Exercise 13.1
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. Which method did you use for finding the mean, and why?
| Number of plants | 0 - 2 | 2 - 4 | 4 - 6 | 6 - 8 | 8 - 10 | 10 - 12 | 12 - 14 |
|---|---|---|---|---|---|---|---|
| Number of houses | 1 | 2 | 1 | 5 | 6 | 2 | 3 |
Solution:
| No.of plants | Mid value( | No.of Houses( | |
|---|---|---|---|
| 0 - 2 | 1 | ||
| 2 -4 | 2 | ||
| 4 - 6 | 1 | ||
| 6 - 8 | 5 | ||
| 8 - 10 | 6 | ||
| 10 - 12 | 2 | ||
| 12 - 14 | 3 | ||
| Σ | Σ |
We know that mean , X =
=
Hence, mean = 8.1
Direct method is
2. Consider the following distribution of daily wages of 50 workers of a factory.
| Daily wages(in ₹) | 500 - 520 | 520 - 540 | 540 - 560 | 560 - 580 | 580 - 600 |
|---|---|---|---|---|---|
| Number of workers | 12 | 14 | 8 | 6 | 10 |
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 followig relation is used.
Taking 55o as assured mean (a),
| Daily wages(in Rs) | Number of workers( | ||||
|---|---|---|---|---|---|
| 500 - 520 | 12 | ||||
| 520 - 540 | 14 | ||||
| 540 - 560 | 8 | ||||
| 560 - 580 | 6 | ||||
| 580 - 600 | 10 | ||||
| Total | 50 |
From the table, it can be observed that
Mean x = a +
=
=
= 550 -
Therefore, the mean daily wage of the workers of the factory is Rs 545.20.
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.
| Daily pocket allowance(in ₹) | 11 - 13 | 13 - 15 | 15 - 17 | 17 - 19 | 19 - 21 | 21 - 23 | 23 - 25 |
|---|---|---|---|---|---|---|---|
| Number of children | 7 | 6 | 9 | 13 | f | 5 | 4 |
Solution :
Here, mean = x = 18
| Class interval | Number of children( | Class-mark( | |
|---|---|---|---|
| 11 - 13 | 7 | ||
| 13 - 15 | 6 | ||
| 15 - 17 | 9 | ||
| 17 - 19 | 13 | ||
| 19 - 21 | f | ||
| 21 - 23 | 5 | ||
| 23 - 25 | 4 | ||
| Total | Σ | Σ |
Mean = x =
18 =
18(44 + f) = (752 + 20f)
792 +
792 - 752 = 20f - 18f
f =
Hence,the missing frequency f = 20.
4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.
| Number of heartbeats per minute | 65 - 68 | 68 - 71 | 71 - 74 | 74 - 77 | 77 - 80 | 80 - 83 | 83 - 86 |
|---|---|---|---|---|---|---|---|
| Number of women | 2 | 4 | 3 | 8 | 7 | 4 | 2 |
Solution :
Let us assume the mean as A = 75.5
Class size (h) = 3
| Class Interval | Number of women ( | Mid-point( | ||
|---|---|---|---|---|
| 65 - 68 | 2 | |||
| 68 - 71 | 4 | |||
| 71 - 74 | 2 | |||
| 74 - 77 | 8 | |||
| 77 - 80 | 7 | |||
| 80 - 83 | 4 | |||
| 83 - 86 | 2 | |||
| Sum | Sum |
Mean = x = A + h
75.5 =
75.5 +
75.5 +
Therefore, the mean heartbeats 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.
| Number of mangoes | 50 - 52 | 53 - 55 | 56 - 58 | 59 - 61 | 62 - 64 |
|---|---|---|---|---|---|
| Number of boxes | 15 | 110 | 135 | 115 | 25 |
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution :
We may observe that class intervals are not continuous. There is a gap of 1 between two class intervals.So we have to add
Class size h of thid data = 3
Now, taking 57 as assumed mean(a), we may calculate
| Class interval | ||||
|---|---|---|---|---|
| 49.5 - 52.5 | 15 | 51 | -6 | -90 |
| 52.5 - 55.5 | 110 | |||
| 55.5 - 58.5 | 135 | |||
| 58.5 - 61.5 | 115 | |||
| 61.5 - 64.5 | 25 | |||
| Total |
Now, we may observe that:
Mean x = a +
= 57 +
= 57 +
=
Clearly, mean number of mangoes kept in a packing boc is 57.2.
