Question

Suppose that the miles-per-gallon (mpg) rating of passenger cars is normally distributed with a mean and a standard deviation of 31.6 and 4.9 mpg, respectively.

a. What is the probability that a randomly selected passenger car gets more than 35 mpg? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

b. What is the probability that the average mpg of two randomly selected passenger cars is more than 35 mpg? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

c. If two passenger cars are randomly selected, what is the probability that all of the passenger cars get more than 35 mpg? (Round “z” value to 2 decimal places, and final answer to 4 decimal places.)

Answer 1

This is a normal distribution question with

P(x > 35.0)=?
The z-score at x = 35.0 is,

z = 0.6939
This implies that
P(x > 35.0) = P(z > 0.6939) = 1 - 0.7561275382003682


b) Sample size (n) = 2
Since we know that

P(x > 35.0)=?
The z-score at x = 35.0 is,

z = 0.9813
This implies that
P(x > 35.0) = P(z > 0.9813) = 1 - 0.8367775886235855


c) This is a binomial distribution question with
n = 2
p = 0.2439
q = 1 - p = 0.7561
where

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