Question

Missy Walters owns a mail-order business specializing in baby clothes. Missy is confident the dollar amounts of all her orders are normally distributed or nearly so. Assume she knows the mean and standard deviation are $320 and $36, respectively, for all orders she receives.

a. Describe the sampling distribution of x, where x is the mean dollar-amount of an order for a sample of 15 orders.

b. What is the probability that a simple random sample of 28 orders will provide an estimate of the population mean dollar-amount of an order that is within plus or minus $8 of the actual population mean?

c. What happens to the sampling distribution of x when the sample size is increased from 28 to 70? With a sample size of 70, what is the probability that x will be between $260 and $379?

Answer 1

a)
E(x) =    320      
SD(x) =    36      
n=   15      
E(xbar) =    E(X)    320  
SD(Xbar) =   SD(X)/sqrt(n) =   36/SQRT(15) =   9.295
Xbar follows Normal with mean = 320 and sd = 9.295

b)
E(x) =    320      
SD(x) =    36      
n=   28      
E(xbar) =    E(X)    320  
SD(Xbar) =   SD(X)/sqrt(n) =   36/SQRT(28) =   6.803

P(312 < X < 328) = P(X<328) - P(X<312) = ?
I know that, z = (X-mean)/(sd)  
z1 = (312-320)/6.80336051416609) = -1.1759
z2 = (328-320)/6.80336051416609) = 1.1759
  
hence, P(312 < X < 328) = P(Z<1.1759) - P(Z<-1.1759) = NORMSDIST(1.1759) - NORMSDIST(-1.1759) = 0.7604

c)
with the increase in sample size from 28 to 70, the standard deviation of the sampling distribution decreases.
E(x) =    320      
SD(x) = 36      
n=   70      
E(xbar) =    E(X) = 320  
SD(Xbar) =   SD(X)/sqrt(n) =   36/SQRT(70) =   4.303

P(260 < X < 379) = P(X<379) - P(X<260) = ?
  
I know that, z = (X-mean)/(sd)  
z1 = (260-320)/4.30282299360382) =   -13.9443
z2 = (379-320)/4.30282299360382) =   13.7119
  
hence,  
P(260 < X < 379) = P(Z<13.7119) - P(Z<-13.9443) = NORMSDIST(13.7119) - NORMSDIST(-13.9443) = 1