The scores of 12th-grade students on the National Assessment of Educational Prog

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)]

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