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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.)

Respuesta :

Answer:

a) 0.9998      

b) 0.9264        

Step-by-step explanation:

Given:

Annual expenditures for cellular phone in 2001 = $237

Annual expenditures for cellular phone in 2007 = $634

standard deviation of annual cellular expenditure in 2001 = $52

standard deviation of annual cellular expenditure in 2007 = $207

a) 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 deviation

n = sample size

Thus,

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 deviation

n = sample size

Thus,

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 ]

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