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 2626 thousand dollars and 2 22 thousand dollars, respectively. Suppose we select one of these cars at random. Let X = X=X, equals the sale price (in thousands of dollars) for the selected car.

Complete 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

0.48

Step-by-step explanation:

Mean, m = 26000 ; Standard deviation, s = 2000

P(x < x)

Zscore = (x - m) / s

P(26< X<30) ;

P[(26 - 26 / 2) < (30 - 26) / 2)]

P[(0 < Z < 2)]

P(Z < 2) - P(Z <0)

Using a z table of z calculator, we can obtain the probabilties of the Z value :

0.97725 - 0.5

= 0.47725

= 0.48 ( 2 decimal places)

b) 36

step-by-step explanation:

18 times 2 is 36