### Chapter 6 * Normal Probability Distributions

```Chapter 6 – Normal Probability
Distributions
Normal Distribution
If a continuous random variable has a distribution with a graph
that is symmetric and bell-shaped and it satisfies the equation
y
e
1  x 
 

2  
2
 2
we say that it has a normal distribution.
***No! We will never use this formula!
Density Curves
A density curve links a probability to an area under the curve and
above the x-axis.
The total area under the curve is 1.
Standard Normal Distribution
• Graph is bell-shaped
• Mean is equal to 0   0
• Standard deviation is equal to 1.   1
z-score Table
Appendix A
Table A - 2
z-score are given to two decimal places
Probabilities are given to four decimal places
Negative z-scores
Numbers inside the table represent areas to
Positive z-scores
the left
of the z-score.
z-score Table
z-score Table
P = 0.5871
z = 0.22
P = 0.3632
z = -0.35
P = 0.0047
z = -2.60
P = 0.7389
z = 0.64
Find the z-score
P = 0.9678
P = 0.1781
z = 1.85
z = -0.92
P = 0.8588
P = 0.0134
z = 1.08
z = -2.22
Critical Values
A critical value is a z-score on the borderline separating the z-scores that are likely to
occur from those that are unlikely to occur.
Common critical values are z = -1.96 and z = 1.96
Recall the Empirical Rule:
Above 2 or below -2 were
considered unusual.
The notation
z denotes the z-score with an area of  in its tail.
Always rewrite  to four decimal places!
Critical Values
z0.07  z0.0700   1.48
The area in the tail is 0.0700.
That means the area above the positive z-score is
0.0700 and
the area below the negative z-score is 0.0700.
The total area under the curve is 1.
1 – 0.0700 = 0.9300
Critical Values
z0.10  z0.1000   1.28
The area in the tail is 0.1000.
That means the area above the positive z-score is
0.1000 and
the area below the negative z-score is 0.1000.
The total area under the curve is 1.
1 – 0.1000 = 0.9000
Find the Probability
P( 0.58  z  0.58)  0.4380
The area to the left of z = 0.58
0.7190
0.2810
The area to the left of z = -0.58
Find the Probability
P( 1.35  z  0.02)  0.4195
The area to the left of z = 0.02
0.5080
0.0885
The area to the left of z = -1.35
Heights
Heights Men’s heights are normally distributed with mean 68.9 in
and standard deviation of 2.8 in. Women’s heights are normally
distributed with mean 63.5 in and standard deviation of 2.5 in. The
standard doorway height is 80 in.
a. What percentage of men are too tall to fit through a standard
doorway without bending, and what percentage of women are too
tall to fit through a standard doorway without bending?
M  68.9
 M  2.8
z
x

80  68.9

2.8
 3.96
P( height  80)  P( z  3.96)
1  0.9999  0.0001
0.01%
68.9 71.7 74.5 77.3 80.1
Heights
Women’s heights are normally distributed with mean 63.5 in and
standard deviation of 2.5 in. The standard doorway height is 80 in.
a. What percentage of women are too tall to fit through a standard
doorway without bending?
W  63.5
 W  2 .5
z
x

80  63.5

2.5
 6.6
P( height  80)  P( z  6.6)
1  0.9999  0.0001
0.01%
63.5
66
68.5
71 73.5
Heights
If a statistician designs a house so that all of the doorways have
heights that are sufficient for all men except the tallest 5%, what
doorway height would be used?
M  68.9
 M  2.8
P  0.0500
z0.0500  1.645
1  0.0500  0.9500
z
x

x  68.9
1.65 
2.8
4.62  x  68.9
x  73.506
73.5 in
Heart Rates
Suppose that heart rates are normally distributed with a mean of 61.4 beats per
minute and a standard deviation of 3.7 beats per minute.
a. If one person is selected randomly, find the probability that their heart rate
is less than 62 beats per minute
  61.4
  3.7
62  61.4
z
 0.16
3.7
P( heart rate  62)  P(z  0.16) 
0.5636
50.3
54
57.7 61.4
65.1 68.8
72.5
Heart Rates
Suppose that heart rates are normally distributed with a mean of 61.4 beats per
minute and a standard deviation of 3.7 beats per minute.
b. What is the probability that a person has a heart rate greater than 65 beats
per minute?
  61.4
  3.7
65  61.4
z
 0.97
3.7
P( heart rate  65)
 P(z  0.97)
 1  0.8340
 0.1660
50.3
54
57 .7 61 .4
65.1 68.8
72.5
Heart Rates
Suppose that heart rates are normally distributed with a mean of 61.4 beats per
minute and a standard deviation of 3.7 beats per minute.
c. What is the probability that a person has a heart rate between 57 and 62
beats per minute?
