Question

Assume the average weight of an American adult male is 180 pounds with a standard deviation of 34 pounds. The distribution of weights follows a normal distribution. What is the probability that a man weighs somewhere between 120 and 155 pounds?

Answer 1

Explanation:

Find a-score of both

z-score = (x-mean)/SD

for 120

z =( 120- 180)/34 = -1.765

For 155

z = (155-180)/34 = -0.735

The probability to look for using z-score table is;

P(-1.765<z<-0.735) = 0.19239