A population of unknown shape has a mean of 75.  You select a sample
of 40.  The standard deviation of the sample is 5.  Compute the
probability the sample mean is:

Less than 74.
Between 74 and 76.
Between 76 and 77.
Greater than 77.

Hi!!

To answer this question we must apply 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. 

See for more references "Central Limit Theorem":
http://bcs.whfreeman.com/pbs/cat_050/pbs/CLT-SampleMean.html


Now we can answer the question assuming that Central Limit Theorem holds:

Use the following tables for calculations:
"z-distribution":
This one for P[Z < A] with A<0, for example P[Z < -1.26].
http://www.math2.org/math/stat/distributions/z-dist.htm

"Public Domain Normal Distribution Table":
This one for P[Z < B] with B>0, for example P[Z < 1.26].
http://www.math.unb.ca/~knight/utility/NormTble.htm


For this problem the STD of the sample mean is 5/sqrt(40) = 0.7906 .
So we have a normal distributed ramdom variable X with mean Mu = 75
and standard deviation StD = 0.7906


-Compute the probability the sample mean is less than 74:

The first step is to normalize the variable X = 74:
Z = (X-Mu)/STD = (74-75)/0.7906 = -1.26

Then P[X < 74] = P[Z < -1.26] = 0.10383



-Compute the probability the sample mean is between 74 and 76:

Normalize the variable X = 76:
Z = (X-Mu)/STD = (76-75)/0.7906 = 1.26
Then:
P[74 < X < 76] = P[-1.26 < Z < 1.26] = 
               = P[Z < 1.26] - P[Z < -1.26] =
               = 0.8962 - 0.1038 =
               = 0.7924



-Compute the probability the sample mean is between 76 and 77:

Normalize the variable X = 77:
Z = (X-Mu)/STD = (77-75)/0.7906 = 2.53
Then:
P[76 < X < 77] = P[1.26 < Z < 2.53] = 
               = P[Z < 2.53] - P[Z < 1.26] =
               = 0.9943 - 0.8962 =
               = 0.0981



-Compute the probability the sample mean is greater than 77:

P[X > 77] = P[Z > 2.53] = 1 - P[Z < 2.53] =
                        = 1 - 0.9943 =
                        = 0.0057


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Search strategy:
The problem was solved using my own knowledge on the subject, to find
the references and tables I used the following keywords at Google.com:
"central limit theorem"
Normal Distribution Table


I hope that this helps you. If you find something unclear or wrong in
this answer feel free to request for a clarification.


Best regards,
livioflores-ga