NCERT Solutions for Class 10 Maths Ch 14 Statistics
Here, we have provided NCERT Solutions for Class 10 Maths Chapter 14 Statistics which will be helpful in attaining good marks in the examinations. These NCERT Solutions for Class 10 are prepared by Studyrankers experts who have great experience in teaching students. You can also check your answers with these NCERT Solutions and complete your homework in no time.
Study Materials for Class 10 Maths Chapter 14 Statistics 


Exercise 14.1
1. 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  02  24  46  68  810  1012  1214 
Number of Houses  1  2  1  5  6  2  3 
Answer
No. of plants (Class interval) 
No. of houses (f_{i})  Midpoint (x_{i})  f_{i}x_{i } 
02  1  1  1 
24  2  3  6 
46  1  5  5 
68  5  7  35 
810  6  9  54 
1012  2  11  22 
1214  3  13  39 
Sum f_{i }= 20_{}  Sum f_{i}x_{i} = 162 
Mean = x̄ = ∑f_{i}x_{i} /∑f_{i }= 162/20 = 8.1_{}
We would use direct method because the numerical value of f_{i} and x_{i} are small.
2. Consider the following distribution of daily wages of 50 workers of a factory.
Daily wages (in Rs.)  100120  120140  140160  160180  180200 
Number of workers  12  14  8  6  10 
Answer
Here, the value of midpoint (x_{i}) is very large, so assumed mean A = 150 and class interval is h = 20.
So, u_{i }= (x_{i}  A)/h = u_{i }= (x_{i}  150)/20
Daily wages (Class interval) 
Number of workers frequency (f_{i}) 
Midpoint (x_{i})  u_{i }= (x_{i}  150)/20  f_{i}u_{i } 
100120  12  110  2  24 
120140  14  130  1  14 
140160  8  150  0  0 
160180  6  170  1  6 
180200  10  190  2  20 
Total  Sum f_{i }= 50_{}  Sum f_{i}u_{i} = 12 
Thus, mean daily wage = 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.
Answer
Here, the value of midpoint (x_{i}) mean x̄ = 18
Class interval  Number of children (f_{i})  Midpoint (x_{i})  f_{i}x_{i } 
1113  7  12  84 
1315  6  14  84 
1517  9  16  144 
1719  13  18 = A  234 
1921  f  20  20f 
2123  5  22  110 
2325  4  24  96 
Total  f_{i} = 44+f_{}  Sum f_{i}x_{i} = 752+20f 
Mean = x̄ = ∑f_{i}x_{i} /∑f_{i }= (752+20f)/(44+f)
⇒ 18 = (752+20f)/(44+f)
⇒ 18(44+f) = (752+20f)
⇒ 792+18f = 752+20f
⇒ 792+18f = 752+20f
⇒ 792  752 = 20f  18f
⇒ 40 = 2f
⇒ f = 20
Page No: 271
4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
Answer
x_{i }= (Upper limit + Lower limit)/2
Class size (h) = 3
Assumed mean (A) = 75.5
Class Interval  Number of women (f_{i})  Midpoint (x_{i})  u_{i} = (x_{i}  75.5)/h  f_{i}u_{i} 
6568  2  66.5  3  6 
6871  4  69.5  2  8 
7174  3  72.5  1  3 
7477  8  75.5  0  0 
7780  7  78.5  1  7 
8083  4  81.5  3  8 
8386  2  84.5  3  6 
Sum f_{i}= 30  Sum f_{i}u_{i }= 4 
Mean = x̄ = A + h∑f_{i}u_{i} /∑f_{i }= 75.5 + 3×(4/30) = 75.5 + 4/10 = 75.5 + 0.4 = 75.9
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.
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Answer
Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit.
Here, assumed mean (A) = 57
Class size (h) = 3
Class Interval  Number of boxes (f_{i})  Midpoint (x_{i})  d_{i} = x_{i}  A  f_{i}d_{i} 
49.552.5  15  51  6  90 
52.555.5  110  54  3  330 
55.558.5  135  57 = A  0  0 
58.561.5  115  60  3  345 
61.564.5  25  63  6  150 
Sum f_{i} = 400  Sum f_{i}d_{i} = 75 
Mean = x̄ = A + ∑f_{i}d_{i} /∑f_{i }= 57 + (75/400) = 57 + 0.1875 = 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.
Here, assumed mean (A) = 225
Class Interval  Number of households (f_{i})  Midpoint (x_{i})  d_{i} = x_{i}  A  f_{i}d_{i} 
100150  4  125  100  400 
150200  5  175  50  250 
200250  12  225  0  0 
250300  2  275  50  100 
300350  2  325  100  200 
Sum f_{i} = 25  Sum f_{i}d_{i} = 350 
Mean = x̄ = A + ∑f_{i}d_{i} /∑f_{i }= 225 + (350/25) = 225  14 = 211
The mean daily expenditure on food is 211
7. 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:
Answer
Concentration of SO_{2 }(in ppm)  Frequency (f_{i})  Midpoint (x_{i}) 
f_{i}x_{i}

0.000.04  4  0.02  0.08 
0.040.08  9  0.06  0.54 
0.080.12  9  0.10  0.90 
0.120.16  2  0.14  0.28 
0.160.20  4  0.18  0.72 
0.200.24  2  0.20  0.40 
Total  Sum f_{i} = 30  Sum (f_{i}x_{i}) = 2.96 
Mean = x̄ = ∑f_{i}x_{i} /∑f_{i}
= 2.96/30 = 0.099 ppm
Page No. 272
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
 06  610  1014  1420  2028  2838  3840 

