Question

example 2: the gas mileage a car gets is thought to have a distribution with μ = 35 and σ = 3.
a) a random sample of 64 cars is taken. what is the probability that the average gas mileage is greater than 36?
b) what is the probability that the average gas mileage for a sample of 200 cars is between 33 and 37?
c) below what mean would the gas mileage have to be in the bottom 30%?

Explanation:

Step1: Calculate the standard error

For part a), the standard error \( \sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}} \). Given \( \sigma = 3 \) and \( n = 64 \), then \( \sigma_{\bar{x}}=\frac{3}{\sqrt{64}}=\frac{3}{8}=0.375 \).
For part b), \( n = 200 \), \( \sigma_{\bar{x}}=\frac{3}{\sqrt{200}}\approx\frac{3}{14.14}\approx0.212 \).
For part c), \( n = 200 \), \( \sigma_{\bar{x}}=\frac{3}{\sqrt{200}}\approx0.212 \).

Step2: Calculate the z - score

Part a)

The z - score formula is \( z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}} \). We want \( P(\bar{X}>36) \), so \( z=\frac{36 - 35}{0.375}=\frac{1}{0.375}\approx2.67 \). Then \( P(\bar{X}>36)=1 - P(Z\leqslant2.67) \). From the standard normal table, \( P(Z\leqslant2.67) = 0.9962 \), so \( P(\bar{X}>36)=1 - 0.9962=0.0038 \).

Part b)

For \( \bar{x}_1 = 33 \), \( z_1=\frac{33 - 35}{0.212}=\frac{- 2}{0.212}\approx - 9.43 \). For \( \bar{x}_2 = 37 \), \( z_2=\frac{37 - 35}{0.212}=\frac{2}{0.212}\approx9.43 \). Then \( P(33<\bar{X}<37)=P(-9.43<Z<9.43)\approx1 \) (since for a standard normal distribution, \( P(- 3<Z<3)\approx0.9973 \) and \( |z| = 9.43\) is far in the tails).

Part c)

We want to find \( \bar{x} \) such that \( P(\bar{X}<\bar{x})=0.3 \). Looking up in the standard normal table, the z - score corresponding to a left - tail probability of \( 0.3 \) is \( z\approx - 0.524 \). Using the formula \( z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}} \), and \( \sigma_{\bar{x}}=\frac{3}{\sqrt{200}}\approx0.212 \), \( - 0.524=\frac{\bar{x}-35}{0.212} \). Then \( \bar{x}=35+( - 0.524)\times0.212=35 - 0.111=34.89 \).

Answer:

a) \( 0.0038 \)
b) \( 1 \)
c) \( 34.89 \)