Answer:
(a)P(x>85.55)=0.02275
(b)[tex]P(x<76.59)=0.84134[/tex]
Step-by-step explanation:
We are given that
Average sales for an online textbook distributor per customer per purchase
,[tex]\mu=[/tex]$67.63
Standard deviation of the amount spent on textbooks,[tex]\sigma=[/tex]$8.96
(a) We have to find probability for a randomly selected customer spent more than $85.55 per purchase.
[tex]P(x>85.55)=P(\frac{x-\mu}{\sigma}>\frac{85.55-67.63}{8.96})[/tex]
=[tex]P(Z>2)[/tex]
=[tex]1-P(Z\leq 2)[/tex]
=[tex]1-0.97725[/tex]
P(x>85.55)=0.02275
(b)We have to find probability for a randomly selected customer spent less than $76.59 per purchase
[tex]P(x<76.59)=P(\frac{x-\mu}{\sigma}<\frac{76.59-67.63}{8.96})[/tex]
[tex]P(x<76.59)=P(Z<1)[/tex]
[tex]P(x<76.59)=0.84134[/tex]
