Statistic problem on z score

Question

Professor smith administer a survey to a group of inmates that is believed to measure antisocial attitudes. The average score on the antisocial attitude index is 100 with a standard deviation of 25.

a. What percent of inmates have a score lower than 50?

b. If the survey is valid, we would expect to find that only 5% of individual score above a score of ____ on the index.

Answer 1

x = score

want to find P(x < 50)

convert inequality:

P(x < 50 ) = P(z < (50 - mean)/SD) = P(z < (50 - 100)/25) = p(z < -2) = 0.0228

find z corresponding 95th percentile = 1.645

1.645 = (x - 100)/25

x = 141.124 ~ 141