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Section 5.4 The Distribution of the Sample Mean
Since the sample mean X is an estimator for the population parameter µ
can provide us with information about µ . Recall that X is a random
variable and hence has a probability distribution. In this section we will
discuss specifically what that distribution looks like.
General Description of the Distribution of X
Let X 1 , X 2 , X 3 ,..., X n be a random sample (again they are iid) from a
distribution with mean µ and standard deviation σ .
Then
1.
2.
In addition with
To = X 1 + X 2 +  + X n
Example 1: The inside diameter of a piston ring is a random variable with
mean value 12 cm and standard deviation .04 cm.
a. If X is the sample mean diameter for a random sample of n =16 rings
where is the sampling distribution of X centered, and what is the
standard deviation of the X distribution?
b. What about if the sample consisted of n = 64 rings?
c. For which of the two random samples the one in part (a) or the one in
part (b) is more likely to be within .01 cm of 12 cm? Explain.
The Case of the Normal Distribution
The simulation experiment we did hinted at the fact that if the underlying
distribution is normal, then the sample mean appeared to also be normal. We
will state that more formally here:
Let X 1 , X 2 , X n be a random sample from a normal distribution with mean µ
and standard deviation σ . Then for any n, then
Example 2:
#52: the lifetime of a certain type of battery is normally distributed with mean
value 10 huors and standard deviation 1 hour. There are 4 batteries in the
package. What lifetime value is such that the total lifetime of all batteries in a a
package exceeds the value for only 5% of all packages?
The Central Limit Theorem
We just learned that when the random sample is iid from a normal distribution,
then regardless of the sample size, the sample mean is also normally distributed.
However, the interesting thing is, if you sample enough, no matter what the
shape of the original distribution, the distribution of X will still be approximately
normal. This is the crux of the Central Limit Theorem
The Central Limit Theorem (CLT) states
Let X 1 , X 2 , X n be a random sample from a distribution with mean µ and
variance σ 2 , then if n is sufficiently large
The CLT is truly powerful because in order to find P ( a ≤ X ≤ b ) you would have
to first actually determine the distribution of the sample mean, but by being able
to assume normality, all you have to do is standardize and use the normal table!
A visual example:
The uniform distribution above is obviously non-Normal. Call that the parent
distribution.
To compute an average, Xbar, two samples are drawn, at random, from the
parent distribution and averaged and plotted.
.
Repeatedly taking four from the parent distribution, and computing the averages
and plotting them, produces the probability density above
.
It should also be noted that the CLT can be used for either discrete or a
continuous .
What does it mean for N to be large enough
Let’s look at an example:
Example 3:
A lightbulb manufacturer claims that the lifespan of its lightbulbs has a mean of
54 months and a st. deviation of 6 months. Your consumer advocacy group tests
50 of them. Assuming the manufacturer's claims are true, what is the probability
that it finds a mean lifetime of less than 52 months?
Example 4:
Let To denote the sum of the variables in a random sample of size 30 from the
uniform distribution on [0, 1]. Find normal approximations to each of the
following:
a. P(13< To <18)
b. The 90th percentile of To