Answer :
The mean and the standard deviation are 16.15 and 12.03, respectively
How to determine the mean and the standard deviation?
The table of values is given as:
Work hours Frequency
0 - <10 45
10 - <20 33
20 - <30 20
30 - <40 7
40 - <50 8
Start by rewriting the table using the midpoint of each class
x f
5 45
15 33
25 20
35 7
45 8
The mean is then calculated as:
[tex]\bar x = \frac{\sum fx}{\sum f}[/tex]
This gives
[tex]\bar x = \frac{5 * 45 + 15 * 33 + 25 * 20 35 * 7 + 45 * 8}{45 + 33 + 20 + 7 + 8}[/tex]
Evaluate the sum and the products
[tex]\bar x = \frac{1825}{113}[/tex]
Divide
[tex]\bar x = 16.15[/tex]
The standard deviation is:
[tex]\sigma =\sqrt{\frac{\sum f(x - \bar x)^2}{\sum f}}[/tex]
This gives
[tex]\sigma= \sqrt{\frac{45 * (5 - 16.15)^2 + 33 * (15 - 16.15)^2 + 20 * (25 - 16.15)^2 + 7 * (35 - 16.15)^2 + 8 * (45 - 16.15)^2 }{45 + 33 + 20 + 7 + 8}}[/tex]
Evaluate the sum and the products
[tex]\sigma= \sqrt{\frac{16350.4425}{113}}[/tex]
Divide
[tex]\sigma= \sqrt{144.694181416}[/tex]
Take the square root
[tex]\sigma= 12.03[/tex]
Hence, the mean and the standard deviation are 16.15 and 12.03, respectively
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