SOLUTION: Hi, I just need help on doing the steps. I figured out it was 30, but do not remember how to do it. Thanks! Assume that body temperatures of healthy adults are normally distrib

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Question 675089: Hi, I just need help on doing the steps. I figured out it was 30, but do not remember how to do it. Thanks!
Assume that body temperatures of healthy adults are normally distributed with a mean of 98.2�F and a standard deviation of 0.62�F. Is P(98 < xbar < 98.4) greater for n =15 or
n = 30?
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Assume that body temperatures of healthy adults are normally distributed with a mean of 98.2�F and a standard deviation of 0.62�F.
Is P(98 < xbar < 98.4) greater for n =15 or n = 30?
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Standard deviation for the distribution of x-bar is s/sqrt(n)
So std is smaller when n is larger.
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But z-value = (x-u)/[s/sqrt(n)]
so z-value and n are directly related;
so z-value is larger as n is larger
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Therefore the probability is greater when n = 30.
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Note:
z(98) = (98-98.2)/[0.62/sqrt(15)] = -1.2493
z(98) = (98-98.2)/[0.62/sqrt(30)] = -1.7668
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Cheers,
Stan H.