Graduate students applying for entrance to many universities must take a Miller
ID: 3131705 • Letter: G
Question
Graduate students applying for entrance to many universities must take a Miller Analogies Test. It is known that the test scores have a mean of 75 and a variance of 16. In 1990, 100 students applied for entrance into graduate school in physics.
(a)[3] Find the mean and standard deviation of the sampling distribution of X .
(b)[2] Find the probability that the average score of this group of students is higher than 76.
(c)[3] Find the probability that the sample mean deviates from the population mean by less than 2.
(d)[4] Construct a 98% confidence interval for , the true mean test score.
Explanation / Answer
a)
By central limit theorem, the mean is still the population mean,
u(X) = 75 [ANSWER]
but the standard deviation is reduced by sqrt(n),
sigma(X) = sigma/sqrt(n) = sqrt(16)/sqrt(100) = 0.4 [ANSWER]
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b)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 76
u = mean = 75
n = sample size = 100
s = standard deviation = 4
Thus,
z = (x - u) * sqrt(n) / s = 2.5
Thus, using a table/technology, the right tailed area of this is
P(z > 2.5 ) = 0.006209665 [ANSWER]
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c)
That means the sample mean is between 73 and 77.
We first get the z score for the two values. As z = (x - u) sqrt(n) / s, then as
x1 = lower bound = 73
x2 = upper bound = 77
u = mean = 75
n = sample size = 100
s = standard deviation = 4
Thus, the two z scores are
z1 = lower z score = (x1 - u) * sqrt(n) / s = -5
z2 = upper z score = (x2 - u) * sqrt(n) / s = 5
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 2.86652E-07
P(z < z2) = 0.999999713
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.999999427 [ANSWER]
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d)
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.01
X = sample mean = 75
z(alpha/2) = critical z for the confidence interval = 2.326347874
s = sample standard deviation = 4
n = sample size = 100
Thus,
Margin of Error E = 0.93053915
Lower bound = 74.06946085
Upper bound = 75.93053915
Thus, the confidence interval is
( 74.06946085 , 75.93053915 ) [ANSWER]
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