RD Sharma Solutions for CBSE Class 10 Mathematics chapter 15  Statistics
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Chapter 15  Statistics Excercise Ex. 15.1
Chapter 15  Statistics Excercise Ex. 15.2
Chapter 15  Statistics Excercise Ex. 15.3
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?
Let us find class marks (xi) for each interval by using the relation.
Now we may compute x_{i} and f_{i}x_{i}as following
Number of plants  Number of houses (f_{i})  x_{i}  f_{i}x_{i} 
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.
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  7174 
74  77  77  80  80  83  83  86 
Number of women  2  4  3  8  7  4  2 
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 d_{i}, u_{i}, f_{i}u_{i} as following.
Number of heart beats per minute  Number of women f_{i}  x_{i}  d_{i} = x_{i} 75.5  f_{i}u_{i}  
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.
Find the mean of the following frequency distribution:
Find the mean of the following frequency distribution:
Find the mean of the following frequency distribution:
Find the mean of the following frequency distribution:
Find the mean of the following frequency distribution:
Find the mean of the following frequency distribution:
Find the mean of the following frequency distribution:
Find the mean of the following frequency distribution:
Find the mean of the following frequency distribution:
For the following distribution, calculate mean using all suitable methods :
Size of item  14  49  916  1627 
Frequency  6  12  26  20 
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.
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?
Number of mangoes  Number of boxes f_{i} 
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 d_{i}, u_{i}, f_{i}u_{i} as follows:
Class interval  f_{i}  x_{i}  d_{i} = x_{i}  57  f_{i}u_{i}  
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.
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.
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 d_{i}, u_{i}, f_{i}u_{i} as follows:
Daily expenditure (in Rs)  f_{i}  x_{i}  d_{i} = x_{i}  225  f_{i}u_{i}  
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 
To find out the concentration of SO_{2} 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 SO_{2} (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 SO_{2} in the air.
Concentration of SO2 (in ppm)  Frequency  Class mark x_{i}  d_{i} = x_{i  0.14}  f_{i}u_{i}  
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 
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 
We may find class mark of each interval by using the relation
Now, taking 17 as assumed mean (a) we may calculate d_{i} and f_{i}d_{i} as follows:
Number of days 
Number of students f_{i} 
x_{i} 
d_{i}= x_{i}  17 
f_{i}d_{i} 
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.
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 
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 d_{i}, u_{i}, and f_{i}u_{i} as follows:
Literacy rate (in %) 
Number of cities f_{i} 
x_{i} 
d_{i}= x_{i}  70 
u_{i} =d_{i}/10 
f_{i}u_{i} 
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%.
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 
Here, we have cumulative frequency distribution less than type. First we convert it into an ordinary frequency distribution.
If the means of the following frequency distribution is 18, find the missing frequency.
Class interval: 
1113 
1315 
1517 
1719 
1921 
2123 
2325 
Frequency: 
3 
6 
9 
13 
f 
5 
4 
Find the missing frequencies in the following distribution, if the sum of the frequencies is 120 and the mean is 50.
Class: 
020 
2040 
4060 
6080 
80100 
Frequency: 
17 
f_{1} 
32 
f_{2} 
19 
The daily income of a sample of 50 employees are tabulated as follows:
Income (in Rs.): 
1200 
201400 
401600 
601800 
No. of employees: 
14 
15 
14 
7 
Find the mean daily income of employees.
Chapter 15  Statistics Excercise Ex. 15.4
The median height of the students is Rs 167.13.
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.
We can find cumulative frequencies with their respective class intervals as below 
Life time 
Number of lamps (f_{i}) 
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.
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 
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.
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 
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 (f_{i}) 
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.
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 f_{i} 
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.
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 f_{i} 
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.
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: 
06 
612 
1218 
1824 
2430 
Frequency: 
4 
x 
5 
y 
1 
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: 
2030 
3040 
4050 
5060 
6070 
7080 
8090 
Frequency: 
p 
15 
25 
20 
q 
8 
10 
Chapter 15  Statistics Excercise Ex. 15.5
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.
We may compute class marks (xi) as per the relation
Now taking 30 as assumed mean (a) we may calculate d_{i} and f_{i}d_{i} as follows.
Age (in years) 
Number of patients f_{i} 
class mark x_{i} 
d_{i}= x_{i}  30 
f_{i}d_{i} 
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 (f_{1}) of modal class = 23
Class size (h) = 10
Frequency (f_{0}) of class preceding the modal class = 21
Frequency (f_{2}) 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.
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.
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 (f_{1}) of modal class = 61
Frequency (f_{0}) of class preceding the modal class = 52
Frequency (f_{2}) of class succeeding the modal class = 38
Class size (h) = 20
So, modal lifetime of electrical components is 65.625 hours.
