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 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 |
Which method did you use for finding the mean, and why?
To find the mean value, we will use the direct method because the numerical value of $f_i$ and $x_i$ are small.
Find the midpoint of the given interval using the formula.
| Midpoint ($x_i$) = $\dfrac{(upper limit + lower limit)}{2}$ |
| No. of plants(Class interval) | No. of houses (Frequency (fi)) | Mid-point ($x_i $) | fi$x_i$ |
|---|---|---|---|
| 0-2 | 1 | 1 | 1 |
| 2- 4 | 2 | 3 | 6 |
| 4-6 | 1 | 5 | 5 |
| 6- 8 | 5 | 7 | 35 |
| 8-10 | 6 | 9 | 54 |
| 10- 12 | 2 | 11 | 22 |
| 12-14 | 3 | 13 | 39 |
| Sum $f_i$ = 20 | Sum$ f_{i}x_i$ = 162 |
The formula to find the mean is:
| Mean = $\overline{x} = \dfrac{∑f{i}x_{i}}{∑f_i}$ |
= $\dfrac{162}{20}$
= 8.1
Therefore, the mean number of plants per house is 8.1.
Consider the following distribution of daily wages of 50 workers of a factory.
| Daily wages (in Rs.) | 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.
Find the midpoint of the given interval using the formula.
| Midpoint $(x_i)$ = $\dfrac{(upper limit + lower limit)}{2}$ |
In this case, the value of Mid-point $(x_i)$ is very large, so let us assume the mean value, a = 550.
Class interval (h) = 20
So, $u_i = \dfrac{(x_i – a)}{h}$
$u_i$ = $\dfrac{(x_i – 550)}{20}$
Substitute and find the values as follows:
| Daily wages(Class interval) | Number of workers (frequency $(f_i)$) | Mid-point $(x_i)$ | $u_i$ = $\dfrac{(x_i – 550)}{20}$ | $f_i u_i$ |
|---|---|---|---|---|
| 500- 520 | 12 | 510 | -2 | -24 |
| 520-540 | 14 | 530 | -1 | - 14 |
| 540-560 | 8 | 550 = a | 0 | 0 |
| 560- 580 | 6 | 570 | 1 | 6 |
| 580- 600 | 10 | 590 | 2 | 20 |
| Total | Sum $f_i$ = 50 | Sum $f_iu_i$ = -12 |
So, the formula to find out the mean is:
| Mean = $\overline{x} = a + h(\dfrac{∑f_iu_i }{∑f_i} )$ |
= 550 + [20 × $(-\dfrac{12}{50})$] = 550 – 4.8 = 545.20
Thus, mean daily wage of the workers = Rs. 545.20
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 c) | 11-13 | 13-15 | 15- 17 | 17-19 | 19-21 | 21-23 | 23-35 |
| Number of children | 7 | 6 | 9 | 13 | f | 5 | 4 |
To find out the missing frequency, use the mean formula.
Given, mean $\overline{x}$ = 18
| Class interval | Number of children $(f_i)$ | Mid-point $(x_i)$ | $f_ix_i $ |
|---|---|---|---|
| 11-13 | 7 | 12 | 84 |
| 13- 15 | 6 | 14 | 84 |
| 15-17 | 9 | 16 | 144 |
| 17- 19 | 13 | 18 | 234 |
| 19-21 | f | 20 | 20f |
| 21- 23 | 5 | 22 | 110 |
| 23-25 | 4 | 24 | 96 |
| Total | $f_i$ = 44+f | Sum$ f_{i}x_i$ = 752+20f |
The mean formula is
| Mean = $\overline{x} = \dfrac{∑f_ix_i }{∑f_i}$ |
= $\dfrac{(752 + 20f)}{(44 + f)}$
Now substitute the values and equate to find the missing frequency (f)
18 = $\dfrac{(752 + 20f)}{(44 + f)}$
18(44 + f) = (752 + 20f)
792 + 18f = 752 + 20f
792 + 18f = 752 + 20f
792 – 752 = 20f – 18f
40 = 2f
f = 20
So, the missing frequency, f = 20.
