Question

The measurements of the diameters (in cms) of the plates prepared in a factory are given below. Find the standard deviation

Diameter (cm)	21 − 24	25 − 28	29 − 32	33 − 36	37 − 40	41 − 44
Number of plates	15	18	20	16	8	7

Answer 1

Assumed mean = 34.5

Diameter (cm)	Number of plates fi	mid value xi	di = xi − A = xi − 34.5	fidi	fidi2
21 − 24	15	22.5	− 12	− 180	2160
25 − 28	18	26.5	− 8	− 144	1152
29 − 32	20	30.5	− 4	− 80	320
33 − 36	16	34.5	0	0	0
37 − 40	8	38.5	4	32	128
41 − 44	7	42.5	8	56	448
	`sumf_"i"` = 84			`sumf_"i"d_"i"` = − 316	`sumf_"i""d"_"i"^2` = 4208

Here `sumf_"i"` = 84, `sumf_"i"d_"i"` = − 316, `sumf_"i""d"_"i"^2` = 4208

N = `sumf_i = 84`

Standard deviation (σ) = `sqrt((sumf_"i""d"_"i"^2)/(sumf_"i") - ((sumf_"i""d"_"i")/(sumf_"i"))^2`

= `sqrt(4208/84 - (316/84)`

= `sqrt(50.095 - (3.76)^2`

= `sqrt(50.1 - 14.14)`

= `sqrt(35.96)`

= 5.996

= 6

∴ Standard deviation (σ) = 6