The following distribution gives the statewise teacherstudent 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 
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 (f_{1}) of modal class = 10
Frequency (f_{0}) of class preceding modal class = 9
Frequency (f_{2}) 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 (f_{i}) 
x_{i} 
d_{i} = x_{i}  32.5 
u_{i} 
f_{i}u_{i} 
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.
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 
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 (f_{1}) of modal class = 20
Frequency (f_{0}) of class preceding modal class = 12
Frequency (f_{2}) of class succeeding modal class = 11
Class size = 10
So mode of this data is 44.7 cars.
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 
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 (f_{1}) of modal class = 40
Frequency (f_{0}) of class preceding modal class = 24
Frequency (f_{2}) 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 d_{i}, u_{i} and f_{i}u_{i} as follows:
Expenditure (in Rs) 
Number of families f_{i} 
x_{i} 
d_{i} = x_{i}  2750 
u_{i} 
f_{i}u_{i}

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.
The given distribution shows the number of runs scored by some top batsmen of the world in oneday 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.
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 (f_{1}) of modal class = 18
Frequency (f_{0}) of class preceding modal class = 4
Frequency (f_{2}) of class succeeding modal class = 9
Class size (h) = 1000
So mode of given data is 4608.7 runs.
The frequency distribution table of agriculture holdings in a village is given below:
Area of land (in hectares): 
13 
35 
57 
79 
911 
1113 
Number of families: 
20 
45 
80 
55 
40 
12 
Find the modal agriculture holdings of the village.
The monthly income of 100 families are given as below:
Income in (in Rs.) 
Number of families 
05000 
8 
500010000 
26 
1000015000 
41 
1500020000 
16 
2000025000 
3 
2500030000 
3 
3000035000 
2 
3500040000 
1 
Calculate the modal income.
Chapter 15  Statistics Excercise Ex. 15.6
The annual rainfall record of a city for 66 days if given in the following table:
Rainfall (in cm): 
010 
1020 
2030 
3040 
4050 
5060 
Number of days: 
22 
10 
8 
15 
5 
6 
Calculate the median rainfall using ogives of more than type and less than type.
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
Chapter 15  Statistics Excercise 15.66
Which of the following is not a measure of central tendency?
(a) Mean
(b) Median
(c) Mode
(d) Standard deviation
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).
The algebraic sum of deviations of a frequency distribution from its mean is
(a) always positive
(b) always negative
(c) 0
(d) a nonzero number
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
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).
Which of the following cannot be determined graphically?
(a) Mean
(b) Median
(c) Mode
(d) None of these
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).
The median of a given frequency distribution is found graphically with the help of
(a) Histogram
(b) Frequency curve
(c) Frequency polygon
(d) Ogive
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).
The mode of a frequency distribution can be detremined graphically from
(a) Histogram
(b) frequency polygon
(c) ogive
(d) frequency curve
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 xcoordinate corresponding to the peak value on y  axis, is the mode.
So, the correct option is (a).
Mode is
(a) least frequent value
(b) middle most value
(c) most frequent value
(d) None of these
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).
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
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
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
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
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).
If the arithmetic mean of x, x +3 , x + 6, x + 9, x + 12 is 10, then x =
(a) 1
(b) 2
(c) 6
(d) 4
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
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
The median of first 10 prime numbers is
(a) 11
(b) 12
(c) 13
(d) 14
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
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).
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
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).
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
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
Chapter 15  Statistics Excercise 15.68
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
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).
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
If the arithmetic mean of 7, 8, x, 11, 14 is x then x =
(a) 1
(b) 9.5
(c) 10
(d) 10.5
If mode of a series exceeds its mean by 12, then mode exceeds the median by
(a) 4
(b) 8
(c) 6
(d) 10
If the mean of first n natural number is 15, then n =
(a) 15
(b) 30
(c) 14
(d) 29
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
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.
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).
In the formula , for finding the mean of grouped frequency distribution u_{i} =
a.
b. h(x_{i}  a)
c.
d.
Chapter 15  Statistics Excercise 15.69
For the following distribution:
Class: 
05 
510 
1015 
1520 
2025 
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
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. 1020
b. 2030
c. 3040
d. 5060
Consider the following frequency distribution:
Class: 
6585 
85105 
105125 
125145 
145165 
165185 
185205 
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
In the formula , for finding the mean of grouped data d_{i}^{s} are deviations from a of
a. Lower limits of classes
b. Upper limits of classes
c. Midpoints of classes
d. Frequency of the class marks
d_{i}'s are the deviations from a of midpoints of classes.
Hence, correct option is (c).
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
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).
Consider the following frequency distribution:
Class: 
05 
611 
1217 
1823 
2429 
Frequency: 
13 
10 
15 
8 
11 
The upper limit of the median class is
a. 17
b. 17.5
c. 18
d. 18.5
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