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. nequals1410, pequals3 divided by 5 muequals nothing (Do not round.)

Answer #1

Solution :

Given that,

p = 3 / 5 = 0.6

q = 1 - p = 1 - 0.6 = 0.4

n = 1410

Using binomial distribution,

= mean = n * p = 1410 * 0.6 = 846

= standard deviation = n * p * q = 1410 * 0.6 * 0.4 = 18.395

minimum usual value = - 2 * = 846 - 2 * 18.395 = 809.21

maximum usual value = + 2 * = 846 + 2 * 18.395 = 882.79

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