6.4 - Triola textbook 8.2 - Sullivan textbook

1. Elementary Statistics
Chapter 6:
Normal Probability
Distributions
6.4 The Central Limit
Theorem
1

2. Chapter 6: Normal Probability Distribution
6.1 The Standard Normal Distribution
6.2 Real Applications of Normal Distributions
6.3 Sampling Distributions and Estimators
6.4 The Central Limit Theorem
6.5 Assessing Normality
6.6 Normal as Approximation to Binomial
2
Objectives:
• Identify distributions as symmetric or skewed.
• Identify the properties of a normal distribution.
• Find the area under the standard normal distribution, given various z values.
• Find probabilities for a normally distributed variable by transforming it into a standard normal variable.
• Find specific data values for given percentages, using the standard normal distribution.
• Use the central limit theorem to solve problems involving sample means for large samples.
• Use the normal approximation to compute probabilities for a binomial variable.

3. Recall: 6.1 The Standard Normal Distribution
Normal Distribution
If a continuous random variable has a distribution with a graph that
is symmetric and bell-shaped, we say that it has a normal
distribution. The shape and position of the normal distribution
curve depend on two parameters, the mean and the standard
deviation.
SND: 1) Bell-shaped 2) µ = 0 3) σ = 1
3
2 2
( ) (2 )
2
x
e
y
 
 
 

2
1
2
:
2
x
e
OR y


 
 
  
 

x
z




TI Calculator:
Normal Distribution Area
1. 2nd + VARS
2. normalcdf(
3. 4 entries required
4. Left bound, Right
bound, value of the
Mean, Standard
deviation
5. Enter
6. For −∞, 𝒖𝒔𝒆 − 𝟏𝟎𝟎𝟎
7. For ∞, 𝒖𝒔𝒆 𝟏𝟎𝟎𝟎
TI Calculator:
Normal Distribution: find
the Z-score
1. 2nd + VARS
2. invNorm(
3. 3 entries required
4. Left Area, value of the
Mean, Standard
deviation
5. Enter

4. Key Concept: Work with normal distributions that are not standard: µ ≠ 𝟎 & σ ≠ 𝟏
Converting to a Standard Normal Distribution (SND): Use the formula
The area in any normal distribution bounded by some score x (as in Figure a) is the same as the area
bounded by the corresponding z score in the standard normal distribution (as in Figure b).
1. Sketch a normal curve, label the mean and any specific x values, and then shade the region
representing the desired probability.
2. For each relevant value x that is a boundary for the shaded region, use the formula
3. Use technology (software or a calculator) or z-Table to find the area of the shaded region.
This area is the desired probability.
Recall: 6.2 Real Applications of Normal Distributions
x
z




4

5. Estimator
An estimator is a statistic used to infer (or estimate) the value of a population parameter.
Unbiased Estimator
An unbiased estimator is a statistic that targets the value of the corresponding population
parameter in the sense that the sampling distribution of the statistic has a mean that is equal to
the corresponding population parameter such as:
Proportion: 𝒑
Mean: 𝒙
Variance: s²
Biased Estimator
These statistics are biased estimators. That is, they do not target the value of the corresponding
population parameter:
• Median
• Range
• Standard deviation s
5
Recall: 6.3 Sampling Distributions and Estimators

6. Key Concept: the central limit theorem.
The central limit theorem allows us to use a normal distribution for some very
meaningful and important applications.
A sampling distribution of sample means is a distribution obtained by using the means
computed from random samples of a specific size taken from a population.
Sampling error is the difference between the sample measure and the corresponding
population measure due to the fact that the sample is not a perfect representation of the
population.
Conclusion:
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 𝑥 = 𝜇 .
3. The standard deviation of all sample means is 𝜎 𝑥= 𝜎/ 𝑛
6
6.4 The Central Limit Theorem

7. Central Limit Theorem and the Sampling Distribution 𝒙
A sampling distribution of sample means is a distribution obtained by using the means computed from random samples of a
specific size taken from a population. Sampling error is the difference between the sample measure and the corresponding
population measure due to the fact that the sample is not a perfect representation of the population.
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.
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.)
Conclusion:
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 𝑥 = 𝜇 .
3. The standard deviation of all sample means ( also called the standard error of the mean) is 𝜎 𝑥= 𝜎/ 𝑛
1. For samples of size n > 30, the distribution of the sample means can be approximated reasonably well by a normal distribution.
The approximation becomes 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).
Requirements: Population has a normal distribution or n > 30:
7/
x
x
x x
z
n
 
 
 
 

