Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2?
-n = 1104, p = 0.93
A)Minimum: 1009.76; maximum: 1043.68
B)Minimum: 1043.68; maximum: 1009.76
C)Minimum: 1014.73; maximum: 1038.71
D)Minimum: 1018.24; maximum: 1035.2

Question

Quizplus · Accepted Answer

The minimum usual value (µ - 2σ) and the maximum usual value (µ + 2σ) for a binomial distribution can be found using the formulas µ = np and σ = sqrt(np(1-p)). Here, n = 1104 and p = 0.93, so µ = 1104 * 0.93 = 1026.72 and σ = sqrt(1104 * 0.93 * 0.07) ≈ 8.48. Therefore, µ - 2σ ≈ 1026.72 - 2*8.48 ≈ 1009.76 and µ + 2σ ≈ 1026.72 + 2*8.48 ≈ 1043.68.

Use the Given Values of N and P to Find