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It is important to understand when to use the CLT . Use the CLT for means or averages when you are asked to find the probability for a sample average or mean, or when working with percentiles for sample averages. (If youare being asked to find the probability or percentile of a sum or total, use the CLT for sums.)
The Law of Large Numbers says that if you take samples of larger and larger size from any population, then the mean of the sample gets closer and closer to . From the Central Limit Theorem, we know that as gets larger and larger, the sample averages follow a normal distribution. The larger n gets, the smaller thestandard deviation gets. (Remember that the standard deviation for is .) This means that the sample mean must be close to the population mean . We can say that is the value that the sample averages approach as gets larger. The Central Limit Theorem illustrates the Law of Large Numbers.
A study involving stress is done on a college campus among the students. The stress scores follow a continuous uniform distribution with the lowest stress score equal to 1 and the highest equal to 5. Using a sample of 75 students, find:
Let = the stress score for one individual student
The individual stress scores follow a continuous uniform distribution, ~ where and (See the chapter on Continuous Random Variables ).
Problems a and b ask you to find a probability or a percentile for an average or mean . The sample size, , is equal to 75.
Let = the average stress score for the 75 students.
For the average stress score, use the CLT which tells us that ~
~ where .
Find . Draw the graph.
The probability that the average stress score is lessthan 2 is about 0.
normalcdf
Find the 90th percentile for the sample average of 75 stress scores. Draw a graph.
Let = the 90th precentile. Find where .
using
invNorm
The 90th percentile for the sample average of 75 scores is about 3.17. This means that 90% of all the averages of samples of 75 stress scores are at most 3.17 and 10% of the sample averages are at least3.17 .
Suppose that a market research analyst for a cell phone company conducts a study of their customers who exceed the time allowance included on their basic cell phone contract; the analyst finds that for those people who exceed the time included in their basic contract, the excess time used follows an exponential distribution with a mean of 22 minutes.
Consider a random sample of 80 customers who exceed the time allowance included in their basic cell phone contract.
Let = the excess time used by one INDIVIDUAL cell phone customer who exceeds his contracted time allowance.
~ From Chapter 5, we know that and .
Let = the AVERAGE excess time used by a sample of customers who exceed their contracted time allowance.
~ by the CLT for Sample Means or Averages
using
normalcdf
The probability is 0.7919 that the average excess time used is morethan 20 minutes, for a sample of 80 customers who exceed their contracted time allowance.
EE
key for E. Or just use 10^99 instead of 1E99.
P(X>20) = e^(–(1/22)*20) or e^(–.04545*20) = 0.4029
Let = the 95th percentile. Find where
using
invNorm
The 95th percentile for the sample average excess time used is about 26.0 minutes for random samples of 80 customers who exceed their contractual allowed time.
95% of such samples would have averages under 26 minutes; only 5% of such samples would have averages above 26 minutes.
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