TOPICS
Exercise - 14.1

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

Solution :-
  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2
    
Number of plantsNumber of houses (fi)xifixi
0 - 2111
2 - 4236
4 - 6155
6 - 85735
8 - 106954
10 - 1221122
12 - 1431339
TotalΣfi = 20Σfixi = 162
Now, x = Σfixi/Σfi = 162/20 = 8.1 Therefore, the 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.

Question-2 :-  Consider the following distribution of daily wages of 50 workers of a factory Statistics Find the mean daily wages of the workers of the factory by using an appropriate method

Solution :-
  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2
  Class size (h) = 20
  Taking assumed mean (a) = 150
    
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
TotalΣfi = 50Σfiui = -12
Now, x = a + (Σfiui/Σfi) x h = 150 + (-12/50) x 20 = 150 - 24/5 = 150 - 4.8 = 145.2 Therefore, the mean daily wage of the workers of the factory is Rs. 145.20.

Question-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. Statistics

Solution :-
  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2

  Given that the mean pocket allowance x = Rs. 18
  Taking assumed mean (a) = 18
    
Daily pocket allowance (in Rs.)Number of children (fi)xidi = xi - 18fidi
11 - 13712-6-42
13 - 15614-4-24
15 - 17916-2-18
17 - 19131800
19 - 21f2022f
21 - 23522420
23 - 25424624
TotalΣfi = 44 + fΣfidi = 2f - 40
Now, x = a + (Σfidi/Σfi) 18 = 18 + (2f - 40)/(44 + f) 18 - 18 = (2f - 40)/(44 + f) 0 = (2f - 40)/(44 + f) 0 = 2f - 40 40 = 2f f = 40/2 f = 20 Therefore, the missing frequency f is 20.

Question-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. Statistics

Solution :-
  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2
  Class size (h) = 3
  Taking assumed mean (a) = 75.5
    
Number of heart beats (per min.)Number of women (fi)xidi = xi - 75.5ui = 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
TotalΣfi = 30Σfiui = 4
Now, x = a + (Σfiui/Σfi) x h = 75.5 + (4/30) x 3 = 75.5 + 4/10 = 75.5 + 0.4 = 75.9 Therefore, the mean hear beats per minute for these women are 75.9 beats per minute.

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

Solution :-
  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 upper class limit and 1/2 has to be substracted from the lower class limit of each interval.
  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2
  Class size (h) = 3
  Taking assumed mean (a) = 57
    
Number of mangoesNumber of boxes (fi)xidi = 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
TotalΣfi = 400Σfiui = 25
Now, x = a + (Σfiui/Σfi) x h = 57 + (25/400) x 3 = 57 + 3/16 = 57 + 0.1875 = 57.1875 = 57.19 (approx.) Therefore, the mean number of mangoes kept in a packing box is 57.19.

Question-6 :- The table below shows the daily expenditure on food of 25 households in a locality. Statistics Find the mean daily expenditure on food by a suitable method.

Solution :-
  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2
  Class size (h) = 50
  Taking assumed mean (a) = 225
    
Daily expenditure (in Rs.)Number of households (fi)xidi = xi - 225ui = di/50fiui
100 - 1504125-100-2-8
150 - 2005175-50-1-5
200 - 25012225000
250 - 30022755012
300 - 350232510024
TotalΣfi = 25Σfiui = -7
Now, x = a + (Σfiui/Σfi) x h = 225 + (-7/25) x 50 = 225 - 14 = 211 Therefore, the mean daily expenditure on food is Rs. 211.

Question-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: Statistics Find the mean concentration of SO2 in the air.

Solution :-
  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2
  Class size (h) = 0.04
  Taking assumed mean (a) = 0.14
    
Concentration of SO2 (in ppm)Frequency (fi)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 - 0.2040.180.414
0.20 - 0.2420.220.824
TotalΣfi = 30Σfiui = -31
Now, x = a + (Σfiui/Σfi) x h = 0.14 + (-31/30) x 0.04 = 0.14 - 0.04133 = 0.09867 = 0.099 ppm (approx.) Therefore, the mean concentration of SO2 in the air is 0.099 ppm.

Question-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. Statistics

Solution :-
  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2
  Taking assumed mean (a) = 17
    
Number of daysNumber of students (fi)xidi = xi - 17fidi
0 - 6113-14-154
6 - 10108-9-90
10 - 14712-535
14 - 2041700
20 - 28424728
28 - 383331648
38 - 401392222
TotalΣfi = 40Σfidi = -181
Now, x = a + (Σfidi/Σfi) = 17 + (-181/40) = 17 - 4.525 = 12.475 = 12.48 (approx.) Therefore, the mean number of days is 12.48 days for which a student was absent.

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

Solution :-
  From given table, to find the class mark (xi) for each inerval, the following relation is used.
  Class mark (xi) = (upper class limit + lower class limit)/2
  Class size (h) = 10
  Taking assumed mean (a) = 70
    
Literacy rate (in %)Number of cities (fi)xidi = xi - 70ui = di/10fiui
45 - 55350-20-2-6
55 - 651060-10-1-10
65 - 751170000
75 - 858801018
85 - 953902026
TotalΣfi = 35Σfiui = -2
Now, x = a + (Σfiui/Σfi) x h = 70 + (-2/35) x 10 = 70 - 4/7 = 70 - 0.57 = 69.43 Therefore, the mean literacy rate is 69.43%.
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