Question

Approximate the population mean and standard deviation of age for males.

Age Lower Limit Upper Limit Males Females
0-9 0 9 11 9
10-19 10 19 10 9
20-29 20 29 12 13
30-39 30 39 16 16
40-49 40 49 20 24
50-59 50 59 25 24
60-69 60 69 17 18
70-79 70 79 13 12

Answer 1

The population mean and standard deviation of age for males are 43.21 and 20.83, respectively

How to determine the population mean and the standard deviation?

The frequency table is given as:

Age Lower Limit Upper Limit Males Females

0-9 0 9 11 9

10-19 10 19 10 9

20-29 20 29 12 13

30-39 30 39 16 16

40-49 40 49 20 24

50-59 50 59 25 24

60-69 60 69 17 18

70-79 70 79 13 12

Rewrite the frequency table to show the class midpoints and the male frequency, only.

Midpoint (x)            Male Frequency (f)

4.5                          11

14.5                          10

24.5                          12

34.5                          16

44.5                          20

54.5                          25

64.5                          17

74.5                          13

The population mean is then calculated using:

$\mu =\frac{\sum fx}{\sum f}$

This gives

$\mu = \frac{11 * 4.5 + 10 * 14.5 + 12 * 24.5 + 16 * 34.5 + 20 * 44.5 + 25* 54.5 + 17 * 64.5 + 13 * 74.5}{11 + 10 + 12 + 16 + 20 + 25 + 17 + 13}$

Evaluate

$\mu = 43.21$

The standard deviation is calculated using:

$\sigma = \sqrt{\frac{\sum f(x - \mu)^2}{\sum f}}$

So, we have:

$\sigma = \sqrt{\frac{11 * (4.5 -43.21)^2+....................+ 13 * (74.5-43.21)^2}{11 + 10 + 12 + 16 + 20 + 25 + 17 + 13}}$

Evaluate the expression

$\sigma = 20.83$

Hence, the population mean and standard deviation of age for males are 43.21 and 20.83, respectively

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