

5



Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution 9

We may compute class marks (xi) as per the relation

 
Now taking 30 as assumed mean (a) we may calculate di and fidi as follows.


Age (in years)

Number of patients
fi

class mark
xi

di= xi - 30

fidi

5 - 15

6

10

-20

-120

15 - 25

11

20

-10

-110

25 - 35

21

30

0

0

35 - 45

23

40

10

230

45 - 55

14

50

20

280

55 - 65

5

60

30

150

Total

80





430



From the table we may observe that


Clearly, mean of this data is 35.38. It represents that on an average the age of a patient admitted to hospital was 35.38 years.
As we may observe that maximum class frequency is 23 belonging to class interval 35 - 45.
So, modal class = 35 - 45
Lower limit (l) of modal class = 35
Frequency (f1) of modal class = 23
Class size (h) = 10
Frequency (f0) of class preceding the modal class = 21
Frequency (f2) of class succeeding the modal class = 14


Clearly mode is 36.8.It represents that maximum number of patients admitted in hospital were of 36.8 years.

Question 10

The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:


Lifetimes (in hours)

0 - 20

20 - 40

40 - 60

60 - 80

80 - 100

100 - 120

Frequency

10

35

52

61

38

29



Determine the modal lifetimes of the components.

Solution 10

From the data given as above we may observe that maximum class frequency is 61 belonging to class interval 60 - 80.
So, modal class = 60 - 80
Lower class limit (l) of modal class = 60
Frequency (f1) of modal class = 61
Frequency (f0) of class preceding the modal class = 52
Frequency (f2) of class succeeding the modal class = 38
Class size (h) = 20


 
So, modal lifetime of electrical components is 65.625 hours.

Question 11

Solution 11

Question 12

The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.


Number of students
per teacher

Number of
states/U.T

15 - 20

3

20 - 25

8

25 - 30

9

30 - 35

10

35 - 40

3

40 - 45

0

45 - 50

0

50 - 55

2
Solution 12

We may observe from the given data that maximum class frequency is 10 belonging to class interval 30 - 35.
So, modal class = 30 - 35
Class size (h) = 5
Lower limit (l) of modal class = 30
Frequency (f1) of modal class = 10
Frequency (f0) of class preceding modal class = 9
Frequency (f2) of class succeeding modal class = 3


It represents that most of states/U.T have a teacher - student ratio as 30.6

Now we may find class marks by using the relation


Now taking 32.5 as assumed mean (a) we may calculate di, ui and fiui as following.


Number of students
per teacher

Number of states/U.T
(fi)

xi

di = xi - 32.5

 ui

fiui

15 - 20

3

17.5

-15

-3

-9

20 - 25

8

22.5

-10

-2

-16

25 - 30

9

27.5

-5

-1

-9

30 - 35

10

32.5

0

0

0

35 - 40

3

37.5

5

1

3

40 - 45

0

42.5

10

2

0

45 - 50

0

47.5

15

3

0

50 - 55

2

52.5

20

4

 


8

Total

35

 

     
-23





So mean of data is 29.2
It represents that on an average teacher - student ratio was 29.2.

Question 13

Solution 13

Question 14

A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:


Number
of cars

0 - 10

10 - 20

20 - 30

30 - 40

40 - 50

50 - 6

60 - 70

70 - 80

Frequency

7

14

13

12

20

11

15

8
Solution 14

From the given data we may observe that maximum class frequency is 20 belonging to 40 - 50 class intervals.
So, modal class = 40 - 50
Lower limit (l) of modal class = 40
Frequency (f1) of modal class = 20
Frequency (f0) of class preceding modal class = 12
Frequency (f2) of class succeeding modal class = 11
Class size = 10

So mode of this data is 44.7 cars.

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.


Expenditure
(in Rs)

Number of families

 


1000 - 1500

24

1500 - 2000

40

2000 - 2500

33

2500 - 3000

28

3000 - 3500

30

3500 - 4000

22

4000 - 4500

16

4500 - 5000

7
Solution 18

We may observe from the given data that maximum class frequency is 40 belonging to 1500 - 2000 intervals.
So, modal class = 1500 - 2000
Lower limit (l) of modal class = 1500
Frequency (f1) of modal class = 40
Frequency (f0) of class preceding modal class = 24
Frequency (f2) of class succeeding modal class = 33
Class size (h) = 500

 

So modal monthly expenditure was Rs. 1847.83
Now we may find class mark as


Class size (h) of give data = 500
Now taking 2750 as assumed mean (a) we may calculate di, ui and fiui as follows:


Expenditure
(in Rs)

Number of families

fi


xi

di = xi - 2750

 ui

fiui

 


1000 - 1500

24

1250

-1500

-3

-72

1500 - 2000

40

1750

-1000

-2

-80

2000 - 2500

33

2250

-500

-1

 

 

-33


2500 - 3000

28

2750

0

0

0

3000 - 3500

30

3250

500

1

30

3500 - 4000

22

3750

1000

2

44

4000 - 4500

16

4250

1500

3

48

4500 - 5000

7

4750

2000

4

28

Total

200

 

 

 

-35



Now from table may observe that


So, mean monthly expenditure was Rs. 2662.50.

Question 19

The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scored
No of batsman

3000 - 4000

4

4000 - 5000

18

5000 - 6000

9

6000 - 7000

7

7000 - 8000

6

8000 - 9000

3

9000 -10000

1

10000 - 11000

1



Find the mode of the data.

