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 $88 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? (Round your answer to four decimal places.)

(b) What is the probability that he has spent between $94 and $109 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 $109 at the mall? (Round your answer to four decimal places.)

a) P(X>94)= 0.38209

b) P(94≤X≤109)= 0.23523

c) The probability that the two shoppers spending more than $109 is 0.0225

Step-by-step explanation:

Hello!

X: Amount spends by shoppers between 4 P.M. and 6 P.M. on a Sunday.

This variable has a normal distribution with mean μ= $88 and standard deviation σ= $20

A shopper is selected at random:

a) What is the probability that he has spent more than $94 at the mall?

Symbolically:

P(X>94)

To calculate this probability you have to work using the standard normal distribution Z= (X-μ)/σ ~N(0;1)

First, you have to "transform" the value of X to a value of Z (standardize the value of X) using the formula.

P(X>94)= P(Z>(94-88)/20)= P(Z>0.3)

Now you have to use the tables of the standard normal distribution to reach the corresponding value of probability. These tables show the accumulated probability for the distribution P(Z<z₀)= 1 - α, so to know the probability under the curve for values greater than 0.3, you have to do the following calculation:

1 - P(Z≤0.3)= 1 - 0.61791= 0.38209

b) What is the probability that he has spent between $94 and $109 at the mall?

Symbolically:

P(94≤X≤109)

You can also express it as:

P(X≤109) - P(X≤94)

Now you standardize both terms of the interval and look for the corresponding probabilities in the table:

P(Z≤(109-88)/20)= P(Z≤1.05)= 0.85314

P(Z≤94)= P(Z≤(94-88)/20)= P(≤0.3)= 0.61791

0.85314-0.61791= 0.23523

c) If two shoppers are randomly selected, what is the probability that both shoppers have spent more than $109 at the mall?

First, let's use the known distribution to calculate the probability of one shopper spending more than $109:

P(X>109)= 1 - P(X≤109)= 1 - P(Z≤(109-88)/20)= 1 - P(Z≤1.05)= 1 - 0.85314= 0.14686

Let Ai be the event of "the shopper spent more than $109 at the mall"

Then P(Ai)= 0.14686 ≅ 0.15

If two independent shoppers are selected, the probability of both spending more than $109 is:

P(A₁∩A₂) = P(A₁)*P(A₂)= 0.15*0.15= 0.0225

I hope it helps!

answer: a video game has three difficulty levels: easy, medium, and expert. the game's scoring works in such a way that a player has to score a fixed number of points to advance to the next level, and then the scoring starts over. the total score needed to clear all three difficulty levels is 7,500. two times the number of points needed to clear the easy level is 500 more than the number of points needed to clear the medium level. twice the number of points needed to clear the medium difficulty level is 1,500 more than the number of points needed to clear the expert level. if x, y, and z represent the easy, medium, and expert levels, respectively, and a is the coefficient matrix of the system of equations modeling this situation, identify the inverse matrix, a-1, and the solution matrix, x, of the system of equations.

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