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. nequals150​, pequals0.75 muequals ​(Do not​ round.)

Answer 1

n = 150 , p = 0.75

= np = 150 * 0.75 = 112.5

= sqrt ( n p (1 - p) ) = sqrt ( 150 * 0.75 ( 1 - 0.75) ) = 5.3033

minimum usual value = - 2

= 112.5 - 2 * 5.3033

= 101.8934

maximum usual value = + 2

= 112.5 + 2 * 5.3033

= 123.1066