MATH SOLVE

2 months ago

Q:
# Data from the Bureau of Labor Statistics’ Consumer Expenditure Survey show that annual expenditures for cellular phone services per consumer unit increased from $237 in 2001 to $634 in 2007. Let the standard deviation of annual cellular expenditure be $52 in 2001 and $207 in 2007. a. What is the probability that the average annual expenditure of 125 cellular customers in 2001 exceeded $220? (Round answer to 4 decimal places.) b. What is the probability that the average annual expenditure of 125 cellular customers in 2007 exceeded $607? (Round answer to 4 decimal places.)

Accepted Solution

A:

Answer:a) 0.9998 b) 0.9264 Step-by-step explanation:Given:Annual expenditures for cellular phone in 2001 = $237Annual expenditures for cellular phone in 2007 = $634standard deviation of annual cellular expenditure in 2001 = $52standard deviation of annual cellular expenditure in 2007 = $207a) P( average annual expenditure of 125 cellular customers in 2001 exceeded $220)= P(X > $220)Now,Z value = [tex]\frac{X-Mean}{\frac{\sigma}{\sqrt(n)}}[/tex]Here,σ = standard deviationn = sample sizeThus,P(X > $220) = [tex]P(Z >\frac{220-237}{\frac{52}{\sqrt(125)}})[/tex]or= [tex]P(Z >\frac{220-237}{\frac{52}{\sqrt(125)}})[/tex]or= P( Z > -3.65 )= 0.9998 [From z table ]b) P( average annual expenditure of 125 cellular customers in 2007 exceeded $607)= P(X > $220)Now,Z value = [tex]\frac{X-Mean}{\frac{\sigma}{\sqrt(n)}}[/tex]Here,σ = standard deviationn = sample sizeThus,P(X > $220) = [tex]P(Z >\frac{607-634}{\frac{207}{\sqrt(125)}})[/tex]or= [tex]P(Z >\frac{-27}{\frac{207}{\sqrt(125)}})[/tex]or= P( Z > -1.458 ) = 0.9264 [From z table ]