Solution 19

From the given data we may observe that maximum class frequency is 18 belonging to class interval 4000 - 5000.
So, modal class = 4000 - 5000
Lower limit (l) of modal class = 4000
Frequency (f1) of modal class = 18
Frequency (f0) of class preceding modal class = 4
Frequency (f2) of class succeeding modal class = 9
Class size (h) = 1000

So mode of given data is 4608.7 runs.

Question 20

The frequency distribution table of agriculture holdings in a village is given below:

 

Area of land (in hectares):

1-3

3-5

5-7

7-9

9-11

11-13

Number of families:

20

45

80

55

40

12

Find the modal agriculture holdings of the village.

Solution 20

Question 21

The monthly income of 100 families are given as below:

 

Income in (in Rs.)

Number of families

0-5000

8

5000-10000

26

10000-15000

41

15000-20000

16

20000-25000

3

25000-30000

3

30000-35000

2

35000-40000

1

Calculate the modal income.

Solution 21

Chapter 15 - Statistics Excercise Ex. 15.6

Question 1

Solution 1



Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6















Question 7

Solution 7

Question 8

The annual rainfall record of a city for 66 days if given in the following table:

 

Rainfall (in cm):

0-10

10-20

20-30

30-40

40-50

50-60

Number of days:

22

10

8

15

5

6

Calculate the median rainfall using ogives of more than type and less than type.

 

Solution 8

Less Than Series:

Class interval

Cumulative Frequency

Less than 10

22

Less than 20

32

Less than 30

40

Less than 40

55

Less than 50

60

Less than 60

66

 

We plot the points (10, 22), (20, 32), (30, 40), (40, 55), (50, 60) and (60, 66) to get 'less than type' ogive.

 

More Than Series:

Class interval

Frequency

More than 0

66

More than 10

44

More than 20

34

More than 30

26

More than 40

11

More than 50

6

 

We plot the points (0, 66), (10, 44), (20, 24), (30, 26), (40, 11), and (50, 6) to get more than ogive.

 

From the graph, median = 21.25 cm

Question 9

Solution 9

Question 10

Solution 10

Chapter 15 - Statistics Excercise 15.66

Question 1

Which of the following is not a measure of central tendency?

(a) Mean

(b) Median

(c) Mode

(d) Standard deviation

Solution 1

There are three main measure of central tendency the mode, the median and the mean.

Each of these measures describes a different indication of the typical or central value in the distribution.

The mode is the most commonly occuring value in a distribution.

Median is middle value of distribution.

While standard deviation is a measure of dispersion of a set of data from its mean.

So, the correct option is (d).

Question 2

The algebraic sum of deviations of a frequency distribution from its mean is

(a) always positive

(b) always negative

(c) 0

(d) a non-zero number

Solution 2

begin mathsize 12px style Let space the space mean space be space space straight x with bar on top space for space any space distribution
If space straight n space terms space are space straight a subscript 1 comma space straight a subscript 2 space comma space.......... straight a subscript straight n
deviation space of space data space from space mean space colon space open parentheses straight a subscript 1 space minus space straight x with bar on top close parentheses comma space open parentheses straight a subscript 2 space minus space straight x with bar on top close parentheses comma space......... open parentheses straight a subscript straight n space minus space straight x with bar on top close parentheses
algebraic space sum space equals space open parentheses straight a subscript 1 space minus space straight x with bar on top close parentheses space plus space open parentheses straight a subscript 2 space minus space straight x with bar on top close parentheses space plus space space......... open parentheses straight a subscript straight n space minus space straight x with bar on top close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open parentheses straight a subscript 1 space plus space straight a subscript 2 space plus space........ straight a subscript straight n close parentheses space minus space straight n straight x with bar on top space space space space space space space space space space space space space....... open parentheses 1 close parentheses
We space also space know comma space space space straight x with bar on top space space equals space fraction numerator open parentheses straight a subscript 1 space plus space straight a subscript 2 space plus space....... straight a subscript straight n close parentheses over denominator straight n end fraction space space space space space space space space space space space space space space space....... open parentheses 2 close parentheses
from space open parentheses 1 close parentheses space & space open parentheses 2 close parentheses
algebraic space sum space equals space 0
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 3

begin mathsize 12px style The space arithmetic space mean space of space 1 comma space 2 comma space 3 comma space..... space straight n space is space
open parentheses straight a close parentheses space fraction numerator straight n space plus space 1 over denominator 2 end fraction
left parenthesis straight b right parenthesis space fraction numerator straight n space minus space 1 over denominator 2 end fraction
left parenthesis straight c right parenthesis space straight n over 2
left parenthesis straight d right parenthesis space straight n over 2 space plus space 1 end style

Solution 3

begin mathsize 12px style arithmetic space mean space equals space fraction numerator sum space of space all space numbers over denominator total space numbers end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 1 space plus space 2 space plus space 3 space plus space...... space straight n over denominator straight n end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator begin display style fraction numerator straight n open parentheses straight n space plus space 1 close parentheses over denominator 2 end fraction end style over denominator straight n end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator straight n space plus space 1 over denominator 2 end fraction
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 4

For a frequency distribution, mean, median and mode are connected by the relation

(a) Mode = 3 mean - 2 median

(b) Mode = 2 median - 3 mean

(c) Mode = 3 median - 2 mean

(d) Mode = 3 median + 2 mean

Solution 4

It is well known that relation between mean, median and mode 1 s

3 median = mode + 2 mean

Mode = 3 median - 2 mean

So, the correct option is (a).

