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 2...
Question 1 Report
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?
Answer Details
Class width \(=5\); mid-values \(19,24,\dots,54\). Code with \(u=\dfrac{x-39}{5}\), \(A=39\).
| Age | Mid \(x\) | \(f\) | \(u\) | \(fu\) | \(fu^2\) |
|---|---|---|---|---|---|
| 17-21 | 19 | 12 | -4 | -48 | 192 |
| 22-26 | 24 | 24 | -3 | -72 | 216 |
| 27-31 | 29 | 30 | -2 | -60 | 120 |
| 32-36 | 34 | 37 | -1 | -37 | 37 |
| 37-41 | 39 | 45 | 0 | 0 | 0 |
| 42-46 | 44 | 25 | 1 | 25 | 25 |
| 47-51 | 49 | 10 | 2 | 20 | 40 |
| 52-56 | 54 | 7 | 3 | 21 | 63 |
| Total | 190 | -151 | 693 |
(a)(i) Mean.\[\bar{x}=39+\frac{\sum fu}{\sum f}\times 5=39+\frac{-151}{190}\times 5=39-3.97=35.0\text{ years}\]
(a)(ii) Standard deviation.\[\sigma=c\sqrt{\frac{\sum fu^2}{\sum f}-\left(\frac{\sum fu}{\sum f}\right)^2}=5\sqrt{\frac{693}{190}-\left(\frac{-151}{190}\right)^2}\]\[=5\sqrt{3.647-0.632}=5\sqrt{3.015}=5(1.737)=8.68\text{ years}\]
(b) Probability at most 36 years old. Ages \(\le 36\) cover the classes \(17-21,\,22-26,\,27-31,\,32-36\), giving \(12+24+30+37=103\) workers out of 190.\[P(\text{at most }36)=\frac{103}{190}=0.542\]
Mean \(\approx\) 35.0 years, standard deviation \(\approx\) 8.68 years, and \(P=0.542\).
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