Statistics- Exercise 14.1 – Class 10

  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 plants0-12-44-66-108-1010-1212-14
number of houses1215623

Which method did you use for finding the mean, and why?

Solution:

To find the class mark (xi) for each interval, the following relation is used.

Class mark 𝒙𝒊 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕+𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐

xi and fixi can be calculated as follows.

Number of plantsnumber of housesxifixi
0 – 2111
2 – 4232 x 3 = 6
4 – 6151 x 5 = 5
6 – 8575 x 7 = 35
8 – 10696  x 9 = 54
10 – 122112 x 11 = 22
12 – 143133 x13 = 39
Total20 162

From the table, it can be observed that

∑fi = 20

∑fixi = 162

Statistics- Exercise 14.1 – Class 10

= 162/20 = 8.1

Therefore, mean number of plants per house is 8.1.

Here, direct method has been used as the values of class marks (xi) and fi are small.


  1. Consider the following distribution of daily wages of 50 worker of a factory.
Daily wages(in Rs)100-120120-140140-160160-180180-200
number of workers12148610

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

Solution:

To find the class mark for each interval, the following relation is used.

𝒙𝒊 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐

Class size (h) of this data = 20

Taking 150 as assured mean (a), di, ui, and fiui can be calculated as follows.

Daily wages(in  Rs)Number of workers(fi)xidi =  xi  – 150ui=di/20fiui
100 – 12012110-40-2-24
120 -14014130-20-1-14
140 – 1608150000
160  – 18061702016
180 – 2001019040220
total50-12

From the table, it can be observed that

∑fi = 50

∑fixi = -12

Statistics- Exercise 14.1 – Class 10

 

=150+(-12/50)20

 = 150 – 24/5

= 150 – 4.8 = 145.2

Therefore, mean number of plants per house is 145.20


  1. 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 – 1313 – 1515 – 1717 – 19 19 – 2121 – 2323 – 25
Number of workers76913f54

Solution:

To find the class mark (xi) for each interval, the following relation is used.

𝒙𝒊 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐

Given that, mean pocket allowance,

Taking 18 as assured mean (a), di and fidi are calculated as follows.

Daily pocket allowance(in  Rs)Number of children (fi)class mark xidi =  xi  – 18fidifiui
11 – 13712-6– 42-24
13 -15614-4-24-14
15 – 17916-2-180
17  – 191318006
19 – 21f2022f20
21 – 23522420-12
23 – 2542462424
totalfi = 44 + f

 

2f – 40

From above,

fi = 44 + f

fiui = 2f – 40

 

Statistics- Exercise 14.1 – Class 10

18 = 18 +(2f-40)/(44+f)

2f – 40 = 0

2f = 40

f = 20

Hence, the missing frequency, f, is 20.


4: 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 minute65 – 6868 – 7171 – 7474 – 7777 – 8080 – 8383 – 86
Number of women2438742

Solution:

To find the class mark of each interval (xi), the following relation is used.

𝒙𝒊 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐

Class size, h, of this data = 3

Taking 75.5 as assumed mean (a), di, ui, fiui are calculated as follows.

Number of heart beats per minuteNumber of women fixidi = xi − 75𝒖𝒊 = 𝒅𝒊/𝟑fiui
65 – 68266.5-9-3-6
68 – 71469.5-6-2-8
71 – 74372.5-3-1-3
74 – 77875.5000
77 – 80778.5317
80 – 83481.5628
83 – 86284.5936
total304

From the table

fi = 30

fiui = 4

Statistics- Exercise 14.1 – Class 10

= 75.5+(4/30)x3

= 75.5 + 4/30 x 3

= 75.5 – 0.4

= 75.9

Therefore, mean hear beats per minute for these women are 75.9 beats per minute.


  1. 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 mangoes50 – 5253 – 5556 – 5859 – 6162 – 64
Number of boxes1511013511525

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Solution:

Number of mangoesNumber of boxes fi
50 – 5215
53 – 55110
56 – 58135
59 – 61115
62 – 6425

It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals. Therefore, 1/2 has to be added to the upper class limit and 1/2 has to be subtracted from the lower class limit of each interval.

