Use the given values of n and p to find the minimum usual value µ - 2? and the maximum usual value µ + 2?
-$$n = 2277 , p = \frac { 3 } { 4 }$$
A)Minimum: 1598.41; maximum: 1817.09
B)Minimum: 1666.43; maximum: 1749.07
C)Minimum: 1749.07; maximum: 1666.43
D)Minimum: 1687.09; maximum: 1728.41

Question

Quizplus · Accepted Answer

To find the minimum and maximum usual values, we first calculate the mean (µ) and standard deviation (σ) of the binomial distribution. The mean of a binomial distribution is given by µ = n * p, and the standard deviation is given by σ = sqrt(n * p * (1 - p)). For n = 2277 and p = 3/4, µ = 2277 * (3/4) = 1707.75, and σ = sqrt(2277 * (3/4) * (1/4)) ≈ 20.66. The minimum usual value is µ - 2σ = 1707.75 - 2(20.66) = 1666.43, and the maximum usual value is µ + 2σ = 1707.75 + 2(20.66) = 1749.07.

Use the Given Values of N and P to Find n=2277,p=34n = 2277 , p = \frac { 3 } { 4 }n=2277,p=43​

Use the Given Values of N and P to Find $n = 2277 , p = \frac { 3 } { 4 }$