Question

Different dealers may sell the same car for different prices. The sale prices for a particular car are normally distributed with a mean and standard deviation of 26 thousand dollars and 2 thousand dollars, respectively. Suppose we select one of these cars at random. Let X represent the sale price (in thousands of dollars) for the selected car.1. Find P(26< X<30) You may round your answer to two decimal places

Answer 1

Answer:

0.48

Step-by-step explanation:

P(26<X<30)=?

This probability can be calculated using normal distribution.

Here mean=μ=26

standard deviation=σ=2

So, P(26<X<30)=P(26-26/2<X-μ/σ<30-26/2)

P(26<X<30)=P(0<Z<2)

P(26<X<30)=0.4772=0.48

Hence P(26<X<30)=0.48=48%