Homework SourseThe distribution of body weights for Caucasian men in a part
The distribution of body weights for Caucasian men in a particular age group has a normal distribution with µ = 190 pounds and = 24 pounds.
1. For randomly selected samples of n = 16 men, what is the expected value of the sampling distribution of the sample mean?
2. For randomly selected samples of n = 16 men, what is the standard error of the sampling distribution of the sample mean?
3. What would be the minimum sample size required to make the standard error of the sampling distribution of the 4. sample mean less than 2.4? (Round to a whole number.)
4. If I randomly select one man, what is the probability that his weight is 195 pounds or more?
5. If I randomly select 100 men, what is the probability that their average weight is 195 pounds or more?
Solution
1.
It is the population mean itself,
u(X) = 190 lbs [ANSWER]
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2.
SE = sigma/sqrt(n)
= 24/sqrt(16)
= 6 [ANSWER]
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3.
SE = sigma/sqrt(n) < 4
24/sqrt(n) < 4
sqrt(n) > 24/4
sqrt(n) > 6
n > 36
Thus, the smallest number greater than 36 is
n = 37 [ANSWER]
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4.
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 195
u = mean = 190
s = standard deviation = 24
Thus,
z = (x - u) / s = 0.208333333
Thus, using a table/technology, the right tailed area of this is
P(z > 0.208333333 ) = 0.417484353 [ANSWER]
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5.
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 195
u = mean = 190
n = sample size = 100
s = standard deviation = 24
Thus,
z = (x - u) * sqrt(n) / s = 2.083333333
Thus, using a table/technology, the right tailed area of this is
P(z > 2.083333333 ) = 0.018610425 [ANSWER]
