1. A survey was conducted by a group of students as a part of their environment awareness program, 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 |
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Answer :
In order to find the mean value, we will use direct method because the numerical value of
The midpoint of the given interval is found by formula:
No. of plants (Class interval) | No. of houses (Frequency |
Mid-point |
|
0-2 | 1 | 1 | 1 |
2-4 | 2 | 3 | 6 |
4-6 | 1 | 5 | 5 |
6-8 | 5 | 7 | 35 |
8-10 | 6 | 9 | 54 |
10-12 | 2 | 11 | 22 |
12-14 | 3 | 13 | 39 |
2. Consider the following distribution of daily wages of 50 workers of a factory.
Daily wages (in Rs.) | 100-120 | 120-140 | 140-160 | 160-180 | 180-200 |
Number of workers | 12 | 14 | 8 | 6 | 10 |
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Answer :
The midpoint of the given interval is found by formula:
In this case, the value of mid-point (
So,
Daily wages (Class interval) | Number of workers (frequency ( |
Mid-point ( |
|
|
100-120 | 12 | 110 | -2 | -24 |
120-140 | 14 | 130 | -1 | -14 |
140-160 | 8 | 150 | 0 | 0 |
160-180 | 6 | 170 | 1 | 6 |
180-200 | 10 | 190 | 2 | 20 |
Total | |
So, the formula to find out the mean is :
? Mean daily wage of the workers = Rs. 145.20
3. 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 Allowence(in Rs) | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-35 |
Number of children | 7 | 6 | 9 | 13 | f | 5 | 4 |
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Answer :
Given,
Class interval | Number of children (frequency ( |
Mid-point |
|
11-13 | 7 | 12 | 84 |
13-15 | 6 | 14 | 84 |
15-17 | 9 | 16 | 144 |
17-19 | 13 | 18 | 234 |
19-21 | f | 20 | 20f |
21-23 | 5 | 22 | 110 |
23-25 | 4 | 24 | 96 |
Total | |
Thus,
So, the missing frequency, f = 20.
4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
Number of heartbeats 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 |
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Answer :
From the given data, let us assume the mean as A = 75.5
Class size (h) = 3
So,
Class Interval | Number of women ( |
Mid-point ( |
||
65-68 | 2 | 66.5 | -3 | -6 |
68-71 | 4 | 69.5 | -2 | -8 |
71-74 | 3 | 72.5 | -1 | -3 |
74-77 | 8 | 75.5 | 0 | 0 |
77-80 | 7 | 78.5 | 1 | 7 |
80-83 | 4 | 81.5 | 3 | 8 |
83-86 | 2 | 84.5 | 3 | 6 |
|
? The mean heart beats per minute for these women is 75.9
5. 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 |
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Answer :
Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.5 from the lower limit as the gap between two intervals are 1
Here, assumed mean (A) = 57, Class size (h) = 3
Here, the step deviation is used because the frequency values are big.
Class Interval | Number of boxes |
Mid-point |
||
49.5-52.5 | 15 | 51 | -6 | 90 |
52.5-55.5 | 110 | 54 | -3 | -330 |
55.5-58.5 | 135 | 57 | 0 | 0 |
58.5-61.5 | 115 | 60 | 3 | 345 |
61.5-64.5 | 25 | 63 | 6 | 150 |
The formula to find out the Mean is:
? The mean number of mangoes kept in a packing box is 57.19
6. The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.
Daily expenditure(in Rs) | 100-150 | 150-200 | 200-250 | 250-300 | 300-350 |
Number of households | 4 | 5 | 12 | 2 | 2 |
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Answer :
Let us assume the mean (A) = 225, Class size (h) = 50
Class Interval | Number of households |
Mid-point |
|
|
|
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 |
|
? The mean daily expenditure on food is 211
7. 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 |
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Answer :
Calculate mean by direct method:
Concentration of SO2 (in ppm) | Frequency |
Mid-point |
|
0.00 - 0.04 | 4 | 0.02 | 0.08 |
0.04 - 0.08 | 9 | 0.06 | 0.54 |
0.08 - 0.12 | 9 | 0.10 | 0.90 |
0.12 - 0.16 | 2 | 0.14 | 0.28 |
0.16 - 0.20 | 4 | 0.18 | 0.72 |
0.20 - 0.24 | 2 | 0.20 | 0.40 |
Total |
The formula to find out the mean is
? The mean concentration of SO2 in air is 0.099 ppm.
8. 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 |
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Answer :
Class interval | Frequency |
Mid-point |
|
0-6 | 11 | 3 | 33 |
6-10 | 10 | 8 | 80 |
10-14 | 7 | 12 | 84 |
14-20 | 4 | 17 | 68 |
20-28 | 4 | 24 | 96 |
28-38 | 3 | 33 | 99 |
38-40 | 1 | 39 | 39 |
The mean formula is,
? The mean number of days a student was absent = 12.48.
9. 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 | 85-98 |
Number of cities | 3 | 10 | 11 | 8 | 3 |
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Answer :
In this case, the value of mid-point (xi) is very large,
so let us assume the mean value, A = 70 and class interval is h = 10.
