Answer to Question #279792 in Statistics and Probability for nana

Question

The population average fuel efficiency of U.S. light vehicles for 2005

was 21 mpg. If the standard deviation of the population was 2.9 and

the gas ratings were normally distributed, what is the probability that

the mean for a random sample of 25 light vehicles is under 20 mpg?

(Must show your work)

Answer 1

Let $X=$ the mean for a random sample: $X\sim N(\mu, \sigma^2/n).$

Given $\mu=21mpg, \sigma=2.9mpg, n=25.$

$$P(X<20)=P(Z<\dfrac{20-\mu}{\sigma/\sqrt{n}})$$

$$=P(Z<\dfrac{20-21}{2.9/\sqrt{25}})\approx P(Z<-1.724138)$$

$$\approx 0.042341$$