Class 10 RD SHARMA Solutions Maths Chapter 15  Statistics
Statistics Exercise Ex. 15.1
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Statistics Exercise Ex. 15.2
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Statistics Exercise Ex. 15.3
Solution 1
Solution 2
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.
Solution 3
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 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.
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
Solution 18
Solution 19
Solution 20
Solution 21
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.
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 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 
Solution 23
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 
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 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%.
Solution 26
Here, we have cumulative frequency distribution less than type. First we convert it into an ordinary frequency distribution.
Solution 27
Solution 28
Solution 29
Solution 5
Let the assumed mean be A = 8
Here, h = 2
Class interval 
Mid value x_{i} 
d_{i} = x_{i}  8 

f_{i} 
f_{i}u_{i} 
3  5 5  7 7  9 9  11 11  13 
4 6 8 10 12 
4 2 0 2 4 
2 1 0 1 2 
5 10 10 7 8 
10 10 0 7 16 




N = 40 

Solution 24 (i)
We may find class mark of each interval by using the relation
Number of days 
Number of students f_{i} 
x_{i} 
d_{i} = x_{i}  21 
f_{i}d_{i} 
0  6 6  12 12  18 18  24 24  30 30  36 36  42 
10 11 7 4 4 3 1 
3 9 15 21 27 33 39 
18 12 6 0 6 12 18 
180 132 42 0 24 36 18 

40 



Assumed mean A = 21
Hence, the number of days a student was absent is 14.1.
Solution 30
We may find class mark of each interval by using the relation
Marks 
Number of students f_{i} 
x_{i} 
d_{i} = x_{i}  47.5 
f_{i}d_{i} 
30  35 35  40 40  45 45  50 50  55 55  60 60  65 
14 16 28 23 18 8 3 
32.5 37.5 42.5 47.5 52.5 57.5 62.5 
15 10 5 0 5 10 15 
210 160 140 0 90 80 45 

110 



Assumed mean A = 47.5
Hence, mean marks of the students is 44.81.
Statistics Exercise Ex. 15.4
Solution 1
Solution 2
The median height of the students is Rs 167.13.
Solution 3
Solution 5
Solution 6
Solution 7
Solution 8
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.
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.
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15(i)
Solution 15(ii)
Solution 16
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 (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.
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 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.
Solution 19
Solution 20
Solution 21
Solution 4
The cumulative frequency table is
Salary 
Frequency f_{i} 
Cumulative frequency (c.f.) 
5  10 10  15 15  20 20  25 25  30 30  35 35  40 40  45 45  50 
49 133 63 15 6 7 4 2 1 
49 182 245 260 266 273 277 279 280 
The cumulative frequency greater than and nearest to 140 is 182.
So, median class is 10  15
Lower limit (l) = 10, class size (h) = 5, c.f. = 49
Frequency of median class = 133
Hence, mean median salary is Rs. 13421.
Statistics Exercise Ex. 15.5
Solution 1
Solution 2
Solution 3(i)
Solution 3(ii)
Solution 3(iii)
Solution 4
Solution 5
Solution 6
Solution 7
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.
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 (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.
Solution 9
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 (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.
Solution 11
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 (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.
Solution 13
Solution 14
Solution 15
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 (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.
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 (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.
Solution 18
Solution 19
Solution 3 (iv)
The cumulative frequency table is
Salary 
Frequency f_{i} 
0  10 10  20 20  30 30  40 40  50 50  60 60  70 
8 10 10 16 12 6 7 
Modal class is 30  40
Lower limit (l) = 30, class size (h) = 10, f = 16, f_{1} = 10 and f_{2} = 12
Hence, the mode is 36.
Statistics Exercise Ex. 15.6
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
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
Solution 9
More than type frequency distribution is given by
Class interval 
Frequency f_{i} 
Cumulative frequency 
More than 20 More than 30 More than 40 More than 50 More than 60 More than 70 More than 80 
10 8 12 24 6 25 15 
100 90 82 70 46 40 15 
So, more than type frequency curve is
Solution 10
Less than type frequency distribution is given by
Class interval 
Frequency f_{i} 
Cumulative frequency 
Less than 220 Less than 240 Less than 260 Less than 280 Less than 300 
12 14 8 6 10 
12 26 34 40 50 
So, less than type frequency curve is
Statistics Exercise 15.66
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).
Solution 2
Solution 3
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).
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).
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).
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 xcoordinate corresponding to the peak value on y  axis, is the mode.
So, the correct option is (a).
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).
Solution 9
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).
Statistics Exercise 15.67
Solution 11
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).
Solution 13
Solution 14
Solution 15
Solution 16
Solution 17
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).
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).
Solution 20
Solution 21
Solution 22
Solution 23
Solution 24
Statistics Exercise 15.68
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).
Solution 26
Solution 27
Solution 28
Solution 29
Solution 30
Solution 31
Solution 32
Solution 33
Solution 34
Solution 35
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).
Solution 37
Statistics Exercise 15.69
Solution 38
Solution 39
Solution 40
Solution 41
d_{i}'s are the deviations from a of midpoints of classes.
Hence, correct option is (c).
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).
Solution 43