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 = 21.50 mpg and a standard deviation of σ = 2.75 mpg.

(a) What is the standard error of X⎯⎯⎯ , the mean from a random sample of 9 fill-ups by one driver? (Round your answer to 4 decimal places.)

(b) Within what interval would you expect the sample mean to fall, with 95 percent probability? (Round your answers to 4 decimal places.)

Answer 1

Solution:

Part a

Standard error = σ/Sqrt(n)

We are given

σ = 2.75

n = 9

Standard error = 2.75/sqrt(9) = 0.916667

Standard error = 0.9167

Part b

Here, we have to find that within what interval the sample mean to fall, that is, we have to find the margin of error with 95 percent confidence level.

Margin of error = Z*σ/sqrt(n)

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

σ = 2.75

n = 9

Margin of error = Z*σ/sqrt(n)

Margin of error = 1.96*2.75/sqrt(9)

Margin of error = 1.96*0.916667

Margin of error =1.7967

We would expect the sample mean fall by 1.7967 approximately.