Question 5

Which of the following cannot be determined graphically?

(a) Mean

(b) Median

(c) Mode

(d) None of these

Solution 5

Mean is an average value of any given data which cannot be determined by a graph.

Value of median and mode can easily be calculated by graph.

Median is middle value of a distribution and mode is highest frequent value of a given distribution.

So, the correct option is (a).

Question 6

The median of a given frequency distribution is found graphically with the help of

(a) Histogram

(b) Frequency curve

(c) Frequency polygon

(d) Ogive

Solution 6

The median of a series may be determined through the graphical presentation of data in the forms of Ogives.

Ogive is a curve showing the cummulative frequency for a given set of data.

To get the median we present the data graphically in the form of 'less than' ogive  or 'more than' ogive

Then the point of intersection of the two graphs gives the value of the median.

So, the correct option is (d).

Question 7

The mode of a frequency distribution can be detremined graphically from

(a) Histogram

(b) frequency polygon

(c) ogive

(d) frequency curve

Solution 7

Histogram is used to plot the distribution of numerical data or frequency of occurrences of data.

Mode is the most commonly occurring value in the data.

So in distribution or Histogram, the value of the x-coordinate corresponding to the peak value on y - axis, is the mode.

So, the correct option is (a).

Question 8

Mode is

(a) least frequent value

(b) middle most value

(c) most frequent value

(d) None of these

Solution 8

Mode is the most frequent value in the data.

Mode is the value which occurs the most number of times.

So, the correct option is (c).

Question 9

begin mathsize 12px style The space mean space of space straight n space observation space is space straight x with bar on top. space If space the space first space item space is space increased space by space 1 comma space second space by space 2 space and space so space on comma space then space the space new space mean space is
left parenthesis straight a right parenthesis space straight x with bar on top space plus space straight n
left parenthesis straight b right parenthesis space straight x with bar on top space plus space straight n over 2
left parenthesis straight c right parenthesis space straight x with bar on top space plus space fraction numerator straight n plus 1 over denominator 2 end fraction
left parenthesis straight d right parenthesis space None space of space these end style

Solution 9

begin mathsize 12px style text Let sum of observation is s end text
table attributes columnalign left end attributes row cell rightwards double arrow straight space straight x with bar on top equals straight s over straight n end cell row cell rightwards double arrow space straight s equals space straight x with bar on top straight n space space space space space........ left parenthesis 1 right parenthesis end cell row cell If space the space first space item space is space increased space by space 1 comma space second space by space 2 space and space so space on comma space Then end cell row cell sum space of space new space straight n space obervations space equals space straight s space plus space 1 space plus space 2 space plus 3...... plus straight n end cell row cell                                                equals space straight s space plus space fraction numerator straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction end cell row cell New space mean space space space space equals space fraction numerator straight s space plus begin display style fraction numerator straight n left parenthesis straight n space plus space 1 right parenthesis over denominator 2 end fraction end style straight space over denominator straight n end fraction    end cell row cell                           =   straight s over straight n  +  fraction numerator straight n plus 1 over denominator 2 end fraction end cell row cell                           =   straight x with bar on top   +   fraction numerator straight n plus 1 over denominator 2 end fraction end cell end table
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 10

One of the methods of determining mode is

(a) Mode = 2 median - 3 Mean

(b) Mode = 2 Median + 3 Mean

(c) Mode = 3 Median - 2 Mean

(d) Mode = 3 Median + 2 Mean

Solution 10

We know that the relation between mean, median & mode is

3 Median = Mode + 2 Mean

Hence, Mode = 3 Median - 2 Mean

So, the correct option is (c).

Chapter 15 - Statistics Excercise 15.67

Question 1

If the mean of the following distribution is 2.6, then the value of y is

Variable (y) : 1   2   3   4   5

Frequency :   4   5    y   1   2

(a) 3    (b) 8     (c) 13    (d) 24

Solution 1

begin mathsize 12px style mean space equals space fraction numerator begin display style sum from straight i space equals space 1 to 5 of straight x subscript 1 straight f subscript straight i end style over denominator begin display style sum from straight i space equals space 1 to 5 of straight f subscript straight i end style end fraction
space space space space space space space space space space space space space equals space fraction numerator open parentheses 1 space cross times space 4 close parentheses space plus space open parentheses 2 space cross times space 5 close parentheses space plus space open parentheses 3 space cross times space straight y close parentheses space plus space open parentheses 4 space cross times space 1 close parentheses space plus space open parentheses 5 space cross times space 2 close parentheses over denominator 4 space plus space 5 space plus space straight y space plus space 1 space plus space 2 end fraction
space space space space space space 2.6 equals space fraction numerator 3 straight y space plus space 28 over denominator straight y space plus space 12 end fraction
rightwards double arrow space 2.6 space straight y space plus space 31.2 space equals space 3 straight y space plus space 28
rightwards double arrow space 3.2 space equals 0.4 straight y
rightwards double arrow space straight y space equals space 8
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 2

The relationship between mean, median and mode for a moderately skewed distribution is

(a) Mode = 2 Median - 3 Mean

(b) Mode = Median - 2 Mean

(c) Mode = 2 Median - Mean

(d) Mode = 3 Median - 2 mean

Solution 2

We know that the relation between mean, median & mode is

3 Median = mode + 2 Mean

Hence, mode = 3 Median - 2 Mean

So, the correct option is (d).

