The scores of 12thgrade students on the National Assessment

The scores of 12th-grade students on the National Assessment of Educational Progress year 2000 mathematics test have a distribution that is approximately Normal with mean = 329 and standard deviation = 38 .

Choose one 12th-grader at random. What is the probability (±0.1) that his or her score is higher than 329 ? Higher than 443 (±0.001)?

Now choose an SRS of 4 twelfth-graders and calculate their mean score x. If you did this many times, what would be the mean of all the x-values?

What would be the standard deviation (±0.1) of all the x-values?

What is the probability that the mean score for your SRS is higher
than 329 ? (±0.1) Higher than 443 ? (±0.0001)

Solution

Normal Distribution
Mean ( u ) =329
Standard Deviation ( sd )=38
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X > 329) = (329-329)/38
= 0/38 = 0
= P ( Z >0) From Standard Normal Table
= 0.5                  
P(X > 443) = (443-329)/38
= 114/38 = 3
= P ( Z >3) From Standard Normal Table
= 0.0013                  
b)
SRS of 4 twelfth-graders, Mean ( u ) =329

c)
Mean ( u ) =329
Standard Deviation ( sd )= 38/ Sqrt ( 4 ) = 19

d)
P(X > 329) = (329-329)/38/ Sqrt ( 4 )
= 0/19= 0
= P ( Z >0) From Standard Normal Table
= 0.5                  
P(X > 443) = (443-329)/38/ Sqrt ( 4 )
= 114/19= 6
= P ( Z >6) From Standard Normal Table
= 0