Class 10 NCERT Solutions Maths Chapter 13  Statistics
Statistics Exercise Ex. 13.1
Solution 1
Let us find class marks (x_{i}) 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— 1 = 1 
2  4  2  3  2 — 3 = 6 
4  6  1  5  1 — 5 = 5 
6  8  5  7  5 — 7 = 35 
8  10  6  9  6 — 9 = 54 
10  12  2  11  2 —11 = 22 
12  14  3  13  3 — 13 = 39 
Total  20  162 
From the table, we may observe that
So, the mean number of plants per house is 8.1.
We have used here direct method as values of class marks (x_{i}) and f_{i} are small.
Solution 2
Let us find class mark for each interval by using the relation.
Class size (h) of this data = 20
Now taking 550 as assured mean (a) we may calculate di, u_{i} and f_{i}u_{i}as following.
Daily wages (in Rs)  Number of workers (f_{i})  x_{i}  d_{i} = x_{i } 550  f_{i}u_{i }  
500  520  12  510   40  2   24 
520  540  14  530   20  1   14 
540  560  8  550  0  0  0 
560  580  6  570  20  1  6 
580  600  10  590  40  2  20 
Total  50  12 
From the table we may observe that
So mean daily wages of the workers of the factory is Rs. 545.2.
Solution 3
We may find class mark (x_{i}) for each interval by using the relation.
Given that mean pocket allowance = Rs.18
Now taking 18 as assured mean (a) we may calculate d_{i} and f_{i}d_{i}as following.
Daily pocket allowance (in Rs.)  Number of children f_{i}  Class mark x_{i}  d_{i} = x_{i}  18  f_{i}d_{i} 
11  13  7  12   6   42 
13  15  6  14   4   24 
15  17  9  16   2   18 
17  19  13  18  0  0 
19  21  f  20  2  2 f 
21  23  5  22  4  20 
23  25  4  24  6  24 
Total  2f  40 
From the table we may obtain
Hence the missing frequency f is 20.
Solution 4
We may find class mark of each interval (x_{i}) 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 heart beats per minute for these women are 75.9 beats per minute.
Solution 5
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 (x_{i}) 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 following 
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 may observe that
Clearly, mean number of mangoes kept in a packing box is 57.19.
We have chosen step deviation method here as values of f_{i}, d_{i} are big and also there is a common multiple between all d_{i}.
Solution 6
We may calculate class mark (x_{i}) 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 following
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  25  7 
Now we may observe that 
Solution 7
Concentration of SO_{2} (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 8
We may find class mark of each interval by using the relation
Now taking 16 as assumed mean (a) we may calculate di and fidi as following
Number of days 
Number of students f_{i} 
x_{i} 
d_{i}= x_{i}  16 
f_{i}d_{i} 
0  6 
11 
3 
13 
143 
6 10 
10 
8 
8 
80 
10  14 
7 
12 
4 
28 
14  20 
4 
16 
0 
0 
20  28 
4 
24 
8 
32 
28  38 
3 
33 
17 
51 
38  40 
1 
39 
23 
23 
Total 
40 
145 
Now we may observe that
So, mean number of days is 12.38 days, for which a student was absent.
Solution 9
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 following
Literacy rate (in %) 
Number of cities f_{i} 
x_{i} 
d_{i}= x_{i}  70 
f_{i}u_{i} 

45  55 

50 
20 
2 
6 
55  65 
10 
60 
10 
1 
10 
65  75 
11 
70 
0 
0 
0 
75  85 
8 
80 
10 
1 

85  95 
3 
90 
20 
2 
6 
Total 
35 
2 
Now we may observe that
So, the mean literacy rate is 69.43%.
Statistics Exercise Ex. 13.2
Solution 1
We may compute class marks (x_{i}) as per the relation
Now taking 30 as assumed mean (a) we may calculate d_{i} and f_{i}d_{i} as following.
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 2
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 3
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
Now we may find classmark 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 following
Expenditure (in Rs) 
Number of families f_{i} 
x_{i} 
d_{i} = x_{i}  2750 
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 the table, we may observe that
So, mean monthly expenditure was Rs.2662.50.
Solution 4
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 d_{i}, u_{i} and f_{i}u_{i} as following.
Number of students per teacher 
Number of states/U.T (f_{i}) 
x_{i} 
d_{i} = x_{i}  32.5 

