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 = 26.50 mpg and a standard deviation of σ = 3.25 mpg.

Required:
a. What is the standard error of X and the mean from a random sample of 25 fill-ups by one driver?
b. Within what interval would you expect the sample mean to fall, with 98 percent probability?

Answer 1

a) 0.65 mpg

b) Between 24.99 mpg and 28.01 mpg.

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation, which is also called standard error,
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

`\mu = 26.50, \sigma = 3.25, n = 25, s = (3.25)/(√(25)) = 0.65`

a. What is the standard error of X and the mean from a random sample of 25 fill-ups by one driver?

s = 0.65 mpg

b. Within what interval would you expect the sample mean to fall, with 98 percent probability?

From the: 50 - (98/2) = 1st percentile

To the: 50 + (98/2) = 99th percentile

1st percentile:

X when Z has a pvalue of 0.01. So X when Z = -2.327.

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`-2.327 = (X - 26.50)/(0.65)`

`X - 26.50 = -2.327*0.65`

`X = 24.99`

99th percentile:

X when Z has a pvalue of 0.99. So X when Z = 2.327.

`Z = (X - \mu)/(s)`

`2.327 = (X - 26.50)/(0.65)`

`X - 26.50 = 2.327*0.65`

`X = 28.01`

Between 24.99 mpg and 28.01 mpg.