Answer:
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
[tex]Z=\frac{x-85}{20}[/tex]
a) the probability that he has spent more than $95 at the mall
=[tex]P(X>95) = P(Z>0.5)=\\0.5-0.1915\\=0.3085[/tex]
b. the probability that he has spent between $95 and $115 at the mall
=[tex]P(95<x<115) \\=P(0.5<z<1.5)\\= 0.4332-0.1915\\=0.2417[/tex]
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
= [tex]{P(X>115)}^2\\= P(Z>1.5)*P(Z>1,5)\\= 0.0668^2\\=0.0045[/tex]