RD SHARMA Solutions for Class 10 Maths Chapter 15 - Statistics

Page / Exercise

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

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 Exercise 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

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 Exercise 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(i)

Solution 15(i)

Question 15(ii)

Solution 15(ii)

Question 16

Solution 16

Question 17

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 17

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 18

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 18

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 19

Solution 19

Question 20

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 20

Question 21

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 21

Chapter 15 - Statistics Exercise Ex. 15.5

Question 1

Solution 1

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

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 7

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 8

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 8

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 9

Solution 9

Question 10

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 10

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 11

Solution 11

Question 12

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 12

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 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

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 16

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 17

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 17

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 18

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 18

Question 19

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 19

Chapter 15 - Statistics Exercise 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 Exercise 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

Question 3

Solution 3

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

Solution 9

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 Exercise 15.67

Question 11

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 11

Question 12

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 12

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 13

Solution 13

Question 14

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 14

Question 15

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 15

Question 16

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 16

Question 17

The median of first 10 prime numbers is

(a) 11

(b) 12

(c) 13

(d) 14

Solution 17

Question 18

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 18

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 19

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 19

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 20

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 20

Question 21

Solution 21

Question 22

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 22

Question 23

Solution 23

Question 24

Solution 24

Chapter 15 - Statistics Exercise 15.68

Question 25

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 25

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 26

Solution 26

Question 27

Solution 27

Question 28

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 28

Question 29

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 29

Question 30

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 30

Question 31

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

(a) 15

(b) 30

(c) 14

(d) 29

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

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 35

Question 36

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 36

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 37

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

a.

b. h(xi - a)

c.

d.

Solution 37

Chapter 15 - Statistics Exercise 15.69

Question 38

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 38

Question 39

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 39

Question 40

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 40

Question 41

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 41

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

Hence, correct option is (c).

Question 42

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 42

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 43

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 43