Assumed mean, A = 19.

Age (years)	Mid-age (x)	Frequency	\(d = x - A\)	\(fd\)	\(fd^{2}\)
12 - 14	13	6	-6	-36	216
15 - 17	16	10	-3	-30	90
18 - 20	19	3	0	0	0
21 - 23	22	2	3	6	18
24 - 26	25	1	6	6	36
		22		-54	360

(a) \(\bar{x} = A + \frac{\sum fd}{\sum f}\)
= \(19 + \frac{-54}{22}\)
= \(19 - 2.455\)
= \(16.545 \approxeq 16.5\) years.
(b) Standard deviation , \(\sigma = \sqrt{\frac{\sum fd^{2}}{\sum f} - (\frac{\sum fd}{\sum f})}\)
= \(\sqrt{\frac{360}{22} - (\frac{-54}{22})}\)
= \(\sqrt{16.364 - (2.45)^{2}}\)
= \(\sqrt{16.364 - 6.025}\)
= \(\sqrt{10.339}\)
= 3.215 \(\approxeq\) 3.22 years.