A standard chemistry examination administered nationally by

A standard chemistry examination administered nationally by the American Chemical Society has a mean of 500 and a standard deviation of 90. What is the probability that the average of a random sample of examination scores of 25 students will be between 450 and 500?

Find a value for the standard normal variable such that the probability of a larger value is equal to .025.

Solution

Normal Distribution
Mean ( u ) =500
Standard Deviation ( sd )=90
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 450) = (450-500)/90
= -50/90 = -0.5556
= P ( Z <-0.5556) From Standard Normal Table
= 0.28926
P(X < 500) = (500-500)/90
= 0/90 = 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(450 < X < 500) = 0.5-0.28926 = 0.2107                  
b)
P ( Z < x ) = 0.025
Value of z to the cumulative probability of 0.025 from normal table is -1.96
P( x-u/s.d < x - 500/90 ) = 0.025
That is, ( x - 500/90 ) = -1.96
--> x = -1.96 * 90 + 500 = 323.6