For the population of adult males in the United States the d

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?

Solution

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]