Proportion Given only Standard Deviation and Mean

Regular-grade gasoline prices for all the gas stations in Lee County have a normal distribution with a mean of $2.95 and

standard deviation of $0.16. Use this information and the standard Normal distribution to answer the next **two** questions.

What proportion of gas stations will charge more than $3.15 per gallon?

A manager of a discount gas station wants to set the price for gasoline at his station where only 10% of the stations in

Lee County will charge **less** than his price? What price should he charge for gasoline at his station?

I have been staring at this question for way too long. I cannot figure out the correct answer to this because I only know how to do it when I am given a population.

Re: Proportion Given only Standard Deviation and Mean

Hey lpg0005.

To start you off if X is Normal with mean mu and variance sigma^2 then Z = [X - mu]/sigma where Z is Normal(0,1) and probability tables can be used to get probabilities for P(Z < z) for various values of z.

So if you want to find P(X < x) then remember that this is the same as finding P(Z < (x-mu)/sigma) where Z is Normal (0,1).

Re: Proportion Given only Standard Deviation and Mean

How many standard deviations above the mean is $3.15?

You can now use the standard normal distribution.

Z~N(0,1)

You need P(Z>1.25). This is equal to 1 - P(Z<1.25). This can be found in a suitable table or maybe on your calculator.

P(Z<1.25)=0.8944

So your answer is 1 - 0.8944 = 0.1056.

Re: Proportion Given only Standard Deviation and Mean

Awesome. Thank you so much. You were a ton of help with this problem!

Re: Proportion Given only Standard Deviation and Mean

But can you help me with the second part of the question?

A manager of a discount gas station wants to set the price for gasoline at his station where only 10% of the stations in

Lee County will charge less than his price? What price should he charge for gasoline at his station?

Re: Proportion Given only Standard Deviation and Mean

If you Have P(X < x) = 0.1 and you know the distribution of X, then you can solve for x using the cumulative distribution.

If X is normal then convert to a standard normal and look up the appropriate value for x* = (x-mu)/sigma and re-arrange to solve for x.