Yearly travel expenses for a family of four are normally distributed with a mean expense equal to $3,000 and a standard deviation equal to $500. What is the probability of travel expenses being greater than $4,000?

The probability is 0.02275

Step-by-step explanation:

We shall first find the z-score here

z-score = (x -mean)/SD

from the question;

x = 4,000

mean = 3,000

SD = 500

z-score = (4,000-3,000)/500 = 1000/500 = 2

So the probability we want to calculate is;

P(z > 2) = 0.02275

146%

step-by-step explanation: