Download The Central Limit Theorem (CLT)

The Central Limit Theorem
Consider the following probability distribution . The random variable X is the
length of stay in the hospital for a randomly selected patient undergoing a certain
medical procedure.
X:
p(x):
2
.1
4
.2
5
.4
6
.3
The mean and standard deviation of this distribution are  = 4.8 and  = 1.166.
Knowing the appropriate model for a population is like knowing the entire
population itself. In a real life situation we would not know which model would be
the appropriate one for the population of interest, and we certainly would not know
the actual value of . The problem then is to ESTIMATE  in some way.
One way to do this is to take a simple random sample from the population and then
use it to calculate a value of the sample mean X .
If we proceeded in this way, we would become aware of the fact that each time we
took a new sample from the population and calculated X we would get a different
value for our estimate of . Thus it is important to try to understand a bit more
about the behaviour of X.
Mean, Variance and Standard Deviation of the Sample Mean X
Given: 1. A population with mean  and standard deviation  .
2. A random sample of size n from the population:
X1 , X2 , … Xn with sample mean X = (X1 +X2 +… Xn)/n
Then:  =  [ the mean of the distribution of X is equal to the population mean
]
X
 =  / n [ the sd of the distribution of X is equal to ( the population sd)/n]
X
2
X
=  2 /n [ the variance of the distribution of X is equal to
(the population variance)/n ]
(83)
The Central Limit Theorem (CLT) is stated as follows:
Given a large random sample of size n from a population with mean  and
standard deviation , then The sample mean X is approximately normally
distributed with mean and standard deviation given by

X
= 

=  / n
X
Note: 1. n >30 is usually large enough for the CLT to apply.
2. If the population from which we sample is normal thenX is exactly
normally distributed with mean and standard deviation as above for any
sample size.
Example: The population of white males aged 35-44 in Canada has a mean
systolic blood pressure of 130 with a standard deviation of 8. Find the
probability that a random sample of 50 white males has a mean systolic blood
pressure
(a) below 133, (b) above 128, (c) within 2 of the population mean
Solution: Here we are asked to calculate probabilities about the sample mean
X. This suggests we use the CLT.
Population: Systolic blood pressures of white Canadian males aged 35-44.
Population Mean  = 130; Population Stdev  = 8.
Since n =50 > 30, the CLT tells us that X is approximately normally
distributed with
 =  = 130
X
(a) P ( X < 133 ) =
(84)
and  =  / n
X
(b) P( X > 128) =
(c) Notice that “ X within 2 of the population mean “ means that X lies
between 128 and 132.
Thus, P ( 128 < X < 132 ) =
(85)
Example: The scores on a Standardized Reading Test are known to be normally
distributed with a mean of  = 10 and standard deviation of  = 1.5.
(a) Find the probability that a single individual selected at random scores
between 9.4 and 10.6 on this test.
(b) Find the probability that a random sample of 25 students has a mean score
between 9.4 and 10.6 on this test.
(c) Find the probability that a random sample of 64 students has a mean score
between 9.4 and 10.6 on this test.
Solution: Let X be the score of the randomly selected individual. Then X is normal
 = 10,
 = 1.5.
(a) P( 9.4 < X < 10.6 ) =
(86)
(b) Since the population the sample came from is normal, then for any n ( hence for
n =25), X is exactly normal with
 = 10
and  = 1 .5/5 =.3
X
X
P( 9.4 < X < 10.6 ) =
(c) For the same reason as in (b) above, for n =64, X is exactly normal ,  =  = 10
X
and 
=  / n = 1.5/8 = 0.1875.
X
Therefore, P( 9.4 < X < 10.6 ) =
(87)