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?
Answer :
To find the class mark (x_{i}) for each interval, the following relation is used.
Class mark (x_{i}) =
x_{i} andf_{i}x_{i} can be calculated as follows.
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, it can be observed that
Mean,
Therefore, mean number of plants per house is 8.1.
Here, direct method has been used as the values of class marks (x_{i}) and f_{i} are small.
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Consider the following distribution of daily wages of 50 worker of a factory.
Daily wages (in Rs) 
100  120 
120  140 
140  1 60 
160  180 
180  200 
Number of workers 
12 
14 
8 
6 
10 
Find the mean daily wages of the workers of the factory by using an appropriate method.
Answer :
To find the class mark for each interval, the following relation is used.
Class size (h) of this data = 20
Taking 150 as assured mean (a), d_{i}, u_{i}, and f_{i}u_{i} can be calculated as follows.
Daily wages (in Rs) 
Number of workers (f_{i}) 
x_{i} 
d_{i} = x_{i}  150 

f_{i}u_{i} 
100  120 
12 
110 
 40 
 2 
 24 
120  140 
14 
130 
 20 
 1 
 14 
140  160 
8 
150 
0 
0 
0 
160  180 
6 
170 
20 
1 
6 
180  200 
10 
190 
40 
2 
20 
Total 
50 
 12 
From the table, it can be observed that
Therefore, the mean daily wage of the workers of the factory is Rs 145.20.
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The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs.18. Find the missing frequency f.
Daily pocket allowance (in Rs) 
11  13 
13  15 
15  17 
17  19 
19  21 
21  23 
23  25 
Number of workers 
7 
6 
9 
13 
f 
5 
4 
Answer :
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 
Answer :
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?
Answer :
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.
Answer :
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.
Answer :
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 
Answer :
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 
Answer :
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.
Answer :
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.
Answer :
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 
Answer :
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 
Answer :
The given distribution shows the number of runs scored by some top batsmen of the world in oneday international cricket matches.
Runs scored 
Number of batsmen 
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.
Answer :
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  60 
60  70 
70  80 
Frequency 
7 
14 
13 
12 
20 
11 
15 
8 
Answer :
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
Monthly consumption (in units) 
Number of consumers 
65  85 
4 
85  105 
5 
105  125 
13 
125  145 
20 
145  165 
14 
165  185 
8 
185  205 
4 
Answer :
If the median of the distribution is given below is 28.5, find the values of x and y.
Class interval 
Frequency 
0  10 
5 
10  20 
x 
20  30 
20 
30  40 
15 
40  50 
y 
50  60 
5 
Total 
60 
Answer :
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 
Answer :
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.
(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5  126.5, 126.5  135.5… 171.5  180.5)
Answer :
Find the following table 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.
Answer :
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:
Number of letters 
1  4 
4  7 
7  10 
10  13 
13  16 
16  19 
Number of surnames 
6 
30 
40 
6 
4 
4 
Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Answer :
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 
Answer :
The following distribution gives the daily income of 50 workers of a factory.
Daily income (in Rs) 
100  120 
120  140 
140  160 
160  180 
180  200 
Number of workers 
12 
14 
8 
6 
10 
Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Answer :
During the medical checkup of 35 students of a class, their weights were recorded as follows:
Weight (in kg) 
Number of students 
Less than 38 
0 
Less than 40 
3 
Less than 42 
5 
Less than 44 
9 
Less than 46 
14 
Less than 48 
28 
Less than 50 
32 
Less than 52 
35 
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph verify the result by using the formula.
Answer :
The following table gives production yield per hectare of wheat of 100 farms of a village.
Production yield (in kg/ha) 
50  55 
55  60 
60  65 
65  70 
70  75 
75  80 
Number of farms 
2 
8 
12 
24 
38 
16 
Change the distribution to a more than type distribution and draw ogive.
Answer :
Maths : CBSE NCERT Exercise Solutions for Class 10th for Statistics ( Exercise 14.1, 14.2, 14.3, 14.4 ) will be available online in PDF book form soon. The solutions are absolutely Free. Soon you will be able to download the solutions.
Exercise 14.1 
Question 1 
Question 2 
Question 3 
Question 4 
Question 5 
Question 6 
Question 7 
Question 8 
Question 9 
Exercise 14.2 
Question 1 
Question 2 
Question 3 
Question 4 
Question 5 
Question 6 
Exercise 14.3 
Question 1 
Question 2 
Question 3 
Question 4 
Question 5 
Question 6 
Question 7 
Exercise 14.4 
Question 1 
Question 2 
Question 3 