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

Question

1. 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. (2 points)
b. Determine the value of the standard deviation of the distribution of the sample mean
of all possible samples of size 30. (2 points)
c. Determine the probability that the sample of 30 will have a sales average less than
$400,000. (4 points)
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. (6
points)
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). (4 points)

I have to answer questions in PHstat. So please tell me also in phstat what will I use

Explanation / Answer

let X be the random variable denoting the sales figures of a randomly chosen sales person.

Last year the overall
company average was $417,330 with a standard deviation of $45,285.

assumption is that X follows a normal distribution.

hence X~N(417330,452852)

a random sample of n=30 sales persons X1,X2,...,X30 are chosen. let Xbar be the sample mean

a) since X follows a normal distribution and Xbar=(X1+X2+...+X30)/30 is a linear combination of normal variables.

hence Xbar also follows a normal distribution.

hence the shape of the distribution of Xbar is bell shaped. [answer]

b) so E[Xbar]=E[(X1+X2+...+X30)/30]=30*417330/30=$417330

V[Xbar]=V[(X1+X2+...+X30)/30]=30*452852/302=452852/30

hence the standard deviation of Xbar is SD(Xbar)=sqrt[452852/30]=45285/sqrt(30)=$8267.87   [answer]

hence Xbar~N(417330,8267.872)

c) the required probability is

P[Xbar<400000]=P[(Xbar-417330)/8267.87<(400000-417330)/8267.87]=P[Z<-2.0961] where Z~N(0,1)

                       =0.0180367 [answer] [using minitab]

d) if instead of a sample of size 30, a sample of size 60 is considered . let the new sample mean be Xbar"

then there would be no change in part a) because it does not depends on the sample size.

for part b) the standard deviation would then be SD(Xbar")=45285/sqrt(60)=$5846.268

and E[Xbar"]=$417330 as it does not depend on the sample size.

for part c) P[Xbar"<400000]=P[(Xbar"-417330)/5846.268<(400000-417330)/5846.268]=P[Z<-2.95915] where Z~N(0,1)

                                        =0.0015424   [answer]   [suing minitab]

e) let the required sales level is k.

90the percentile means that value of sales under which 90% of the total probability occurs

so P[X<k]=0.90

or P[(X-417330)/45285<(k-417330)/45285]=0.90

or, P[Z<(k-417330)/45285]=0.90=P[Z<1.28155]    where Z~N(0,1) [uding minitab]

hence (k-417330)/45285=1.28155   or, k=417330+45285*1.28155=$475364.9918 [answer]

hence the smallest sales level is $475364.9918 for the sales personnel who were awarded a trip to Hawaii last year.

any sales value greater than this value is also eligible for the Hawaii trip.

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