Question

A racing car consumes a mean of 98 gallons of gas per race with the standard deviation of 3 gallons. If 42 racing cars are randomly selected what is the probability that the sample mean will be greater than 98.6 gallons? Round your answer to four decimal places

Answer 1

Given:

Mean= 98

Standard Deviation= 3

Sample = 42

Observed value = 98.6

To determine the probability that the sample mean will be greater than 98.6 gallons, we first note the z-score formula:

$\begin{gathered} z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}} \\ where: \\ x=observed\text{ value} \\ \mu=mean \\ \sigma=standard\text{ deviation} \\ n=sample \end{gathered}$

Next, we plug in what we know:

$\begin{gathered} z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}} \\ z=\frac{98.6-98}{\frac{3}{\sqrt{42}}} \\ Calculate \\ z=1.2961 \end{gathered}$

Then, we also note that:

$P(x>98.6)=P(z>1.2961)=0.0975$

Therefore, the probability that the sample mean will be greater than 98.6 gallons is: 0.0975