A recent study by the Greater Los Angeles Taxi Drivers Assocation
showed that the mean
fare charged for service from Hermosa Beach to the Los Angeles
International Airport
is $18.00 and the standard deviation is $3.50. We select a sample of 15 fares.		
   a. What is the likelihood that the sample mean is between $17.00 and $20.00?		
		
   b. What must you assume to make the above calculation?

Hi ml123!!

To answer this question I must use the Central Limit Theorem:
"The central limit theorem states that given a distribution with a
mean m and variance s2, the sampling distribution of the mean
approaches a normal distribution with a mean (m) and a variance s2/N
as N, the sample size, increases.
The central limit theorem explains why many distributions tend to be
close to the normal distribution."
"Central Limit Theorem":
http://www.isixsigma.com/dictionary/Central_Limit_Theorem-177.htm


Sample sizes larger than 30 are usually considered large enough for
the sample mean to have an almost normal distribution. In this case
the sample size is 15, so we need to assume that the sample mean will
follow a normal distribution. This answers part b of the problem. Note
that you will get a good aproximation, see "Central Limit Theorem":
Just drag the slider to 15.
http://bcs.whfreeman.com/pbs/cat_050/pbs/CLT-SampleMean.html


a. What is the likelihood that the sample mean is between $17.00 and $20.00?

The STD of the sample mean is $3.5/sqrt(15) = 0.9037 .

Now we have a normal distributed ramdom variable X with mean Mu =
$18.00 and standard deviation StD = $0.9037

We want to know the likelihood that X is between $17.00 and $20.00 .

The first step is to normalize the variable X=$17.00 :
Z1 = (X-Mu)/STD = (17.00-18.00)/0.9037 = -1.11

Now we must to normalize the variable X=$20.00 :
Z2 = (X-Mu)/STD = (20.00-18.00)/0.9037 = 2.21

We will use the following table for calculations:
"Normal Distribution Table"
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/normaltable.html


Now using the table we can find the probabilities:
P(-1.11 < Z < 0) = P(0 < Z < 1.11) = 0.3665      (due simmetry)

P(0 < Z < 2.21) = 0.4861

P(-1.11 < Z < 2.21) = 0.3665 + 0.4861 = 0.8526


The likelihood that the sample mean is between $17.00 and $20.00 is
aproximately 85.26% .


I hope that this helps you. Feel free to request for a clarification
if you need it before rate this answer.

Regards.
livioflores-ga

Thanks for a great answer!