The scores of 12th-grade students on the National Assessment of Educational Prog
ID: 3176441 • Letter: T
Question
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 = 293 and standard deviation = 32. Choose one 12th-grader at random. What is the probability (±±0.1) that his or her score is higher than 293? Higher than 389 (±±0.001)? Now choose an SRS of 4 twelfth-graders and calculate their mean score x¯¯¯x¯. If you did this many times, what would be the mean of all the x¯¯¯x¯-values? What would be the standard deviation (±±0.1) of all the x¯¯¯x¯-values? What is the probability that the mean score for your SRS is higher than 293? (±±0.1) Higher than 389? (±±0.0001)
Explanation / Answer
Solution
Back-up Theory
If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then
Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)
P(X or t) = P[{(X - µ)/ } or {(t - µ)/ }] = P[Z or {(t - µ)/ }] .………(2)
X bar ~ N(µ, 2/n),…………………………………………………………….…….(3),
where X bar is average of a sample of size n from population of X.
So, P(X bar or t) = P[Z or {(n)(t - µ)/ }] …………………………………(4)
Probability values for the Standard Normal Variable, Z, can be directly read off from
Standard Normal Tables or using Excel Function……………………………………..(5)
Now, to work out solution,
Let X = score of Twelfth-grader in Math test. Then, we are given X ~ N(293, 322).
Part (a)
Probability that score of one 12th-grader chosen at random is higher than 293 =
P(X > 293) = P(Z > {(293 – 293)/32} [vide (2) under Back-up Theory]
= P(Z > 0) = 0.5 [using Excel Function] ANSWER
[Note: The above probability could have been directly written as 0.5 since for Normal distribution, mean = median]
Part (b)
Probability that score of one 12th-grader chosen at random is higher than 389 =
P(X > 389) = P(Z > {(389 – 293)/32} [vide (2) under Back-up Theory]
= P(Z > 3) = 0.00135 [using Excel Function] ANSWER
[Note: The above probability also could have been directly written as 0.00135 since for Normal distribution, probability more than 3 sigma from the mean = 0.00135 is a known important property.]
Part (c)
[vide (3) under Back-up Theory], mean of the sample averages = 293 and Standard deviation = 32/2 = 16 ANSWER
Part (d)
[vide (3) and (4) under Back-up Theory], probability that the mean score for SRS is higher than 293 = P(Xbar > 293) = 0.5 ANSWER [exactly same logic as in Part (a)]
Part (e)
[vide (3) and (4) under Back-up Theory], probability that the mean score for SRS is higher than 389 = P(Xbar > 389) = P(Z > 6) = 0 ANSWER [exactly same logic as in Part (b)]