The table below shows the distribution of ages of workers in a company. Age/ yr...
The table below shows the distribution of ages of workers in a company.
| Age/ yr | 17 - 21 | 22 - 26 | 27 - 31 | 32 - 36 | 37 - 41 | 42 - 46 | 47 - 51 | 52 - 56 |
| Workers | 12 | 24 | 30 | 37 | 45 | 25 | 10 | 7 |
(a) Using an assumed mean of 39, calculate the (i) mean (ii) standard deviation; of the distribution.
(b) If a worker is selected at random from the company for an award, what is the probability that he is at most 36 years old?
Explanation
Age (in years) | No of workers | Mid-point (x) | \(d = x - A\) | \(d^{2}\) | \(fd^{2}\) | \(fd\) |
| 17 - 21 | 12 | 19 | -20 | 400 | 4800 | -240 |
| 22 - 26 | 24 | 24 | -15 | 225 | 5400 | -360 |
| 27 - 31 | 30 | 29 | -10 | 100 | 3000 | -300 |
| 32 - 36 | 37 | 34 | -5 | 25 | 925 | -185 |
| 37 - 41 | 45 | 39 | 0 | 0 | 0 | 0 |
| 42 - 46 | 25 | 44 | 5 | 25 | 625 | 125 |
| 47 - 51 | 10 | 49 | 10 | 100 | 1000 | 100 |
| 52 - 56 | 7 | 54 | 15 | 225 | 1575 | 105 |
| Total | 190 | 17325 | -755 |
(a)(i) Mean, \(\bar{x}\) = \(A + \frac{\sum fd}{\sum f}\)
= \(39 + \frac{-755}{190}\)
= \(39 - 3.974 = 35.026\)
(ii) Standard deviation \(SD = \sqrt{\frac{\sum fd^{2}}{N} - (\frac{\sum fd}{N})^{2}}\)
= \(\sqrt{\frac{17325}{190} - (\frac{-755}{190})^{2}}\)
= \(\sqrt{91.184 - (-3.974)^{2}}\)
\(\sqrt{91. 184 - 15.794} = \sqrt{75.39}\)
= \(8.68\)
(b) Total number of workers = 190
No of workers who are at most 36 years old = 12 + 24 + 30 + 37 = 103
p(award to a person at most 36 years old) = \(\frac{103}{190}\)
= 0.542