From the data we collected the population mean height all my
Solution
1.
By central limit theorem:
a) It is approximately normally distributed.
b) It has the same mean, u(X) = 68.3 in.
c) It has a standard deviation of sigma/sqrt(n) = 3.7/sqrt(31) = 0.664539617 in.
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2.
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 69
u = mean = 68.3
s = standard deviation = 3.7
Thus,
z = (x - u) / s = 0.189189189
Thus, using a table/technology, the right tailed area of this is
P(z > 0.189189189 ) = 0.42497227 [ANSWER]
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3.
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 69
u = mean = 68.3
n = sample size = 31
s = standard deviation = 3.7
Thus,
z = (x - u) * sqrt(n) / s = 1.053360825
Thus, using a table/technology, the right tailed area of this is
P(z > 1.053360825 ) = 0.146087826 [ANSWER]
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4.
The probability that a sample of 31 students will have a mean of 69 in or more is 0.146088.