So,
Class Interval | Frequency |
|
|||
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 |
So,
? The mean literacy part is 69.42
1. 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 |
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Answer :
The class 35-45 has the maximum frequency, therefore, this is the modal class.
Here l = 35 , h = 10 , f1 = 23 , f0 = 21 , f2 =14
Now, let us substitute these values in the formula
Now, to calculate the Mean,
Class Interval | Frequency ( |
Mid-point ( |
|
5-15 | 6 | 10 | 60 |
15-25 | 11 | 20 | 220 |
25-35 | 21 | 30 | 630 |
35-45 | 23 | 40 | 920 |
45-55 | 14 | 50 | 700 |
55-65 | 5 | 60 | 300 |
? The mean of the given data = 35.37 years
2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
Lifetime (in hours) | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 |
Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Answer :
The class 60-80 has the maximum frequency, therefore, this is the modal class.
Here l = 60 , h = 20 , f1 = 61 , f0 = 52 , and f2 = 38
Now, let us substitute these values in the formula
? The modal lifetimes of the components is 65.625 hours.
3. 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 | 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 |
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Answer :
The class 1500 – 2000 has the maximum frequency. Therefore, it is the modal class.
Here l = 1500, h = 500, f1 =40 , f0 =24 and f2 =33
Now, let us substitute these values in the formula
Mode
? Modal monthly expenditure = Rs. 1847.83
Let the assumed mean be A = 2750 and h = 500.
Class Interval | |
|
|
|
|
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 |
|
4. 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
No 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 |
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Answer :
The class 30-35 has the maximum frequency, therefore, this is the modal class.
Here l = 30, h = 5, f1 = 10, f0 = 9 and f2 = 3
Now, let us substitute these values in the formula
Mode
Calculation of mean :
Class Interval | Frequency |
Mid-point |
|
15-20 | 3 | 17.5 | 52.5 |
20-25 | 8 | 22.5 | 180.0 |
25-30 | 9 | 27.5 | 247.5 |
30-35 | 10 | 32.5 | 325.0 |
35-40 | 3 | 37.5 | 112.5 |
40-45 | 0 | 42.5 | 0 |
45-50 | 0 | 47.5 | 0 |
50-55 | 2 | 52.5 | 105.5 |
5. The given distribution shows the number of runs scored by some top batsmen of the world in one- day international cricket matches.
Run Scored | Number 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 |
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Answer :
The class 4000-5000 has the maximum frequency ,
therefore, this is the modal class.
Here l = 4000, h = 1000, f1 = 18, f0 = 4 and f2 = 9
Now, let us substitute these values in the formula
Mode
? The mode of the given data is 4608.7 runs.
6. 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 | Frequency |
0-10 | 7 |
10-20 | 14 |
20-30 | 13 |
30-40 | 12 |
40-50 | 20 |
50-60 | 11 |
60-70 | 15 |
70-80 | 8 |
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Answer :
The class 40-50 has the maximum frequency, therefore, this is the modal class.
Here l = 40, h =10, f1 = 20, f0 = 12 and f2 = 11
Now, let us substitute these values in the formula
Mode
? Mode = 44.7 cars
1. The following frequency distribution gives the monthly consumption of an electricity of 68 consumers in a locality. Find the median, mean and mode of the data and compare them.
Monthly consumption(in units) | No. of customers |
65-85 | 4 |
85-105 | 5 |
105-125 | 13 |
125-145 | 20 |
145-165 | 14 |
165-185 | 8 |
185-205 | 4 |
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Answer :
Class Interval | Frequency | Cumulative frequency |
65-85 | 4 | 4 |
85-105 | 5 | 9 |
105-125 | 13 | 22 |
125-145 | 20 | 42 |
145-165 | 14 | 56 |
165-185 | 8 | 64 |
185-205 | 4 | 68 |
N=68 |
Class Interval | |||||
65-85 | 4 | 75 | -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 | 1 | 14 |
165-185 | 8 | 175 | 40 | 2 | 16 |
185-205 | 4 | 195 | 60 | 3 | 12 |
Mean = 137.05
In this case, mean, median and mode are more/less equal in this distribution.
2. If the median of a distribution given below is 28.5 then, find the value of an x &y.
Class Interval | Frequency |
0-10 | 5 |
10-20 | x |
20-30 | 20 |
30-40 | 15 |
40-50 | y |
50-60 | 5 |
Total | 60 |
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Answer :
Given that median is 28.5 and
Here n = 60
Since the median is given to be 28.5, thus the median class is 20-30.
? l = 20, h = 10, f = 20 and cf = 5 + x
? median
Also, 45 + x + y = 60 ???? [ As n = 60 ]
? x = 8, y = 7
Q.3 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 holder |
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 |
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Answer :
We are given the cumulative frequency distribution. So, we first construct a frequency table from the given cumulative frequency distribution and then we will make necessary computations to compute median .
Class interval | Frequency | Cumulative frequency |
15-20 | 2 | 2 |
20-25 | 4 | 6 |
25-30 | 18 | 24 |
30-35 | 21 | 45 |
35-40 | 33 | 78 |
40-45 | 11 | 89 |
45-50 | 3 | 92 |
50-55 | 6 | 98 |
55-60 | 2 | 100 |