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 an average teacherstudent ratio was 29.2.
Solution 5
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 6
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.
Statistics Exercise Ex. 13.3
Solution 1
We may find class marks by using the relation
Taking 135 as assumed mean (a) we may find d_{i}, u_{i}, f_{i}u_{i}, according to step deviation method as following
Monthly consumption (in units) 
Number of consumers (f _{i}) 
x_{i} class mark 
d_{i} = x_{i}  135

f_{i}u_{i} 

65  85 
4 

 60 
 3 
 12 
85  105 
5 
95 
 40 
 2 
 10 
105  125 
13 
115 
 20 
 1 
 13 
125  145 
20 
135 
0 
0 
0 
145  165 
14 
155 
20 

14 
165  185 
8 
175 
40 
2 
16 
185  205 
4 
195 
60 
3 
12 
Total 
68 
7 
From the table we may observe that
Now from table it is clear that maximum class frequency is 20 belonging to class interval 125  145.
Modal class = 125  145
Lower limit (l) of modal class = 125
Class size (h) = 20
Frequency (f_{1}) of modal class = 20
Frequency (f_{0}) of class preceding modal class = 13
Frequency (f_{2}) of class succeeding the modal class = 14
We know that
3 median = mode + 2 mean
= 135.76 + 2 (137.058)
= 135.76 + 274.116
= 409.876
Median = 136.625
So median, mode, mean of given data is 136.625, 135.76, 137.05 respectively.
Solution 2
We may find cumulative frequency for the given data as following
Class interval 
Frequency 
Cumulative frequency 
0  10 
5 
5 
10  20 
x 
5 + x 
20  30 
20 
25 + x 
30  40 
15 
40 + x 
40  5 
y 
40 + x + y 
50  60 
5 
45 + x + y 
Total (n) 
60 
It is clear that n = 60
45 + x + y = 60
x + y = 15 (1)
Median of data is given as 28.5 which lies in interval 20  30.
So, median class = 20  30
Lower limit (l) of median class = 20
Cumulative frequency (cf) of class preceding the median class = 5 + x
Frequency (f) of median class = 20
Class size (h) = 10
From equation (1)
8 + y = 15
y = 7
Hence values of x and y are 8 and 7 respectively.
Solution 3
Here the class width is not the same. There is no need to adjust the frequencies according to class intervals. Now given frequency table is of less than type represented with upperclass 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 policyholders (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 the 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, the median age is 35.76 years.
Solution 4
The given data is not having continuous class intervals. We can observe that the difference between the two class intervals is 1. So, we have to add and subtract
to upperclass 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 than
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, the median length of leaves is 146.75 mm.
Solution 5
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, the median lifetime of lamps is 3406.98 hours.
Solution 6
We can find cumulative frequencies with their respective class intervals as below
Number of letters 
Frequency (f_{i})

Cumulative frequency 
1  4 
6 
6 
4  7 
30 
30 + 6 = 36 
7  10 
40 
36 + 40 = 76 
10  13 
16 
76 + 16 = 92 
13  16 
4 
92 + 4 = 96 
16  19 
4 
96 + 4 = 100 
Total (n) 
100 
Now we may observe that cumulative frequency just greater than is 76 belonging to the class interval 7  10.
Median class = 7  10
Lower limit (l) of median class = 7
Cumulative frequency (cf) of class preceding median class = 36
Frequency (f) of median class = 40
Class size (h) = 3
Now we can find class marks of given class intervals by using relation
Taking 11.5 as assumed mean (a) we can find d_{i}, u_{i} and f_{i}u_{i} according to step deviation method as below.
Number of letters 
Number of surnames 
x_{i} 
x_{i}  a 
f_{i}u_{i} 

1  4 
6 
2.5 
9 
3 
18 
4  7 
30 
5.5 
6 
2 
60 
7  10 
40 
8.5 
3 
1 
40 
10  13 
16 
11.5 
0 
0 
0 
13  16 
4 
14.5 
3 
1 
4 
16  19 
4 
17.5 
6 
2 
8 
Total 
100 
106 
f_{i}u_{i} =  106
f_{i} = 100
We know that
3 median = mode + 2 mean
3(8.05) = mode + 2(8.32)
24.15  16.64 = mode
7.51 = mode
So, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.51.
Solution 7
We may find cumulative frequencies with their respective class intervals as below
Weight (in kg)  No. of students  Cumulative frequency (c.f) 
40  45  2  2 
45  50  3  2+3=5 
50  55  8  5+8=13 
55  60  6  13+6=19 
60  65  6  19+6=25 
65  70  3  25+3=28 
70  75  2 
28+2=30 
Total = 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 class preceeding the median class = 13
Class size (h) = 5
So, the median weight is 56.67 kg.