Standard Normal Probability

Question

A group of retailers models that the amount of dollars X that an individual will spend in christmas
shopping has a normal distribution with mean μ = $1100 and standard deviation σ = $330. What
proportion (probability) of shoppers will spend more than $500?

P(500 < X) =

Answer 1

Prob (X>500) = Prob ( z > (500-1100)/330 ) = Prob ( z> -1.81)

                             = 1 - Prob( z< -1.81)

                           = 1 - 0.0281

                          = 0.9719

                          = 97.19%