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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]