SeaFair Fashions relies on its sales force of 400 to do an initial screening of

Question

SeaFair Fashions relies on its sales force of 400 to do an initial screening of all new fashions. The company is presently bringing out a new line of swimwear and has invited 30 salespeople to its Orlando home office. An issue of constant concern to the SeaFair sales office is the volume of orders generated by each salesperson. Last year the overall company average was $417,330 with a standard deviation of $45,285. a. What shape do you think the distribution of all possible sample means of 30 will have? Discuss. b. Determine the value of the standard deviation of the distribution of the sample mean of all possible samples of size 30. c. Determine the probability that the sample of 30 will have a sales average less than $400,000. d. How would the answer to part a, b, and c change if the home office brought 60 salespeople to Orlando? Provide the respective answers for this sample size. e. Each year SeaFair invites the sales personnel with sales above the 90th percentile to enjoy a complimentary vacation in Hawaii. Determine the smallest sales level for the sales personnel who were awarded a trip to Hawaii last year. (Assume the distribution of sales was normally distributed last year).

Explanation / Answer

a. What shape do you think the distribution of all possible sample means of 30 will have? Discuss.

As the sample size is large enough (n = 30), then the sampling distribution of the mean will be approimately normall distirbuted, by central limit theorem. [ANSWER: APPROXIMATELY NORMAL]

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b. Determine the value of the standard deviation of the distribution of the sample mean of all possible samples of size 30.

By central limit theorem,

sigma(X) = sigma/sqrt(n) = 45285/sqrt(30) = $ 8267.872006 [ANSWER]

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c. Determine the probability that the sample of 30 will have a sales average less than $400,000.

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    400000      
u = mean =    417330      
n = sample size =    30      
s = standard deviation =    45285      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -2.096065346      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2.096065346   ) =    0.018038197 [ANSWER]

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d. How would the answer to part a, b, and c change if the home office brought 60 salespeople to Orlando? Provide the respective answers for this sample size.

For a, it will be closer in shape to a normal distribution.

For b, it will be smaller, becasue standard error decreases with increasing sample size.

For c, it will be a smaller probability, because of the less variation, it is harder to get a sample mean far from the population mean.

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e. Each year SeaFair invites the sales personnel with sales above the 90th percentile to enjoy a complimentary vacation in Hawaii. Determine the smallest sales level for the sales personnel who were awarded a trip to Hawaii last year. (Assume the distribution of sales was normally distributed last year).

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,          
          
where          
          
u = mean =    417330      
z = the critical z score =    1.281551566      
s = standard deviation =    45285      
          
Then          
          
x = critical value = $ 475365.0626   [ANSWER]  

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