Question

This is a 5 Part Hw.

A local hardware store’s receipt’s show that on Saturday customer purchases have a skewed distribution with a mean of $56 and a standard deviation of $36.

a) Explain why you cannot determine the probability that the next Saturday customer will spend at least $65.

b) Can you estimate that the next 10 customers on Saturday will spend an average of at least $65? Explain your reasoning.

c) What is the probability that the next 50 customers on a Saturday will spend an average of at least $65?

d) This Saturday, the hardware store had 350 customers. Estimate the probability that the store’s revenues were at least $20,000.

e) If, on a typical Saturday, the store serves 350 customers, how much does the store take in on the worst 10% of such days?

Answer 1

a)

Since the distribution is skewed and not normal it would not be possible to predict the case

b)

Still the sample size is 10 which is lesser than 30 to get the normal distribution and hence we cannot estimate the

probability.

c)

µ =    56                                  
σ =    36                                  
n=   50                                  
                                      
X =   65                                  
                                      
Z =   (X - µ )/(σ/√n) = (   65   -   56   ) / (    36   / √   50   ) =   1.768
                                      
P(X ≥   65   ) = P(Z ≥   1.77   ) =   P ( Z <   -1.768   ) =    0.0385      

d)

µ =    56                                  
σ =    36                                  
n=   350                                  
                                      
X = 20000/350 = 57.14285714                                  
                                      
Z =   (X - µ )/(σ/√n) = (   57.14285714   -   56   ) / (    36   / √   350   ) =   0.594
                                      
P(X ≥   57.14285714   ) = P(Z ≥   0.59   ) =   P ( Z <   -0.594   ) =    0.2763      

e)

µ =    56                              
σ =    36                              
n=   350                              
proportion=   0.1000                              
                                  
Z value at    0.1   =   -1.282   (excel formula =NORMSINV(   0.10   ) )          
z=(x-µ)/(σ/√n)                                  
so, X=z * σ/√n +µ=   -1.282   *   36   / √    350   +   56   =   53.53

= 53.53 * 350

= 18737

THANKS

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