An article in American Demographics claims that more than twice as many shoppers

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 $84 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 $92 at the mall? (Round your answer to four decimal places.)


(b) What is the probability that he has spent between $92 and $112 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 $112 at the mall? (Round your answer to four decimal places.)

Explanation / Answer

NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 84
standard Deviation ( sd )= 20
a.
P(X > 92) = (92-84)/20
= 8/20 = 0.4
= P ( Z >0.4) From Standard Normal Table
= 0.3446
b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 92) = (92-84)/20
= 8/20 = 0.4
= P ( Z <0.4) From Standard Normal Table
= 0.6554
P(X < 112) = (112-84)/20
= 28/20 = 1.4
= P ( Z <1.4) From Standard Normal Table
= 0.9192
P(92 < X < 112) = 0.9192-0.6554 = 0.2638
c.
mean ( u ) = 84
standard Deviation ( sd )= 20/ Sqrt ( 2 ) =14.1421
sample size (n) = 2

P(X > 112) = (112-84)/20/ Sqrt ( 2 )
= 28/14.142= 1.9799
= P ( Z >1.9799) From Standard Normal Table
= 0.0239

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