An article in american demographics claims that more than twice as many shoppers are out shopping on the weekends that 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 $85 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 $95 at the mall?
b. what is the probability that he has spent between $95 and $115 at the mall?
c. if two shoppers are randomly selected, what is the probability that both shoppers have spent more than $115 at the mall?

0.3085,0.2417,0.0045

Step-by-step explanation:

Given that X, the amount of money spent at shopping centers between 4 P.M. and 6 P.M. on Sundays has a normal distribution with mean $85 and with a standard deviation of $20.

X is N(85, 20)

To convert into std normal variate we use the following formula

a) the probability that he has spent more than $95 at the mall

=

b.  the probability that he has spent between $95 and $115 at the mall

=

c. If two shoppers are randomly selected, what is the probability that both shoppers have spent more than $115 at the mall

=product of two probabilities since independent

=

x = ± 6i

step-by-step explanation:

x^2 = -36

take the square root of each side

sqrt(x^2) = sqrt(-36)

we know sqrt(ab) = sqrt(a) sqrt(b)

x = ±sqrt(36) sqrt(-1)

we know that sqrt (-1) = i

x = ± 6i

maybe i can