Question

# Assume that a procedure yields a binomial distribution with n trials and the probability of success...

Assume that a procedure yields a binomial distribution with n trials and the probability of success for one trial is p. Use the given values of n and p to find the mean μ and standard deviation σ. ​Also, use the range rule of thumb to find the minimum usual value μ−2σ and the maximum usual value μ+2σ.

n=1405, p= 2 / 5

Solution :

Given that,

p = 2 / 5 = 0.4

q = 1 - p = 1 - 0.4 = 0.6

n = 1405

Using binomial distribution,

Mean = = n * p = 1405 * 0.4 = 562

Standard deviation = = n * p * q = 1405 * 0.4 * 0.6 = 9.0

minimum usual value = μ−2σ

minimum usual value = 562 - 2 * 9

minimum usual value = 544

maximum usual value = μ+2σ

maximum usual value = 562 + 2 * 9

maximum usual value = 580

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