For the population of adult males in the United States, the distribution of weig
ID: 3377125 • Letter: F
Question
For the population of adult males in the United States, the distribution of weights is approximately normal with mean mu = 172.2 pounds and standard deviation sigma = 29.8 pounds [10]. Describe the distribution of means of samples of size 25 that arc drawn from this population. What is the upper bound for 90% of the mean weights of samples of size 25? What is the lower bound for 80% of the mean weights? Suppose that you select a single random sample of size 25 and find that the mean weight for the men in the sample is = 190 pounds. How likely is this result? What would you conclude?Explanation / Answer
a)
1. It will be normally distirbuted, by central limit theorem.
2. It will have the same mean, ux = 172.2.
3. It will have a standard deviation of sigma/sqrt(n) = 29.8/sqrt(25) = 5.96.
b)
First, we get the z score from the given left tailed area. As
Left tailed area = 0.9
Then, using table or technology,
z = 1.281551566
As x = u + z * s / sqrt(n)
where
u = mean = 172.2
z = the critical z score = 1.281551566
s = standard deviation = 29.8
n = sample size = 25
Then
x = critical value = 179.8380473 [ANSWER]
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C)
First, we get the z score from the given left tailed area. As
Left tailed area = 0.2
Then, using table or technology,
z = -0.841621234
As x = u + z * s / sqrt(n)
where
u = mean = 172.2
z = the critical z score = -0.841621234
s = standard deviation = 29.8
n = sample size = 25
Then
x = critical value = 167.1839374 [ANSWER]
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C)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 190
u = mean = 172.2
n = sample size = 25
s = standard deviation = 29.8
Thus,
z = (x - u) * sqrt(n) / s = 2.986577181
Thus, using a table/technology, the right tailed area of this is
P(z > 2.986577181 ) = 0.001410598 [answer, how likely]
As this is very rare, then we might conclude that the mean is actually greater than 172.2. [conclusion]