NCERT Solutions for Class 10th: Ch 14 Statistics Maths

NCERT Solutions for Class 10th: Ch 14 Statistics Maths


NCERT Solutions for Class 10th: Ch 14 Statistics Maths

Page No: 270

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   0-2       2-4       4-6      6-8       8-10       10-12       12-14   
Number of Houses    1215623
Which method did you use for finding the mean, and why?

Answer

No. of plants
   (Class interval)  
No. of houses (fi)Mid-point (xi)    fixi    
0-2111
2-4236
4-6155
6-85735
8-106954
10-1221122
12-1431339

Sum f= 20

   Sum fixi = 162    

Mean = x̄ = ∑fixi /f= 162/20 = 8.1

We would use direct method because the numerical value of fi and xi are small.

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    12148610
Find the mean daily wages of the workers of the factory by using an appropriate method.

Answer

Here, the value of mid-point (xi) is very large, so assumed mean A = 150 and class interval is h = 20.
So, u= (xi - A)/h = u= (xi - 150)/20

Daily wages
   (Class interval)  
Number of workers
frequency (fi)
Mid-point (xi)u= (xi - 150)/20    fiui    
100-12012110-2-24
120-14014130-1-14
140-160815000
160-180617016
180-20010190220
TotalSum f= 50


Sum fiui = -12  
Mean = x̄ = A + h∑fiui /f=150 + (20 × -12/50) = 150 - 4.8 = 145.20
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 mid-point (xi)  mean  = 18

Class intervalNumber of children (fi)Mid-point (xi)    fixi    
11-1371284
13-1561484
15-17916144
17-191318 = A234
19-21f2020f
21-23522110
23-2542496
Totalfi = 44+f

Sum fixi = 752+20f 

Mean = x̄ = ∑fixi /f= (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= (Upper limit + Lower limit)/2
Class size (h) = 3
Assumed mean (A) = 75.5 

Class IntervalNumber of women (fi)Mid-point (xi)ui = (xi - 75.5)/hfiui
65-68266.5-3-6
68-71469.5-2-8
71-74372.5-1-3
74-77875.500
77-80778.517
80-83481.538
83-86284.536
Sum fi= 30Sum fiu= 4

Mean = x̄ = A + h∑fiui /f= 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 IntervalNumber of boxes (fi)Mid-point (xi)di = xi - Afidi
49.5-52.51551-690
52.5-55.511054-3-330
55.5-58.513557 = A00
58.5-61.5115603345
61.5-64.525636150
Sum fi = 400Sum fidi = 75

Mean = x̄ = A + ∑fidi /f= 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.

Answer

Here, assumed mean (A) = 225

Class IntervalNumber of households (fi)Mid-point (xi)di = xi - Afidi
100-1504125-100-400
150-2005175-50-250
200-2501222500
250-300227550100
300-3502325100200
Sum fi = 25Sum fidi = -350

Mean = x̄ = A + ∑fidi /f= 225 + (-350/25) = 225 - 14 = 211
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:
Find the mean concentration of SO2 in the air.

Answer

Concentration of SO(in ppm)Frequency (fi)Mid-point (xi)
fixi
0.00-0.0440.020.08
0.04-0.0890.060.54
0.08-0.1290.100.90
0.12-0.1620.140.28
0.16-0.2040.180.72
0.20-0.2420.200.40
TotalSum fi = 30Sum (fixi) = 2.96

Mean = x̄ = ∑fixi /fi
= 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
0-66-1010-1414-2020-2828-3838-40
Number of students111074431

Answer

Class intervalFrequency (fi)
Mid-point (xi)
fixi
0-611333
6-1010880
10-1471284
14-2041768
20-2842496
28-3833399
38-4013939
Sum fi = 40Sum fixi = 499

Mean = x̄ = ∑fixi /fi
= 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 %)45-5555-6565-7575-8585-98
Number of cities3101183

Answer

Class IntervalFrequency (fi)(xi)di = xi - aui = di/h
fiui
45-55350-20-2-6
55-651060-10-1-10
65-751170000
75-858801018
85-953902026
Sum fi  = 35Sum fiui  = -2

Mean = x̄ = a + (∑fiui /fi) х h
= 70 + (-2/35) х 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)5-1515-2525-3535-4545-5555-65
Number of patients6112123145

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, fm = 23, f1 = 21 and f2 = 14

Calculation of Mean:

Class IntervalFrequency (fi)Mid-point (xi)fixi
5-1561060
15-251120220
25-352130630
35-452340920
45-551450700
55-65560300
Sum fi = 80Sum fixi = 2830

Mean = x̄ = ∑fixi /fi
= 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)0-2020-4040-6060-8080-100100-120
Frequency103552613829

Determine the modal lifetimes of the components.

Answer

Modal class of the given data is 60–80.
Modal class = 60-80, l = 60, fm = 61, f1 = 52, f2 = 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 :


ExpenditureNumber of families
1000-150024
1500-200040
2000-250033
2500-300028
3000-350030
3500-400022
4000-450016
4500-50007

Answer

Modal class = 1500-2000, l = 1500, fm = 40, f1 = 24, f2 = 33 and h = 500

Calculation for mean:

Class Intervalfixidi = xi - aui = di/hfiui
1000-1500241250-1500-3-72
1500-2000401750-1000-2-80
2000-2500332250-500-1-33
2500-3000282750000
3000-3500303250500130
3500-40002237501000244
4000-45001642501500348
4500-5000747502000428
fi = 200fiui = -35

Mean = x̄ = a + (∑fiui /fi) х h
= 2750 + (35/200) х 500
= 2750 - 87.50 = 2662.50