Question

The weight of football players is normally distributed with a mean of 190 pounds and a standard deviation of 20 pounds.

Answer the following questions rounding your solutions to 4 decimal places.

What is the minimum weight of the middle 95% of the players?

Answer 1

Given that,

mean = = 190

standard deviation = = 20

middle 95% of score is

P(-z < Z < z) = 0.95

P(Z < z) - P(Z < -z) = 0.95

2 P(Z < z) - 1 = 0.95

2 P(Z < z) = 1 + 0.95 = 1.95

P(Z < z) = 1.95 / 2 = 0.975

P(Z <1.96 ) = 0.975

z  ± 1.96 using standard normal (Z) table

Using z-score formula  

x= z * +

x= -1.96*20+190

x= 150.8000

minimum value =151