Homework Sourse1 The scores of 12thgrade students on the National Assessmen
1. 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 = 325 and standard deviation = 37. Choose one 12th-grader at random. What is the probability (±0.1) that his or her score is higher than 325? Higher than 436 (±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 325? (±0.1) Higher than 436? (±0.0001)
Solution
a)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 325
u = mean = 325
s = standard deviation = 37
Thus,
z = (x - u) / s = 0
Thus, using a table/technology, the right tailed area of this is
P(z > 0 ) = 0.5 [ANSWER]
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b)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 436
u = mean = 325
s = standard deviation = 37
Thus,
z = (x - u) / s = 3
Thus, using a table/technology, the right tailed area of this is
P(z > 3 ) = 0.001349898 [ANSWER]
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c)
By central limit theorem, it will have the same mean,
Mean = 325 [ANSWER]
Their standard deviation would be
s(X) = s/sqrt(n) = 37/sqrt(4) = 18.5 [ANSWER]
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d)
As 325 is the mean, half of the population is greater than it,
P(x>325) = 0.5 [ANSWER]
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e)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 436
u = mean = 325
n = sample size = 4
s = standard deviation = 37
Thus,
z = (x - u) * sqrt(n) / s = 6
Thus, using a table/technology, the right tailed area of this is
P(z > 6 ) = 0.0000 [ANSWER]
