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