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.)

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 =

Here,

σ = standard deviation

n = sample size

Thus,

P(X > $220) =

or

=

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 =

Here,

σ = standard deviation

n = sample size

Thus,

P(X > $220) =

or

=

or

= P( Z > -1.458 )            

= 0.9264         [From z table ]

i think it is

step-by-step explanation:

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step-by-step explanation: