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 $94 at the mall?=.3632 (b) What is the probability that he has spent between $94 and $114 at the mall?=.2747 and I couldn't figure out the third part. If two shoppers are randomly selected, what is the probability that both shoppers have spent more than $114 at the mall? (Round your answer to four decimal places.)

Answer 1

c. P(X[bar]>114) = 0.017

Explanation:

Hello!

It is like this, to calculate the probabilities in a. and b. you used the distribution of the variable X~N(μ;δ²) with the standardization Z= (x-μ)/δ²

In c. since there is a random sample involved, you need to work using the distribution of the sample mean X[bar]~N(μ;δ²/n)

With this in mind,

P(X[bar]>114) = 1 - P(X[bar]≤114) =1 - P(Z≤(114-84)/(20/√2))

⇒1 - P(Z≤2.12)= 1- 0.98300 = 0.017

I've checked your answers:

a. P(X>94) = 1-P(X≤94) = 1-P(Z≤(94-84)/20) = 1- P(Z≤0.5) = 1- 0.69146 = 0.30854 Correct

b. P(94<X<114) = P(X<114)-P(X<94) = P(Z<(114-84)/20)-P(Z<(94-84)/20)

⇒P(Z<(1.5)-P(Z<(0.5) = 0.93319 - 0.69146 = 0.24173 Correct

I hope you have a SUPER day!