Question

The highway fuel economy of a 2016 Lexus RX 350 FWD 6-cylinder 3.5-L automatic 5-speed using premium fuel is a normally distributed random variable with a mean of μ = 20.50 mpg and a standard deviation of σ = 3.00 mpg. (a) What is the standard error of X¯¯¯X¯ , the mean from a random sample of 36 fill-ups by one driver? (Round your answer to 4 decimal places.) Standard error of X¯¯¯X¯ (b) Within what interval would you expect the sample mean to fall, with 99 percent probability? (Round your answers to 4 decimal places.) The interval is from to

Answer 1

Answer:

a) Standard error = 0.5

b) 99% Confidence interval:  (19.2125,21.7875)

Step-by-step explanation:

We are given the following in the question:

Population mean =  20.50 mpg

Sample standard deviation = 3.00 mpg

Sample size , n = 36

Standard Error =

$=\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{3}{\sqrt{36}} = 0.5$

99% Confidence interval:

$\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}$

Putting the values, we get,

$z_{critical}\text{ at}~\alpha_{0.01} = \pm 2.575$

$20.50 \pm 2.575(\displaystyle\frac{3}{\sqrt{36}})= 20.50 \pm 1.2875 = (19.2125,21.7875)$