Question

Use the given values of n and p to find the minimum usual value µ - 2σ and the maximum usual value µ + 2σ.

n = 1130, p = 0.94

Group of answer choices

Minimum: 1050.91; maximum: 1073.49

Minimum: 1046.23; maximum: 1078.17

Minimum: 1054.22; maximum: 1070.18

Minimum: 1078.17; maximum: 1046.23

Answer 1

$\mu$ = n p = 1130 * 0.94 = 1060.2

$\sigma$ = sqrt [ n p ( 1 - p) ]

= sqrt [ 1130 * 0.94 ( 1 - 0.94) ]

Minimum = $\mu$ - 2 $\sigma$ = 1060.2 - 2 *sqrt [ 1130 * 0.94 ( 1 - 0.94) ] = 1046.23

Mximum = $\mu$ + 2 $\sigma$ = 1060.2 + 2 *sqrt [ 1130 * 0.94 ( 1 - 0.94) ] = 1078.17