6. The table below shows the daily expenditure on food of 25 households in a locality.
| Daily expenditure(in ₹) | 100 - 150 | 150 - 200 | 200 - 250 | 250 - 300 | 300 - 350 |
|---|---|---|---|---|---|
| Number of households | 4 | 5 | 12 | 2 | 2 |
Find the mean daily expenditure on food by a suitable method.
Solution :
Let the assumed mean a = 225 and h = 50
| Daily expenditure (in Rs) | |||||
|---|---|---|---|---|---|
| 100 - 150 | 4 | 125 | -100 | -2 | -8 |
| 150 - 200 | 5 | ||||
| 200 - 250 | 12 | ||||
| 250 - 300 | 2 | ||||
| 300 - 350 | 2 | ||||
| Total | Σ | Σ |
Σ
Mean x = a +
=
= 225 -
Therefore, mean daily expenditure on foof is Rs 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.
| Concentration of | Frequency |
|---|---|
| 0.00 - 0.04 | 4 |
| 0.04 - 0.08 | 9 |
| 0.08 - 0.12 | 9 |
| 0.12 - 0.16 | 2 |
| 0.16 - 0.20 | 4 |
| 0.20 - 0.24 | 2 |
Find the mean concentration of
Solution :
Class size of this data = 0.04
Taking 0.14 as assumed mean (a) ,
| Concentration of | Frequency | Class mark | |||
|---|---|---|---|---|---|
| 0.00 - 0.04 | 4 | 0.02 | -0.12 | -3 | -12 |
| 0.04 - 0.08 | 9 | ||||
| 0.08 - 0.12 | 9 | ||||
| 0.12 - 0.16 | 2 | ||||
| 0.16 - 0.20 | 4 | ||||
| 0.20 - 0.24 | 2 | ||||
| Total |
Σ
Mean , x = a +
= 0.14 +
= 0.14 -
Therefore, mean concentration of
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 - 6 | 6 - 10 | 10 - 14 | 14 - 20 | 20 - 28 | 28 - 38 | 38 - 40 |
|---|---|---|---|---|---|---|---|
| Number of students | 11 | 10 | 7 | 4 | 4 | 3 | 1 |
Solution :
| Class interval | Frequenct( | Mid-point( | |
|---|---|---|---|
| 0 - 6 | 11 | ||
| 6 - 10 | 10 | ||
| 10 - 14 | 7 | ||
| 14 - 20 | 4 | ||
| 20 - 28 | 4 | ||
| 28 - 38 | 3 | ||
| 38 - 40 | 1 | ||
| Sum | Sum |
Mean , x =
=
Hence, the mean number of days a student was absent = 12.48.
9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
| Literacy rate (in %) | 45 - 55 | 55 - 65 | 65 - 75 | 75 - 85 | 85 - 95 |
|---|---|---|---|---|---|
| Number of cities | 3 | 10 | 11 | 8 | 3 |
Class mark =
=
Let the assumed mean(a) be 70
The class size(h) is = 55 - 45 = 10
| Class interval | Frequency( | Class mark( | |||
|---|---|---|---|---|---|
| 45 - 55 | 3 | 50 | -20 | -2 | -6 |
| 55 - 65 | 10 | ||||
| 65 - 75 | 11 | ||||
| 75 - 85 | 8 | ||||
| 85 - 95 | 3 | ||||
| Total | Σ | Σ |
We have, a =
Mean = a +
= 70 +
= (70 -