8. 8
An elevator has a sign stating that the maximum capacity is 4000 lb - 27
passengers. This converts to a mean passenger weight of 148 lb when
the elevator is full. Assume a worst-case scenario of the elevator being
filled with 27 adult males. (Adult males have weights that are normally
distributed with a mean of 189 lb and a standard deviation of 39 lb.)
a. Find the probability that 1 randomly selected adult male has a weight
greater than 148 lb.
b. Find the probability that a sample of 27 randomly selected adult males
has a mean weight greater than 148 lb.
Example 1 Normal Distribution x
z




Solution: ND, µ = 189lb & σ = 39lb, n = 1
( 148)P x   ( 1.05)P z  
148 189
1.05
39
x
z


 
   
= 1 − 0.1469 = 0.8531

9. 9
b. Find the probability that a sample of 27 randomly selected adult males
has a mean weight greater than 148 lb.
Example 1 Continued Normal Distribution
ND: µ = 189lb & σ = 39lb, n = 27
( 148)P x   ( 5.46)P z  
148 189
5.46
39 / 27

  
= 1 − 0.0001 = 0.9999
/
X
X
z
n





 X
X
X
z


189
/ 39 / 27
7.51
X
n
 

 


b. Use ND: If the original population is normally
distributed or n > 30.
n < 30: but the original population of weights of males has
a normal distribution, so samples of any size will yield
means that are normally distributed.
Interpretation: There is a 0.8534 probability that
an individual male will weigh more than 148 lb, and
there is a 0.99999998 probability that 27 randomly
selected males will have a mean weight of more than
148 lb. Given that the safe capacity of the elevator is
4000 lb, it is almost certain that it will be overweight
if it is filled with 27 randomly selected adult males.

10. Assume that the population of human body temperatures has a mean of 98.6°F,
as is commonly believed with a standard deviation of 0.62°F. If a sample of size
n = 106 is randomly selected, find the probability of getting a mean of 98.2°F or
lower.
10
Example 2
Solution: µ = 98.6°F, & σ = 0.62°F, n =106 > 30 → 𝐶𝑒𝑛𝑡𝑟𝑎𝑙 𝑇ℎ𝑒𝑜𝑟𝑒𝑚
98.6
/ 0.62 / 106 0.06022
X
n
 

 
 
( 6.64)P z  
98.2 98.6
6.64
0.62 / 106

  
/
X
X
z
n




( 98.2)P x  
0.0001

 X
X
X
z



11. Interpretation
The result shows that if the mean of our body temperatures is really
98.6°F, as we assumed, then there is an extremely small probability of
getting a sample mean of 98.2°F or lower when 106 subjects are randomly
selected. University of Maryland researchers did obtain such a sample
mean, and after confirming that the sample is sound, there are two feasible
explanations:
(1) The population mean really is 98.6°F and their sample represents a
chance event that is extremely rare;
(2) The population mean is actually lower than the assumed value of
98.6°F and so their sample is typical. Because the probability is so
low, it is more reasonable to conclude that the population mean is
lower than 98.6°F. In reality it appears that the true mean body
temperature is closer to 98.2°F!
11

12. Assume that children under the age of 6 watch an average of 25 hours of
television per week. Also, assume the variable is normally distributed and the
standard deviation is 3 hours. If 20 children between the ages of 2 and 5 are
randomly selected, find the probability that the mean of the number of hours
they watch television will be greater than 26.3 hours.
12
Example 3
Solution: ND: µ = 25hrs, & σ = 3, n = 20
( 1.94)P z  /
X
X
z
n




( 26.3)P x  
26.3 25
3 20


1.94
0.0262

 X
X
X
z



13. The average age of a vehicle registered in the United States is 8 years, or
96 months. Assume the standard deviation is 16 months. If a random
sample of 36 vehicles is selected, find the probability that the mean of
their age is between 90 and 100 months.
13
Example 4
Solution: µ = 96 months, & σ = 16, n = 36 > 30 → 𝐶𝑒𝑛𝑡𝑟𝑎𝑙 𝑇ℎ𝑒𝑜𝑟𝑒𝑚
( 2.25 1.5)P z  
(90 100)P x  
90 96
2.25
16 36

  z
100 96
1.50
16 36

 z
0.9332 – 0.0122 = 0.9210

 X
X
X
z



14. A person consumes 218.4 lb of meat per
year on average with a standard deviation of
25 pounds and the distribution is
approximately normal.
a. Find the probability that a person
selected at random consumes less than
224 pounds per year.
b. If a sample of 40 individuals is selected,
find the probability the sample mean
will be less than 224 pounds per year.
14
Example 5
Solution: ND, µ = 218.4 lb, & σ = 25lb
( 0.22)P z ( 224)P x  
( 224)P x  
x
z




224 218.4
0.22
25

 
= 0.5871
X
z
n



 224 218.4
1.42
25 40

 
( 1.42)P z 
= 0.9222