Question

Historical data show that customers who download music from a popular Web service spend approximately $24 per month, with a standard deviation of $3. Assume the spending follows the normal probability distribution. Find the probability that a customer will spend at least $21 per month. How much (or more) do the top 7% of customers spend? What is the probability that a customer will spend at least 521 per month? (Round to four decimal places as needed.) How much do the top 7% of customers spend? $1(Round to the nearest cent as needed.)

Answer 1

Problem 1 Given the distribution is normal with mean (u) 24 and standard deviation (0) 3 a) Probability that a customer spent at least 21 per month P(x>21) = 1-p(x<21) = 1- P(** < 21721) = 1-P(z<-1.00) = 1-0.1587 = 0.8413 b) P(X>x) = 0.07 Since P(X>x)=1-P(X<X) P(X<x)=1-0.07=0.93 P(** <*52-) = 0.93 Z(0.93)=1.48 (Using excel =NORM.S.INV(0.93)=1.48 =1.48 x = 21+1.48*3 = 25.44 Amount that spent to the top 7% is =$25.44 x-21 3

Problem 2 Given distribution is the application of the binomial distribution with n= 300 and p = 0.08 PMF of the binomial distribution is P(X=x)= (?) p* (1-p)** where x=0,1,2,3... P(x> 273) = 1-P(x<273) = 1-BINOMDIST(272,300,0.08, TRUE) (Using excel) = 1.0-1.0 = 0.0000