Question 3

begin mathsize 12px style text The mean of a discrete frequency distribution  xi/ fi; i=1, 2, .........., n is given by end text
open parentheses straight a close parentheses space fraction numerator sum space straight f subscript straight i straight x subscript straight i over denominator sum space straight f subscript straight i end fraction
open parentheses straight b close parentheses space 1 over straight n sum from straight i equals 1 to straight n of straight x subscript straight i space straight f subscript straight i
left parenthesis straight c right parenthesis space fraction numerator begin display style sum from straight i equals 1 to straight n of end style space straight x subscript straight i space straight f subscript 1 over denominator begin display style sum from straight i equals 1 to straight n of end style space straight x subscript straight i end fraction
left parenthesis straight d right parenthesis space fraction numerator begin display style sum from straight i equals 1 to straight n of end style space straight x subscript straight i space straight f subscript straight i over denominator begin display style sum from straight i equals 1 to straight n of end style space straight i end fraction end style

Solution 3

begin mathsize 12px style We space know space for space straight a space discrete space frequency space distribution
Mean space equals space fraction numerator straight x subscript 1 straight f subscript 1 space plus space straight x subscript 2 straight f subscript 2 space plus space.......... space plus space straight x subscript straight n straight f subscript straight n over denominator straight f subscript 1 space plus space straight f subscript 2 space plus space........ space plus space straight f subscript straight n end fraction
space space space space space space space space space space space space space space space equals space fraction numerator begin display style sum from straight i equals 1 to straight n of straight x subscript straight i straight f subscript straight i end style over denominator begin display style sum from straight i equals 1 to straight n of straight f subscript straight i end style end fraction
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 4

If the arithmetic mean of xx +3 , x + 6, x + 9, x + 12 is 10, then x =

(a) 1

(b) 2

(c) 6

(d) 4

Solution 4

begin mathsize 12px style mean space equals space fraction numerator straight x space plus space open parentheses straight x space plus space 3 close parentheses space plus space open parentheses straight x space plus space 6 close parentheses space plus space open parentheses straight x space plus space 9 close parentheses space plus space open parentheses straight x space plus space 12 close parentheses over denominator 5 end fraction
space space space space space space space space space space space space space equals space fraction numerator 5 straight x space plus space 30 over denominator 5 end fraction
space space space space space space space space space space space space space equals space straight x space plus space 6
given space mean space equals space 10
rightwards double arrow space straight x space plus space 6 space equals space 10
rightwards double arrow space straight x space equals space 4
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 5

If the median of the data : 24, 25, 26, x + 2, x + 3, 30, 31, 34 is 27.5 then x =

(a) 27

(b) 25

(c) 28

(d) 30

Solution 5

begin mathsize 12px style There space are space straight a space total space of space 8 space observations
median space is space average space of space 5 to the power of th space and space 4 to the power of th space term
4 to the power of th space term space equals space straight x space plus space 2
5 to the power of th space end exponent term space equals space straight x space plus space 3
rightwards double arrow median space equals space fraction numerator open parentheses straight x space plus space 2 close parentheses space plus space open parentheses straight x space plus space 3 close parentheses over denominator 2 end fraction equals space fraction numerator 2 straight x space plus space 5 over denominator 2 end fraction space equals space straight x space plus space 2.5
given space median space equals space 27.5
Hence space straight x space plus space 2.5 space equals space 27.5
straight x space equals space 25
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 6

If the median of the data: 6, 7, x - 2, x, 17, 20 written in ascending order, is 16. Then x =

(a) 15

(b) 16

(c) 17

(d) 18

 

Solution 6

begin mathsize 12px style Given space data space has space 6 space numbers
so comma space median space is space average space of space 3 to the power of rd space and space 4 to the power of th space term
3 to the power of rd space term space equals space straight x space minus space 2
4 to the power of th space term space equals space straight x
median space equals space fraction numerator straight x space plus space straight x space minus space 2 over denominator 2 end fraction
space space space space space space space space space space space space space space space space space equals fraction numerator 2 straight x space minus space 2 over denominator 2 end fraction
space space space space space space space space space space space space space space space space space equals space straight x space minus space 1
given space median space equals space 16
straight x space minus space 1 space equals space 16
straight x space equals space 17
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 7

The median of first 10 prime numbers is

(a) 11

(b) 12

(c) 13

(d) 14

Solution 7

begin mathsize 12px style First space 10 space prime space numbers space are space 2 comma space 3 comma space 5 comma space 7 comma space 11 comma space 13 comma space 17 comma space 19 comma space 23 comma space 29
Median space is space average space of space 5 to the power of th space and space 6 to the power of th space term
rightwards double arrow space 1 half open parentheses 11 space plus space 13 close parentheses
rightwards double arrow space 12
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 8

If the mode of the data : 64, 60, 48, x, 43, 48, 43, 34 is 43 then x + 3 =

(a) 44

(b) 45

(c) 46

(d) 48

 

Solution 8

Mode is the number in observation data is that which repeats most number of time

In the given data 48 comes twice and 43 comes twice but mode is 43.

Hence if x = 43 then 43 comes thrice.

So x + 3 = 43 + 3 = 46

So, the correct option is (c).

Question 9

If the mode of the data : 16, 15, 17, 16, 15, x, 19, 17, 14 is 15 then x =

(a) 15

(b) 16

(c) 17

(d) 19

Solution 9

In the given data 15, 16, 17 comes twice but given 15 is mode.

