Assume that the weights of a male students X is normally dis

Assume that the weights of a male students, X is normally distributed with a mean weight of 160 pounds and standard deviation of 20 pounds. out of 1000 men, how many students weigh between 150 and 170 pounds?

I know the \"n\" means that S is in the denominator for the Z formula: x-mean/S and S= stand dev/sqr root of n. When I do it for 150 and 170 they become like 15 something value, which does not work with the Standard Normal Table. Please Help. Thanks.

Solution

Normal Distribution
Mean ( u ) =160
Standard Deviation ( sd )=20
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 150) = (150-160)/20
= -10/20 = -0.5
= P ( Z <-0.5) From Standard Normal Table
= 0.30854
P(X < 170) = (170-160)/20
= 10/20 = 0.5
= P ( Z <0.5) From Standard Normal Table
= 0.69146
P(150 < X < 170) = 0.69146-0.30854 = 0.3829