Consider a large population of scores from a bimodal distrib

Consider a large population of scores from a bimodal distribution with mean ? = 78 and standard deviation ? = 7. Imagine that you randomly selected a sample from that population that contained 50 scores. What is the PROBABILITY that the mean of the sample is above 81 or below 75?

Solution

n = 50

mean = 78

s.d = 7

z = (x-mean)/(s.d/sqrt(n))

P(75<X<81) = P((75-78)/(7/sqrt(50) < Z < (81-78)/(7/sqrt(50))

= P(-3.03<Z<3.03)

= 0.99878 - (1-0.99878)

= 0.99756

But question is asking for P(X>81 or X<75) = 1 - P(75<X<81)

= 1 - 0.99756

= 0.00244

So P(X>81 or X<75) = 0.00244