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 heart beats 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 |
From the given data, let us assume the mean as a = 75.5
| $x_i = \dfrac{(Upper limit + Lower limit)}{2}$ |
Class size (h) = 3
Now, find the $u_i$ and $f_i u_i$ as follows:
| Class Interval | Number of women $(f_i)$ | Mid-point $(x_i)$ | $u_i$ = $\dfrac{(x_i – 75.5)}{h}$ | $f_i u_i$ |
|---|---|---|---|---|
| 65-68 | 2 | 66.5 | - 3 | -6 |
| 68-71 | 4 | 69.5 | -2 | -8 |
| 71- 74 | 3 | 72.5 | -1 | -3 |
| 74-77 | 8 | 75.5 = a | 0 | 0 |
| 77-80 | 7 | 78.5 | 1 | 7 |
| 80- 83 | 4 | 81.5 | 2 | 8 |
| 83- 86 | 2 | 84.5 | 3 | 6 |
| Sum $f_i$= 30 | Sum $f_i u_i$ = 4 |
| Mean = $\overline{x} = a + h(\dfrac{∑f_iu_i}{∑f_i}$) |
= 75.5 + 3 × ($\dfrac{4}{30}$)
= 75.5 + ($\dfrac{4}{10}$)
= 75.5 + 0.4
= 75.9
Therefore, the mean heart beats per minute for these women is 75.9
| Number of mangoes | 50-52 | 53-55 | 56-58 | 59- 61 | 62-64 |
| Number of boxes | 15 | 110 | 135 | 115 | 25 |
The given data is not continuous, so we add 0.5 to the upper limit and subtract 0.5 from the lower limit as the gap between two intervals is 1.
Here, assumed mean (a) = 57
Class size (h) = 3
Here, the step deviation is used because the frequency values are big.
| Class Interval | Number of boxes $(f_i)$ | Mid-point $(x_i)$ | $u_i$ = $ \dfrac{(x_i – 57)}{h}$ | $f_i u_i$ |
|---|---|---|---|---|
| 49.5-52.5 | 15 | 51 | - 2 | -30 |
| 52.5-55.5 | 110 | 54 | -1 | -110 |
| 55.5- 58.5 | 135 | 57 = a | 0 | 0 |
| 58.5- 61.5 | 115 | 60 | 1 | 115 |
| 61.5- 64.5 | 25 | 63 | 2 | 50 |
| Sum $f_i$ = 400 | Sum $f_i u_i$ = 25 |
The formula to find out the Mean is:
| Mean = $\overline{x} = a + h(\dfrac{∑f_iu_i }{∑f_i}$) |
= 57 + 3($\dfrac{25}{400}$)
= 57 + 0.1875
= 57.19
Therefore, the mean number of mangoes kept in a packing box is 57.19
| Daily expenditure(in c) | 100-150 | 150-200 | 200- 250 | 250-300 | 300-350 |
| Number of households | 4 | 5 | 12 | 2 | 2 |
Find the midpoint of the given interval using the formula.
| Midpoint $(x_i)$ = $\dfrac{(upper limit + lower limit)}{2}$ |
Let us assume the mean (a) = 225
Class size (h) = 50
| Class Interval | Number of households $(f_i)$ | Mid-point $(x_i)$ | di = $x_i$ – A | $u_i$ =$ \dfrac{di}{50}$ | $f_i u_i$ |
|---|---|---|---|---|---|
| 100-150 | 4 | 125 | - 100 | -2 | -8 |
| 150-200 | 5 | 175 | -50 | -1 | -5 |
| 200- 250 | 12 | 225 = a | 0 | 0 | 0 |
| 250- 300 | 2 | 275 | 50 | 1 | 2 |
| 300- 350 | 2 | 325 | 100 | 2 | 4 |
| Sum $f_i$ = 25 | Sum $f_i u_i$ = -7 |
| Mean = $\overline{x} = a + h(\dfrac{∑f_iu_i }{∑f_i}$) |
= 225 + 50($\dfrac{-7}{25}$)
= 225 – 14
= 211
Therefore, the mean daily expenditure on food is 211.
