Question

An article in American Demographics claims that more than twice as many shoppers are out shopping on the weekends than during the week. Not only that, such shoppers also spend more money on their purchases on Saturdays and Sundays! Suppose that the amount of money spent at shopping centers between 4 p.m. and 6 p.m. on Sundays has a normal distribution with mean $190 and with a standard deviation of $20. A shopper is randomly selected on a Sunday between 4 p.m. and 6 p.m. and asked about his spending patterns. (a) What is the probability that he has spent more than $210 at the mall? (Round your answer to four decimal places.) (b) What is the probability that he has spent between $210 and $230 at the mall? (Round your answer to four decimal places.) (c) If two shoppers are randomly selected, what is the probability that both shoppers have spent more than $230 at the mall? (Round your answer to four decimal places.)

Answer 1

a)

µ = 190, σ = 20

P(X > 210) =

= P( (X-µ)/σ > (210-190)/20)

= P(z > 1)

= 1 - P(z < 1)

Using excel function:

= 1 - NORM.S.DIST(1, 1)

= 0.1587

b)

P(210 < X < 230) =

= P( (210-190)/20 < (X-µ)/σ < (230-190)/20 )

= P(1 < z < 2)

= P(z < 2) - P(z < 1)

Using excel function:

= NORM.S.DIST(2, 1) - NORM.S.DIST(1, 1)

= 0.1359

c) P(X > 230) =

= P( (X-µ)/σ > (230-190)/20)

= P(z > 2)

= 1 - P(z < 2)

Using excel function:

= 1 - NORM.S.DIST(2, 1)

= 0.0228

probability that both shoppers have spent more than $230 at the mall = 0.0228*2 = 0.0456