Question

The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen.

We are interested in the average cost of gasoline for the 16 gas stations.Find the probability that the average price for 30 gas stations is less than $4.55.a. 0.6554b. 0.3446c. 0.0142d. 0.9858e. 0

Answer 1

`P(\bar X\leq 4.55)=0.0142`

c. 0.0142

Explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:

`X \sim N(\mu=4.59,\sigma=0.1)`

And let
`\bar X` represent the sample mean, the distribution for the sample mean is given by:

`\bar X \sim N(\mu,(\sigma)/(√(n)))`

On this case
`\bar X \sim N(4.59,(0.1)/(√(16)))`

2) Solution to the problem

We want this probability:

`P(\bar X\leq 4.55)`

The question on this case is ".Find the probability that the average price for 30 gas stations is less than $4.55". So then our value for n=30.

If we apply the formula for the z score to our probability we got this:

`P(\bar X \leq 4.55)=P(Z\leq (4.55-4.59)/((0.1)/(√(30))))=P(Z<-2.19)`

And using the normal standard table or the following excel code we calculate the probability "=NORM.DIST(-2.19,0,1,TRUE)"

`P(\bar X\leq 4.55)=0.0142`