| Concentration of $SO_2$ ( in ppm) | 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 |
To find out the mean, first find the midpoint of the given frequencies as follows:
| Concentration of $SO_2$ (in ppm) | Frequency $(f_i)$ | Mid-point $(x_i)$ | $f_ix_i$ |
|---|---|---|---|
| 0.00- 0.04 | 4 | 0.02 | 0.08 |
| 0.04-0.08 | 9 | 0.06 | 0.54 |
| 0.08- 0.12 | 9 | 0.10 | 0.90 |
| 0.12-0.16 | 2 | 0.14 | 0.28 |
| 0.16- 0.20 | 4 | 0.18 | 0.72 |
| 0.20- 0.24 | 2 | 0.22 | 0.44 |
| Total | Sum $f_i$ = 30 | Sum $(f_ix_i)$ = 2.96 |
The formula to find out the mean is
| Mean = $\overline{x} = \dfrac{∑f_ix_i }{∑f_i}$ |
= $\dfrac{2.96}{30}$
= 0.099 ppm
Therefore, the mean concentration of $SO_2$ in the air is 0.099 ppm.
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 |
Find the midpoint of the given interval using the formula.
| Midpoint $(x_i)$ = $\dfrac{(upper limit + lower limit)}{2}$ |
| Class interval | Frequency $(f_i)$ | Mid-point $(x_i)$ | $f_ix_i$ |
|---|---|---|---|
| 0-6 | 11 | 3 | 33 |
| 6- 10 | 10 | 8 | 80 |
| 10-14 | 7 | 12 | 84 |
| 14- 20 | 4 | 17 | 68 |
| 20-28 | 4 | 24 | 96 |
| 28- 38 | 3 | 33 | 99 |
| 38-40 | 1 | 39 | 39 |
| Sum $f_i$ = 40 | Sum$ f_{i}x_i$ = 499 |
The mean formula is,
Mean = $\overline{x} = \dfrac{∑f_ix_i}{∑f_i}$
= $\dfrac{499}{40}$
= 12.48 days
Therefore, the mean number of days a student was absent = 12.48.
| Literacy rate (in %) | 45-55 | 55-65 | 65-75 | 75 -85 | 85-98 |
| Number of cities | 3 | 10 | 11 | 8 | 3 |
Find the midpoint of the given interval using the formula.
| Midpoint $(x_i)$ = $\dfrac{(upper limit + lower limit)}{2}$ |
In this case, the value of Mid-point $(x_i)$ is very large, so let us assume the mean value, a = 70.
Class interval (h) = 10
So, $u_i$ = $\dfrac{(x_i – a)}{h}$
$u_i$ = $\dfrac{(x_i – 70)}{10}$
Substitute and find the values as follows:
| Class Interval | Frequency $(f_i)$ | $(x_i)$ | $u_i = \dfrac{(x_i – 70)} {10}$ | $f_i u_i$ |
|---|---|---|---|---|
| 45-55 | 3 | 50 | -2 | - 6 |
| 55-65 | 10 | 60 | -1 | -10 |
| 65-75 | 11 | 70 = a | 0 | 0 |
| 75-85 | 8 | 80 | 1 | 8 |
| 85- 95 | 3 | 90 | 2 | 6 |
| Sum $f_i$ = 35 | Sum $f_i u_i$ = - 2 |
So,
| Mean = $\overline{x} = a + ( \dfrac{∑f_iu_i }{∑f_i}) × h$ |
= 70 + ($\dfrac{-2}{35}$) × 10
= 69.43
Therefore, the mean literacy part = 69.43%