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**

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