Hence 15 comes more than 16, 17.

This is only possible if x = 15.

So, the correct option is (a).

Question 10

The mean of 1, 3, 4, 5, 7, 4 is m. The numbers 3, 2, 2, 4, 3, 3, p have mean m - 1 and median q. Then, p + q =

(a) 4

(b) 5

(c) 6

(d) 7

Solution 10

begin mathsize 12px style mean space of space 1 comma space 3 comma space 4 comma space 5 comma space 7 comma space 4
rightwards double arrow space fraction numerator 1 space plus space 3 space plus space 4 space plus space 5 space plus space 7 space plus space 4 over denominator 6 end fraction space equals space straight m space space space space space space space open parentheses given close parentheses
rightwards double arrow straight m space equals space 24 over 6
rightwards double arrow straight m space equals space 4
mean space of space 3 comma space 2 comma space 2 comma space 4 comma space 3 comma space 3 comma space straight p
rightwards double arrow space fraction numerator 3 space plus space 2 space plus space 2 space plus space 4 space plus space 3 space plus space 3 space plus space straight p over denominator 7 end fraction equals space rightwards double arrow space straight m space minus space 1 space space left parenthesis given right parenthesis
rightwards double arrow fraction numerator 17 space plus space straight p over denominator 7 end fraction equals space 4 space minus space 1
rightwards double arrow 17 space plus space straight p space equals space 21
rightwards double arrow straight p space equals space 4
so space data space is space 2 comma space 2 comma space 3 comma space 3 comma space 3 comma space 4 comma space 4
median space of space this space is space straight q space left parenthesis given right parenthesis
But space for space this space data space median space is space 3
Hence space straight q space equals space 3
rightwards double arrow straight p space plus space straight q space equals space 7
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 11

begin mathsize 12px style If space the space mean space of space straight a space frequency space distribution space is space 8.1 space and space sum for blank of straight f subscript straight i straight x subscript straight i space equals space 132 space plus space 5 straight k comma space sum for blank of straight f subscript straight i space equals space 20 comma space then space straight k space equals space
left parenthesis straight a right parenthesis space 3
left parenthesis straight b right parenthesis space 4
left parenthesis straight c right parenthesis space 5
left parenthesis straight d right parenthesis space 6 end style

Solution 11

begin mathsize 12px style mean space equals space fraction numerator sum straight f subscript straight i straight x subscript straight i over denominator sum straight f subscript straight i end fraction
rightwards double arrow 8.1 space equals space fraction numerator 132 space plus space 5 straight k over denominator 20 end fraction
rightwards double arrow 162 space equals space 132 space plus space 5 straight k
rightwards double arrow space 5 straight k space equals space 30
rightwards double arrow straight k space equals space 6
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 12

If the mean of 6, 7, x , 8, y, 14 is 9, then

(a) x + y = 21

(b) x + y = 19

(c) x - y = 19

(d) x - y = 21

Solution 12

begin mathsize 12px style 9 equals space fraction numerator 6 space plus space 7 space plus space straight x space plus space 8 space plus space straight y space plus space 14 over denominator 6 end fraction
rightwards double arrow 54 space equals space straight x space plus space straight y space plus space 35
rightwards double arrow straight x space plus space straight y space equals space 19
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 13

begin mathsize 12px style The space mean space of space straight n space observation space is space straight x with bar on top. space If space the space first space observation space is space increased space by space 1 comma space the space second space by space 2 comma space the space third space by space 3 comma space and space so space on comma space then space the space new space mean space is
left parenthesis straight a right parenthesis space straight x with bar on top space plus space open parentheses 2 straight n space plus space 1 close parentheses
open parentheses straight b close parentheses space straight x with bar on top space plus space fraction numerator straight n space plus space 1 over denominator 2 end fraction
open parentheses straight c close parentheses space straight x with bar on top space plus space open parentheses straight n space plus space 1 close parentheses
open parentheses straight d close parentheses space straight x with bar on top minus fraction numerator straight n space plus space 1 over denominator 2 end fraction end style

Solution 13

begin mathsize 12px style Let space sum space of space straight n space observation space is space apostrophe straight s apostrophe
rightwards double arrow straight x with bar on top space equals space straight s over straight n
rightwards double arrow straight s space equals space straight x with bar on top straight n space space space space space space space...... left parenthesis 1 right parenthesis
If space the space first space item space is space increased space by space 1 comma space second space by space 2 space and space so space on comma space then space sum space of space new space straight n space observations space
equals space straight s space plus space 1 space plus space 2 space plus space 3 space plus space........... straight n
equals space straight s space plus space fraction numerator straight n open parentheses straight n space plus space 1 close parentheses over denominator 2 end fraction
New space mean space equals space fraction numerator straight s space plus begin display style fraction numerator straight n open parentheses straight n space plus space 1 close parentheses over denominator 2 end fraction end style over denominator straight n end fraction
space space space space space space space space space space space space space space space space space space space space space space space equals space straight s over straight n space plus space fraction numerator straight n space plus space 1 over denominator 2 end fraction
space space space space space space space space space space space space space space space space space space space space space space space equals straight x with bar on top space plus space fraction numerator straight n space plus space 1 over denominator 2 end fraction
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 14

begin mathsize 12px style If space the space mean space of space first space straight n space natural space numbers space is space fraction numerator 5 straight n over denominator 9 end fraction space then space straight n space equals
left parenthesis straight a right parenthesis space 5
left parenthesis straight b right parenthesis space 4
left parenthesis straight c right parenthesis space 9
left parenthesis straight d right parenthesis space 10 end style