Class mark (xi) can be obtained by using the following relation.

𝒙𝒊 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐

Class size (h) of this data = 3

Here we are using Step deviation method

Taking 57 as assumed mean (a), di, ui, fiui are calculated as follows.

class intervalfixidi = xi-57ui = di/3fiui
49.5 – 52.51551-6-2-30
52.5 – 55.511054-3-1-110
55.5 – 58.513557000
58.5 – 61.51156031115
61.5 – 64.525636250
total40025

fi = a + (∑fiui/∑fi)xh

= 57 + (25/400)x3

= 57 + 3/16 = 57 + 0.1875

= 57.1875 = 57.19

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

Daily expenditure(in Rs)100 – 150150 – 200200 – 250250 – 300300 – 350
Number of households451222

Find the mean daily expenditure on food by a suitable method.

Solution:

To find the class mark (xi) for each interval, the following relation is used.

𝒙𝒊 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐

Class size = 50

Taking 225 as assumed mean (a), di, ui, fiui are calculated as follows.

Daily expenditure(in Rs)fixidi = xi – 225ui == di/50fiu­i
100 – 1504125-100-2-8
150 – 2005175-50-1-5
200 – 25012225000
250 – 30022755012
300 – 350232510024
total25-7

fi = 25

fiui = a + (fiui/ ∑fi)xh

= 225 + (-7/25)x50

= 225 – 14

= 211

Therefore, mean daily expenditure on food is Rs 211.


  1. 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 SO2Frequency
00.0 – 0.044
0.04 – 0.089
0.08 – 0.129
0.12 – 0.162
0.16 – 0.204
0.20 – 0.242

Find the mean concentration of SO2 in the air.

Solution:

To find the class marks for each interval, the following relation is used.

𝒙𝒊 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐

Class size of this data = 0.04

Taking 0.14 as assumed mean (a), di, ui, fiui are calculated as follows.

Concentration of SO2frequency ficlass mark xidi = xi – 0.14ui = di/0.04fiui
0.00-0.0440.02-0.12-3-12
0.04-0.0890.06-0.08-2-18
0.08-0.1290.10-0.04-1-9
0.12-0.1620.14000
0.16-0.2040.180.0414
0.20-0.2420.220.0824
total30   -31

∑fi = 30

∑fiui = -31

Statistics- Exercise 14.1 – Class 10

 

=0.14 +(-31/30)(0.04)

=0.14 – 0.04133

=0.09867

=0.099ppm

Therefore, mean concentration of SO2 in the air is 0.099 ppm.


  1. 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 days0-66-1010-1414-2020-2828-3838-49
number of students111074431

Solution:

To find the class mark of each interval, the following relation is used.

𝒙𝒊 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐

Taking 17 as assumed mean (a), di and fidi are calculated as follows.

Number of daysnumber of students fixidi = xi – 17fidi
0-6113-14-154
6-10108-9-90
10-14712-5-35
14-2041700
20-28424728
28-383331648
38-401392222
total40  -181

∑fi = 40

∑fidi = -181

Statistics- Exercise 14.1 – Class 10

= 17 + (-181/40)

= 17 – 4.525

= 12.475

= 12.48

Therefore, the mean number of days is 12.48 days for which a student was absent.


9: The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate(in%)45-5555-6565-7575-8585-95
number of cities3101183

Solution:

To find the class marks, the following relation is used.

𝒙𝒊 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐

Class size (h) for this data = 10 Taking 70 as assumed mean (a), di, ui, and fiui are calculated as follows.

Literacy rate(in%)number of cities fixidi = xi – 17ui = di/10fidi
45-55350-20-2-6
55-651060-10-1-10
65-751170000
75-858801018
85-953902026
total35   -2

∑fi = 35

∑fidi = -2

Statistics- Exercise 14.1 – Class 10

= 70 + (-2/35)x10

= 17 – 20/35 = 17 – 4/7

= 70 – 0.57

= 69.43

Therefore, mean literacy rate is 69.43%


 

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