Question

Use the given values of n an p to find the minimum usual value mean - 2standard deviations and the maximum usual value. N= 1056, p= .80

Answer 1

Answer:

The minimum usual value: 818.803

The maximum usual value: 870.797

The population mean: 844.8

The standard deviation: 12.9985

Step-by-step explanation:

The parameters provided for the Binomial Distribution are:

                                         n=1056, p=0.80

The population mean can be computed as:

                                               μ = n ⋅ p

                                                  = 1056 × 0.80 = 844.8

The population variance is obtained by:

                                                 σ² = n ⋅ p ⋅ (1 − p)

                                                     = 1056 × 0.80 × 0.2 = 168.96

The standard deviation can be computed by taking the squared root of the variance:

                σ = $\sqrt{n.p.(1-p)}$

                   = $\sqrt{168.96}$

                   = 12.9985

The minimum usual value: μ - 2σ = 844.8 - 2(12.9985)

                                                       = 844.8 - 25.997

                                                        = 818.803

The maximum usual value: μ + 2σ = 844.8 + 2(12.9985)

                                                         = 844.8 + 25.997

                                                          = 870.797

Keywords: minimum and maximum usual value, standard deviation, mean

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