Solution 14

begin mathsize 12px style First space straight n space natural space numbers space 1 comma space 2 comma space 3 space.......... straight n
mean space equals space fraction numerator 1 space plus space 2 space plus space 3 space plus space....... plus space straight n over denominator straight n end fraction
mean space equals space fraction numerator begin display style fraction numerator straight n open parentheses straight n space plus space 1 close parentheses over denominator 2 end fraction end style over denominator straight n end fraction
space space space space space space space space space space space space space equals space fraction numerator straight n space plus space 1 over denominator 2 end fraction
given space mean space equals space fraction numerator 5 straight n over denominator 9 end fraction
rightwards double arrow fraction numerator straight n space plus space 1 over denominator 2 end fraction space equals space fraction numerator 5 straight n over denominator 9 end fraction
rightwards double arrow 9 straight n space plus space 9 space equals space 10 straight n
rightwards double arrow space straight n space equals space 9
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Chapter 15 - Statistics Excercise 15.68

Question 1

The arithmetic mean and mode of a data are 24 and 12 respectively, then its median is

(a) 25

(b) 18

(c) 20

(d) 22

Solution 1

We know, 3 Median = Mode + 2 Mean

mean = 24

mode = 12

3 median = 12 + 2 × 24

              = 12 + 48

              = 60

median = 20

So, the correct option is (c).

Question 2

begin mathsize 12px style The space mean space of space first space straight n space odd space natural space number space is
open parentheses straight a close parentheses space fraction numerator straight n space plus space 1 over denominator straight d end fraction
open parentheses straight b close parentheses space straight n over 2
open parentheses straight c close parentheses space straight n
open parentheses straight d close parentheses space straight n squared end style

Solution 2

begin mathsize 12px style First space straight n space odd space natural space number space 1 comma space 3 comma space 5 comma space 7 comma space......
straight S space equals space 1 space plus space 3 space plus space 5 space plus space 7 space plus space........ space straight n space terms
This space is space an space AP space with space straight a space equals space 1 space and space straight d space equals space 2
straight S space equals space straight n over 2 open parentheses 2 space plus space open parentheses straight n space minus space 1 close parentheses space 2 close parentheses
space space space space equals space straight n over 2 open parentheses 2 straight n close parentheses
space space space space equals space straight n squared
mean space equals space straight S over straight n equals straight n squared over straight n equals space straight n
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. space space space space space space space space space space space end style

Question 3

begin mathsize 12px style The space mean space of space first space straight n space odd space natural space numbers space is space straight n squared over 81 comma space then space straight n space equals
open parentheses straight a close parentheses space 9
open parentheses straight b close parentheses space 81
open parentheses straight c close parentheses space 27
open parentheses straight d close parentheses space 18 end style

Solution 3

begin mathsize 12px style First space straight n space odd space natural space number space 1 comma space 3 comma space 5 comma space 7 comma space......
straight S space equals space 1 space plus space 3 space plus space 5 space plus space 7 space plus space........ space straight n space terms
This space is space an space AP space with space straight a space equals space 1 space and space straight d space equals space 2
straight S space equals space straight n over 2 open parentheses 2 space plus space open parentheses straight n space minus space 1 close parentheses space 2 close parentheses
space space space space equals space straight n over 2 open parentheses 2 straight n close parentheses
space space space space equals space straight n squared
mean space equals space straight S over straight n equals space straight n over straight n squared equals space straight n
Mean space of space first space straight n space odd space natural space numbers space equals space straight n
given space straight n squared over 81 space equals space straight n
rightwards double arrow space straight n space equals space 81
So comma space the space correct space option space is space left parenthesis straight b right parenthesis.
end style

Question 4

If the difference of mode and median of a data is 24, then the difference of median and mean is

(a) 12

(b) 24

(c) 8

(d) 36

Solution 4

begin mathsize 12px style We space know comma space 3 space median space equals space mode space plus space 2 space mean
rightwards double arrow 2 space median space minus space 2 space mean space equals space mode space minus space median
rightwards double arrow 2 open parentheses median space minus space mean close parentheses space equals space mode space minus space median
given space mode space minus space median space equals space 24
rightwards double arrow 2 open parentheses median space minus space mean close parentheses space equals space 24
rightwards double arrow median space minus space mean space equals space 12
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 5

If the arithmetic mean of 7, 8, x, 11, 14 is x then x = 

(a) 1

(b) 9.5

(c) 10

(d) 10.5

Solution 5

begin mathsize 12px style mean space equals space fraction numerator 7 space plus space 8 space plus space straight x space plus space 11 space plus space 14 over denominator 5 end fraction
rightwards double arrow space straight x space plus space 40 space equals space 5 straight x
rightwards double arrow 4 straight x space equals space 40
rightwards double arrow straight x space equals space 10
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 6

