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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 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 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. 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. 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. 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. 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: 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. 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. 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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