Number of students  11  10  7  4  4  3  1 
Answer
Class interval  Frequency (f_{i}) 
Midpoint (x_{i})

f_{i}x_{i}

06  11  3  33 
610  10  8  80 
1014  7  12  84 
1420  4  17  68 
2028  4  24  96 
2838  3  33  99 
3840  1  39  39 
Sum f_{i} = 40  Sum f_{i}x_{i} = 499 
Mean = x̄ = ∑f_{i}x_{i} /∑f_{i}
= 499/40 = 12.48 days
9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean
literacy rate.
Literacy rate (in %)  4555  5565  6575  7585  8598 
Number of cities  3  10  11  8  3 
Answer
Class Interval  Frequency (f_{i})  (x_{i})  d_{i} = x_{i}  a  u_{i} = d_{i}/h 
f_{i}u_{i}

4555  3  50  20  2  6 
5565  10  60  10  1  10 
6575  11  70  0  0  0 
7585  8  80  10  1  8 
8595  3  90  20  2  6 
Sum f_{i} = 35  Sum f_{i}u_{i} = 2 
Mean = x̄ = a + (∑f_{i}u_{i} /∑f_{i}) Ñ… h
= 70 + (2/35) Ñ… 10 = 69.42
Page No. 275
Exercise 14.2
1. The following table shows the ages of the patients admitted in a hospital during a year:
Age (in years)  515  1525  2535  3545  4555  5565 
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
Modal class = 35 – 45, l = 35, class width (h) = 10, f_{m} = 23, f_{1} = 21 and f_{2} = 14
Calculation of Mean:
Class Interval  Frequency (f_{i})  Midpoint (x_{i})  f_{i}x_{i} 
515  6  10  60 
1525  11  20  220 
2535  21  30  630 
3545  23  40  920 
4555  14  50  700 
5565  5  60  300 
Sum f_{i} = 80  Sum f_{i}x_{i} = 2830 
Mean = x̄ = ∑f_{i}x_{i} /∑f_{i}
= 2830/80 = 35.37 yr
2. The following data gives the information on the observed lifetimes (in hours) of 225
electrical components :
Lifetime (in hours)  020  2040  4060  6080  80100  100120 
Frequency  10  35  52  61  38  29 
Determine the modal lifetimes of the components.
Answer
Modal class of the given data is 60–80.
Modal class = 6080, l = 60, f_{m} = 61, f_{1} = 52, f_{2} = 38 and h = 20
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 
10001500  24 
15002000  40 
20002500  33 
25003000  28 
30003500  30 
35004000  22 
40004500  16 
45005000  7 
Answer
Modal class = 15002000, l = 1500, f_{m} = 40, f_{1} = 24, f_{2} = 33 and h = 500
Calculation for mean:
Class Interval  fi  xi  di = xi  a  ui = di/h  fiui 
10001500  24  1250  1500  3  72 
15002000  40  1750  1000  2  80 
20002500  33  2250  500  1  33 
25003000  28  2750  0  0  0 
30003500  30  3250  500  1  30 
35004000  22  3750  1000  2  44 
40004500  16  4250  1500  3  48 
45005000  7  4750  2000  4  28 
fi = 200  fiui = 35 
Mean = x̄ = a + (∑f_{i}u_{i} /∑f_{i}) Ñ… h
= 2750 + (35/200) Ñ… 500
= 2750  87.50 = 2662.50
Go Back To NCERT Solutions for Class 10 Maths
NCERT Solutions for Class 10 Maths Ch 14 Statistics
Chapter 14 Statistics Class 10 Maths that carry height weightage in the board examination so you must prepare for it before exams. There are total 5 topics in the chapter which will be useful in learning efficiently.
• Introduction: In earlier class, we have studied the classification of given data into ungrouped as well as grouped frequency distributions. We have also learnt to represent the data pictorially in the form of various graphs such as bar graphs, histograms etc.
• Mean of Grouped Data: The mean of observations is the sum of the values of all the observations divided by the total number of observations. The exercise 14.1 contains questions about finding mean of a given data.
• Mode of Grouped Data: A mode is that value among the observations which occurs most often, that is, the value of the observation having the maximum frequency. We need to find mean and mode of given observations in the exercise.
• Median of Grouped Data: The median is a measure of central tendency which gives the value of the middlemost observation in the data. The exercise 14.3 contains questions about finding means median and mode of given observations.
• Graphical Representation of Cumulative Frequency Distribution: A graphical representation helps us
in understanding given data with a quick look. We are going to represent a cumulative frequency distribution graphically.
We have also provided exercise NCERT Solutions for Class 10 Maths Chapter 14 that are helpful in improving student's experience.
 Exercise 14.1
 Exercise 14.2
 Exercise 14.3
 Exercise 14.4
NCERT Solutions for Class 10 Maths Chapters:
FAQ on Chapter 14 Statistics
How many exercises in Chapter 14 Statistics
There are total 5 exercise in the Chapter 14 Statistics which are very important for the examinations and guide students in better way.
What are the methods to calculate Mean?
Direct Method, Shortcut Method and Step Deviation Method.
Mode = (...........) − 2 (Mean)
Mode = 3 (median) − 2 (Mean)
Write the empirical relation between mean, mode and median.
The three measures i.e., mean, mode and median are connected by the following empirical relation:
Mode = 3 Median – 2 Mean