Question

The manager of an electric supply store measured the diameters of the rolls of wire in the inventory. The diameters of the rolls (in m) are listed below. Calculate the standard deviation.

0.523    0.139    0.466    0.403   0.343    0.13    0.482

Answer 1

Sample mean = $\sum$ X / n

Sample mean, x = Sgma X/n = (0.523 + 0.139 + 0.466 + 0.403 + 0.343 + 0.13 + 0.482)/7 = 2.486/7 = 0.35514 Standard deviation S = Sqrt(Sigma X^2 - nx^2/n - 1) = squareroot [((0.523^2 + 0.139^2 + 0.466^2 + 0.403^2 + 0.343^2 + 0.13^2 + 0.482^2) - 7 * 0.35514^2)/16] = 0.1615

= ( 0.523 + 0.139 + 0.466 + 0.403 + 0.343 + 0.13 + 0.482) / 7

= 2.486 / 7

= 0.35514

Standard deviation S = Sqrt( $\sum$ X² - n² / n-1)

Sample mean, x = Sgma X/n = (0.523 + 0.139 + 0.466 + 0.403 + 0.343 + 0.13 + 0.482)/7 = 2.486/7 = 0.35514 Standard deviation S = Sqrt(Sigma X^2 - nx^2/n - 1) = squareroot [((0.523^2 + 0.139^2 + 0.466^2 + 0.403^2 + 0.343^2 + 0.13^2 + 0.482^2) - 7 * 0.35514^2)/16] = 0.1615

= sqrt [ ( ( 0.523² + 0.139² + 0.466² + 0.403² + 0.343² + 0.13² + 0.482²) - 7 * 0.35514² ) / 6 ]

= 0.1615