If mode of a series exceeds its mean by 12, then mode exceeds the median by

(a) 4

(b) 8

(c) 6

(d) 10

Solution 6

begin mathsize 12px style We space know
3 space median space equals space mode space plus space 2 space mean space space space space space space space space space space space space....... open parentheses 1 close parentheses
Given space mode space minus space mean space equals space 12 space space space space space space space space...... open parentheses 2 close parentheses
from space open parentheses 1 close parentheses space & space open parentheses 2 close parentheses
rightwards double arrow 3 space median space equals space 12 space plus space mean space plus space 2 space mean
rightwards double arrow 3 space median space equals space 12 space plus space 3 space mean
rightwards double arrow 3 open parentheses median close parentheses space minus space 3 space mean space equals space 12
rightwards double arrow 3 open parentheses median close parentheses space minus space 3 space open parentheses mode space minus space 12 close parentheses space equals space 12
rightwards double arrow 3 space median space minus space 3 space mode space plus space 36 space equals space 12
rightwards double arrow 3 space mode space minus space 3 space median space equals space 8
So comma space the space correct space option space is space left parenthesis straight b right parenthesis. end style

Question 7

If the mean of first n natural number is 15, then  n =

(a) 15

(b) 30

(c) 14

(d) 29

Solution 7

begin mathsize 12px style mean space equals space fraction numerator 1 space plus space 2 plus space 3 space plus space.... space plus space straight n over denominator straight n end fraction
space space space space space space space space space space space space space equals space fraction numerator begin display style fraction numerator straight n open parentheses straight n space plus space 1 close parentheses over denominator 2 end fraction end style over denominator straight n end fraction
space space space space space space space space space space space space space equals space space fraction numerator straight n space plus space 1 over denominator 2 end fraction
given space mean space equals space 15
rightwards double arrow fraction numerator straight n space plus space 1 over denominator 2 end fraction equals space 15
rightwards double arrow space straight n space equals space 29
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 8

begin mathsize 12px style If space the space mean space of space observation space straight x subscript 1 comma space straight x subscript 2 comma space...... comma space straight x subscript straight n space is space straight x with bar on top comma space then space the space mean space of space straight x subscript 1 plus straight a subscript 1 comma space straight x subscript 2 plus straight a comma space..... comma space straight x subscript straight n plus straight a space is space
open parentheses straight a close parentheses space straight a straight x with bar on top
open parentheses straight b close parentheses space straight x with bar on top space minus space straight a
open parentheses straight c close parentheses space straight x with bar on top space plus space straight a
open parentheses straight d close parentheses space fraction numerator straight x with bar on top over denominator straight a end fraction end style

Solution 8

begin mathsize 12px style straight x with bar on top space equals space fraction numerator straight x subscript 1 space plus space straight x subscript 2 space plus space..... space plus space straight x subscript straight n over denominator straight n end fraction space space space space space space space space space space space space space..... open parentheses 1 close parentheses
mean space of space straight x subscript 1 space plus space straight a subscript 1 comma......... comma straight x subscript straight n plus space straight a
equals fraction numerator open parentheses straight x subscript 1 space plus space straight a close parentheses space plus space open parentheses straight x subscript 2 space plus space straight a close parentheses space plus space...... space open parentheses straight x subscript straight n space plus space straight a close parentheses over denominator straight n end fraction
equals fraction numerator open parentheses straight x subscript 1 space plus space straight x subscript 2 space plus space straight x subscript 3 space plus space..... space plus space straight x subscript straight n close parentheses space plus space na over denominator straight n end fraction
equals fraction numerator open parentheses straight x subscript 1 space plus space straight x subscript 2 space plus space....... space plus space straight x subscript straight n close parentheses over denominator straight n end fraction space plus space straight a
equals new space mean space equals space straight x with bar on top space plus space straight a
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 9

begin mathsize 12px style Mean space of space straight a space certain space number space of space observations space is space straight x with bar on top. space If space each space observation space isdivided space by space straight m open parentheses straight m space not equal to space 0 close parentheses
and space increased space by space straight n comma space the space the space mean space of space new space observation space is
open parentheses straight a close parentheses space fraction numerator straight x with bar on top over denominator straight m end fraction space plus space straight n
open parentheses straight b close parentheses space fraction numerator straight x with bar on top over denominator straight n end fraction space plus space straight m
open parentheses straight c close parentheses space straight x with bar on top space plus space straight n over straight m
open parentheses straight d close parentheses space straight x with bar on top space plus space straight m over straight n end style

Solution 9

begin mathsize 12px style Let space the space observations space be space straight a subscript 1 comma space straight a subscript 2 space........ space straight a subscript straight n
It space is space given space that space fraction numerator straight a subscript 1 space plus space straight a subscript 2 space plus space...... space plus space straight a subscript straight n over denominator straight n end fraction equals space straight x with bar on top
Also space given space if space each space observation space is space divided space by space straight m space and space increased space by space straight n comma space then space new space observations space are
rightwards double arrow straight a subscript 1 over straight m plus space straight n comma space straight a subscript 2 over straight m space plus space straight n space space........ space straight a subscript straight n over straight m space plus space straight n
rightwards double arrow space mean space equals space fraction numerator open parentheses begin display style straight a subscript 1 over straight m plus straight n end style close parentheses space plus space open parentheses begin display style straight a subscript 2 over straight m space plus space straight n end style close parentheses space plus space........ space open parentheses begin display style straight a subscript straight n over straight m plus space straight n end style close parentheses over denominator straight n end fraction
space space space space space space space space space space space space space space space space space space space equals space fraction numerator begin display style fraction numerator open parentheses straight a subscript 1 space plus space straight a subscript 2 space plus space......... space plus space straight a subscript straight n close parentheses over denominator straight m end fraction end style plus space straight n squared over denominator straight n end fraction space
space space space space space space space space space space space space space space space space space space space equals space open parentheses fraction numerator straight a subscript 1 space plus space straight a subscript 2 space plus space...... space plus space straight a subscript straight n over denominator straight n end fraction close parentheses 1 over straight m space plus space straight n
space space space space space space space space space space space space space space space space space space space equals space fraction numerator top enclose straight x over denominator straight m end fraction plus space straight n
So comma space the space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 10

