Download 6.5 The Central Limit Theorem

6.5 – The Central Limit Theorem
Objectives:
1. Use sampling distribution of the mean.
2. Apply and interpret results of the Central Limit Theorem.
The Central Limit Theorem:
The Central Limit Theorem tells us that for a population with any distribution, the distribution of the
sample means approaches a normal distribution as the sample size increases. The procedure in this
section forms the foundation for estimating population parameters and hypothesis testing.
Given:
1. The random variable x has a distribution (which may or may not be normal) with mean µ and
standard deviation σ .
2. Simple random samples all of size n are selected from the population. (The samples are selected
so that all possible samples of the same size n have the same chance of being selected.)
Conclusions:
1. The distribution of sample x will, as the sample size increases, approach a normal distribution.
2. The mean of the sample means is the population mean µ.
3. The standard deviation of all sample means is
σ
n
Practical Rules Commonly Used:
1. If the original population is not normally distributed then for samples of size n larger than 30,
the distribution of the sample means can be approximated reasonably well by a normal
distribution. The approximation gets closer to a normal distribution as the sample size n
becomes larger.
2. If the original population is normally distributed, then for any sample size n, the sample means
will be normally distributed (not just the values of n larger than 30).
Notation for the Sampling Distribution of x :
If all possible random samples of size n are selected from a population with mean µ and standard
deviation σ , the mean of the sample means is denoted by µ x so
µx = µ
Also, the standard deviation of the sample means is denoted by σ x so
σx =
σ x is called the standard error of the mean.
σ
n
Normal Distribution:
As we proceed from n = 1 to
n = 50, we see that the distribution of sample
means is approaching the shape of a normal
distribution.
As the sample size increases, the sampling
distribution of sample means approaches a
normal distribution.
Uniform Distribution:
As we proceed from n = 1 to
n n= 50, we see that the distribution of sample
means is approaching the shape of a normal
distribution.
As the sample size increases, the sampling
distribution of sample means approaches a normal
distribution.
U-Shaped Distribution:
As we proceed from n = 1 to
n = 50, we see that the distribution of sample
means is approaching the shape of a normal
distribution.
As the sample size increases, the sampling
distribution of sample means approaches a
normal distribution.
Many practical problems can be solved with the central limit theorem. When working with such
problems, remember that if the sample size is greater than 30 or if the original population is normally
distributed, treat the distribution of sample means as if it were a normal distribution.
In the following problems, we use the central limit theorem and sample means to determine probabilities
of a particular event occurring.
Example: Assume the population of weights of men is normally distributed with a mean of 172 lb and a
standard deviation of 29 lb.
a) Find the probability that if an individual man is randomly selected, his weight is greater than
175 lb.
b) Find the probability that 20 randomly selected men will have a mean weight that is greater than
175 lb
Solution:
a) Find the probability that if an individual man is randomly selected, his weight is greater than
175 lb.
z=
x−µ
σ
=
175 − 172
= 0.10
29
P(x > 175) = 0.4602
b) Find the probability that 20 randomly selected men will have a mean weight that is greater than
175 lb (so that their total weight exceeds the safe capacity of 3500 pounds).
z=
x − µx
σx
=
x − µx
σ
n
=
175 − 172
= 0.46
29
20
P(x > 175) = 0.3228
It is much easier for an individual to deviate from the mean than it is for a group of 20 to deviate
from the mean.
Example: Human body temperatures are normally distributed with a mean of 98.2 F and a standard
deviation of 0.62 F. If 19 people are randomly selected, find the probability that their mean body
temperature will be less than 98.5 F?
Solution:
1. Find the appropriate z-score.
z=
x − µx
σx
=
x − µx
=
σ
n
98.5 − 98.2
0.3
=
= 2.11
0.62
0.1422
19
2. Find the corresponding area/probability:
P ( x < 98.5) = 0.9826
The probability that their mean body temperature will be less than 98.5 F is 0.9826.
Example: The amount of snowfall falling in a certain mountain range is normally distributed with a
mean of 94 inches and a standard deviation of 14 inches. What is the probability that the mean annual
snowfall during 49 randomly picked years will exceed 96.8 inches.
Solution:
1. Find the appropriate z-score.
z=
x − µx
σx
=
x − µx
σ
n
=
96.8 − 94 2.8
=
= 1. 4
14
2
49
2. Find the corresponding area/probability:
P ( x > 96.8) = 1 − P ( x < 96.8) = 1 − 0.9192 = 0.0808
The probability that the mean annual snowfall during 49 randomly picked years will exceed 96.8
inches is 0.0808.
Example: The weights of the fish in a certain lake are normally distributed with a mean of 19 lb and a
standard deviation of 6. If 4 fish are randomly selected, what is the probability that the mean weight will
be between 16.6 and 22.6 lb?
Solution:
1. Find the appropriate z-score.
z=
x − µx
σx
=
x − µx
σ
=
n
z=
x − µx
σx
=
x − µx
σ
n
16.6 − 19 − 2.4
=
= −0.8
6
3
4
=
22.6 − 19 3.6
=
= 1. 2
6
3
4
2. Find the corresponding area/probability:
P (16.6 < x < 22.6) = P ( x < 22.6) − P ( x < 16.6) = 0.8849 − 0.2119 = 0.6730
The probability that the mean weight will be between 16.6 and 22.6 lb is 0.6730
Example: A final exam in Math 274 has a mean of 73 with standard deviation 7.8. If 24 students are
randomly selected, find the probability that the mean of their test scores is greater than 78.
Solution:
1. Find the appropriate z-score.
z=
x − µx
σx
=
x − µx
σ
n
=
78 − 73
5
=
= 3.14
7.8
1.5921
24
2. Find the corresponding area/probability:
P ( x > 78) = 1 − P ( x < 78) = 1 − 0.9992 = 0.0008
The probability that the mean of their test scores is greater than 78 is 0.0008.
Correction for a Finite Population:
When sampling without replacement and the sample size n is greater than
han 5% of the finite population of
size N (that is, n > 0.05N ), adjust the standard deviation of sample means by multiplying it by the finite
population correction factor:
Example: In a population of 225 women, the heights of the women are normally distributed with a
mean of 64.5 in. and a standard deviation of 2.9 in. If 25 women are selected at random, find the
probability that their mean height will exceed 66 inches. Assume that the sampling is done without
replacement and use a finite population correction factor with N = 225.
Solution:
1. Find the appropriate z-score.
score.
z=
x − µx
σx
=
x − µx
66 − 64.5
1.5
=
=
= 2.73
2.9 225 − 25 0.5480
σ N −n
n N −1
25 225 − 1
2. Find the corresponding area/probability:
P ( x > 66) = 1 − P ( x < 66) = 1 − 0.9968 = 0.0032
The probability that their mean height will exceed 66 inches is 0.0032.