Question

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 mu and standard deviation sigma. Also, use the range rule of thumb to find the minimum usual value mu minus 2 sigma and the maximum usual value mu plus 2 sigma. nequals50, pequals0.3

mu equals 15 (Do not round.)

sigma equals (Round to one decimal place as needed.)

Answer #1

Solution :

Given that,

p = 0.3

q = 1 - p = 1 - 0.3 = 0.7

n = 50

Using binomial distribution,

Mean = = n * p = 50 * 0.3 = 15

Standard deviation = = n * p * q = 50 * 0.3 * 0.7 = 3.2

minimum usual value = - 2

minimum usual value = 15 - 3.2

minimum usual value = 11.8

maximum usual value = + 2

maximum usual value = 15 + 3.2

maximum usual value = 18.2

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