Question

The weight of male students at a certain university is normally distributed with a mean of 175 pounds with a standard deviation of 7.6 pounds. Find the probabilities.

1. A male student weighs at most 186 pounds.

2. A male students weighs at least 160 pounds.

3. A male student weighs between 165 and 180 pounds.

Please show work. Ideally excel commands would be helpful, but anything would be great!

Answer 1

Given,

= 175 , = 7.6

We convert this to standard normal as

P(X < x) = P(Z < (x - ) / )

a)

P(X <= 186) = P(Z < (186 - 175) / 7.6)

= P(Z < 1.45)

= 0.9265

Excel = NORM.DIST(x , mean , SD, cumulative) = NORM.DIST(186 , 175 , 7.6 , TRUE)

= 0.9261

(Probability differs because excel takes more than 2 decimals for z score probability, manually we use z table so

rounded z - score for 2 decimals)

b)

P(X >= 160) = P(Z > (160 - 175) / 7.6)

= P(Z < -1.97)

= P(Z < 1.97)  

= 0.9756

Excel = 1 - NORM.DIST(160 , 175 , 7.6 , TRUE)

= 0.9758

c)

P(165 < X < 180) = P(X < 180) - P(X < 165)

= P(Z < (180 - 175) / 7.6) - P(Z < (165 - 175) / 7.6)

= P(Z < 0.66) - P(Z < -1.32)

= 0.7454 - 0.0934

= 0.6520

Excel = NORM.DIST(180 , 175 , 7.6 , TRUE) - NORM.DIST(165 , 175 , 7.6 , TRUE)

= 0.7447 - 0.0941

= 0.6506