  61.4
  3.7
z
62  61.4
 0.16
3.7
z
57  61.4
 1.19
3.7
P(57  heart rate  62)  P( 1.19  z  0.16)
 0.5636  0.1170
 0.4466
50.3
54
57 .7 61 .4
65.1 68.8
72.5
Heart Rates
Suppose that heart rates are normally distributed with a mean of 61.4 beats per
minute and a standard deviation of 3.7 beats per minute.
d. If 50 people are randomly selected, find the probability that they have a
mean heart rate less than 62 beats per minute. (Use the Central Limit
Theorem)
  61.4
  3.7
P( mean heart rate  62)
50.3
54
57.7 61.4
65.1 68.8
72.5
Parameter or Statistic
Law of Large Numbers
Draw observations at random from any population with finite
mean  . As we increase the number of observations, the
mean of the observed values x , gets closer to the mean of the
population.
As we increase the sample
size n, the sample mean gets
closer to the population
mean.
As n increases,x  
Estimators
We will use sample statistics to estimate population parameters.
Some sample statistics that we might use to estimate population parameters:
  population mean
s  sample standard deviation   population standard deviation
2
2

 population variance
s  sample variance
x  sample mean
^
p  sample proportion
sample median
sample range
p  population proportion
population median
population range
Estimators
An estimator is unbiased if it targets the population parameter.
An estimator is biased if it systematically under estimates or over estimates the
population parameter.
  population mean
s  sample standard deviation   population standard deviation
2
2

 population variance
s  sample variance
x  sample mean
^
p  sample proportion
sample median
sample range
p  population proportion
population median
population range
Since the bias associated with the sample standard deviation is very small, we
WILL still use it to estimate the population standard deviation.
Sampling Distributions
The population distribution is the distribution of the values of
the variable among all the individuals of a population.
The sampling distribution of a statistic is a theoretical
distribution of values taken by the statistic in all possible
samples of the same size from the same population.
A population distribution shows how the
value of an individual varies in a
population.
A sampling distribution shows how the
value of a statistic varies from all possible
samples of a certain size.
Sampling Distribution of x
sampling distribution of the mean
The mean of the sampling distribution or
The mean of the means
x
Sampling Distribution of x
An ice cream company was in business for only three
days. Here are the numbers of phone calls received
on each of those days: 12, 10, 5. Assume that
samples of size 2 are randomly selected with
replacement from this population of three values.
a. List the different population samples and find the
mean of each of them.
Sample
x
Sample
x
Sample
x
12 – 10
11
10 – 5
7.5
5 – 12
8.5
12 – 5
8.5
10 – 12
11
5 – 10
7.5
12 – 12
12
10 – 10
10
5–5
5
Sampling Distribution of x
b. Identify the probability of each sample.
Sample
x
Sample
x
Sample
x
12 – 10
11
10 – 5
7.5
5 – 12
8.5
12 – 5
8.5
10 – 12
11
5 – 10
7.5
12 – 12
12
10 – 10
10
5–5
5
Px
12
1
9
2
9
1
9
11
c. Calculate the mean of the sampling distribution.
11  8.5  12  7.5  11  10  8.5  7.5  5
x 
9
9
1
2
1
2
1 1
 x  12  11  10  8.5  7.5  5  9
9
9
9
9
9 9

x
10
8.5
7.5
5
2
9
1
9
1
9
Sampling Distribution of x
d. Is the mean from the sampling distribution from part (c) the same as the
mean of the population of the three listed values? If so, are those means always
equal?
x  9
12  10  5

9
3
Since the mean is an unbiased estimator, it will target the population mean. Yes,
the mean of the sampling distribution will always be the same as the mean of the
population.
x  
Notation
Notation for the sampling distribution of x :
If all possible random samples of size n are selected from a population with
mean  and standard deviation  , the mean of the sample means is denoted
by  x , and
x 
Also the standard deviation of the sample means is denoted by  x , and

x 
n
 x is called the standard error of the mean.
Central Limit Theorem
The Central Limit Theorem allows us to use the Normal probability
observations even when the population distribution is NOT normal.
(For strongly skewed distributions, n > 30 is a good rule of thumb)
xx
z

n
As the sample size increase
- Standard error decreases
- the z score gets larger
Heart Rates
Suppose that heart rates are normally distributed with a mean of 61.4 beats per
minute and a standard deviation of 3.7 beats per minute.
d. If 50 people are randomly selected, find the probability that they have a
mean heart rate less than 62 beats per minute. (Use the Central Limit
Theorem)
  61.4
  3.7
n  50
P( mean heart rate  62)
z
62  61.4
z
 1.15
3.7
50
x  x
x

x

n
50.3
( z  1.15)  0.8749
P(z
54
57.7 61.4
65.1 68.8
72.5
Heart Rates
Suppose that heart rates are normally distributed with a mean of 61.4 beats per
minute and a standard deviation of 3.7 beats per minute.
e. If 100 people are randomly selected, find the probability that they have a
mean heart rate less than 62 beats per minute. (Use the Central Limit
Theorem)
  61.4
  3.7
n  100
P( mean heart rate  62)
z
62  61.4
z
 1.62
3.7
100
x  x
x

x

n
50.3
P((zz  1.62)  0.9474
54
57.7 61.4
65.1 68.8
72.5
```