begin mathsize 12px style If space straight u subscript straight i space equals space fraction numerator straight x subscript straight i space minus space 25 over denominator 10 end fraction comma space sum for blank of straight f subscript straight i straight u subscript straight i space equals space 20 comma space sum for blank of straight f subscript straight i space end subscript space equals space 100 comma space then space straight x with bar on top equals
open parentheses straight a close parentheses space 23
open parentheses straight b close parentheses space 24
open parentheses straight c close parentheses space 27
open parentheses straight d close parentheses space 25 end style

Solution 10

begin mathsize 12px style straight x with bar on top space equals space fraction numerator begin display style sum for blank of straight f subscript straight i straight x subscript straight i end style over denominator begin display style sum for blank of straight f subscript straight i end style end fraction
given space straight u subscript straight i space equals space fraction numerator straight x subscript straight i space minus space 25 over denominator 10 end fraction
rightwards double arrow straight x subscript straight i space equals space 10 straight u subscript straight i space plus space 25
rightwards double arrow straight x with bar on top space equals space fraction numerator begin display style sum for blank of straight f subscript straight i space open parentheses 10 straight u subscript straight i space plus space 25 close parentheses end style over denominator begin display style sum for blank of straight f subscript straight i end style end fraction
rightwards double arrow straight x with bar on top space equals fraction numerator 10 begin display style sum for blank of end style straight f subscript straight i straight u subscript straight i space plus space 25 begin display style sum for blank of end style straight f subscript straight i over denominator begin display style sum for blank of straight f subscript straight i end style end fraction
given space space sum for blank of straight f subscript straight i straight u subscript straight i space equals space 20 space and space sum for blank of straight f subscript straight i space equals space 100
rightwards double arrow straight x with bar on top space equals space fraction numerator 10 space cross times space 20 space plus space 25 space cross times space 100 over denominator 100 end fraction
space space space space space space space space equals space 27 space
So comma space the space correct space option space is space left parenthesis straight c right parenthesis. end style

Question 11

If 35 is removed from the data: 30, 34, 35, 36, 37, 38, 39, 40, then the median increases by

(a) 2

(b) 1.5

(c) 1

(d) 0.5

Solution 11

begin mathsize 12px style Initially space there space are space 8 space observations
Hence comma space median space equals space fraction numerator 4 to the power of th space plus space 5 to the power of th space term over denominator 2 end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 36 space plus space 37 over denominator 2 end fraction space equals space 73 over 2
After space removal space of space 35
median space equals space 4 to the power of th space end exponent space term
space space space space space space space space space space space space space space space space equals space 37
difference space equals space 37 space minus space 73 over 2 space equals space 0.5
So comma space the space correct space option space is space left parenthesis straight d right parenthesis. end style

Question 12

While computing mean of grouped data, we assume that the frequencies are

 

a. evenly distributed over all the classes.

b. centred at the class marks of the classes.

c. centred at the upper limit of the classes.

d. centred at the lower limit of the classes.

Solution 12

While computing the mean of the grouped data, we assume that the frequencies are centred at the class marks of the classes.

Hence, correct option is (b).

Question 13

In the formula  , for finding the mean of grouped frequency distribution ui =

 

 

a. 

b. h(xi - a)

c. 

d. 

Solution 13

Chapter 15 - Statistics Excercise 15.69

Question 1

For the following distribution:

 

Class:

0-5

5-10

10-15

15-20

20-25

Frequency:

10

15

12

20

9

The sum of the lower limits of the median and modal class is

 

a. 15

b. 25

c. 30

d. 35

 

Solution 1

Question 2

For the following distribution:

 

Below:

10

20

30

40

50

60

Number of students:

3

12

27

57

75

80

 

The model class is

 

a. 10-20

b. 20-30

c. 30-40

d. 50-60

 

Solution 2

Question 3

Consider the following frequency distribution:

 

Class:

65-85

85-105

105-125

125-145

145-165

165-185

185-205

Frequency:

4

5

13

20

14

7

4

The difference of the upper limit of the median class and the lower limit of the modal class is

 

a. 0

b. 19

c. 20

d. 38

Solution 3

Question 4

In the formula  , for finding the mean of grouped data dis are deviations from a of

 

a. Lower limits of classes

b. Upper limits of classes

c. Mid-points of classes

d. Frequency of the class marks

Solution 4

di's are the deviations from a of mid-points of classes.

Hence, correct option is (c).

Question 5

The abscissa of the point of intersection of less than type and of the more than type cumulative frequency curves of a grouped data given its

 

a. Mean

b. Median

c. Mode

d. All the three above

Solution 5

The abscissa of the point of intersection of less than type and of the more than type cumulative frequency curves of a grouped data given its median.

Hence, correct option is (b).

 

Question 6

Consider the following frequency distribution:

 

Class:

0-5

6-11

12-17

18-23

24-29

Frequency:

13

10

15

8

11

The upper limit of the median class is

 

a. 17

b. 17.5

c. 18

d. 18.5

 

Solution 6