Chapter 5 Normal Probability Distributions

1
Chapter 5Normal Probability Distributions

• 5-1 Continuous Random Variables
• 5-2 Standard Normal Distribution
• 5-3 Applications of Normal Distributions
• Finding Probabilities
• Finding Values
• 5-4 Determining Normality
• 5-5 The Central Limit Theorem
• 5-6 Normal Distribution as Approximation to
  Binomial Distribution

2
Continuous Random Variables
5-1

• Uniform (rectangular) distribution
• Normal distribution
• Standard Normal distribution
• Note
• Continuous random variables are typically
  represented by curves
• Discrete random variables are typically
  represented by histograms

3
Definitions

• Density Curve (or probability density function)
  is the graph of a continuous probability
  distribution

Properties 1. The total area under the curve
must equal 1. 2. Every point on the curve must
have a vertical height that is 0 or greater.
4

• Two Important Points
• Because the total area under the density curve is
  equal to 1, there is a correspondence between
  area and probability.
• To find probability in a continuous distribution
  just find the area under the curve (Calculus)

5
Uniform Distribution

• Rectangular Distribution
• Simplest type of continuous probability
  distribution
• random variable values are spread evenly over the
  range of  possibilities
• graph results in a rectangular shape.
• Finding probability is the same as finding the
  area of a rectangle

6
Using Area to Find Probability of Class Length
Find P(51.5 lt x lt 52)
7
Definitions

• Uniform Distribution
• Formal definition of density function
• 1

for a lt x lt b
y
b - a
From previous example a 50 and b 52
1
1
y

2
52 - 50
8
Examples

• Let X be a uniform random variable, find the
  density function for each and graph
• If X is between 0 and 5
• Find a. p(x gt 4) b. p(1 lt x lt 5)
• If X is between 6 and 12
• Find a. Find p(x gt 6) b. p(x lt 5) c. p(x gt
  12)
• Go to excel

9
Normal Distribution

• Continuous random variable
• Normal distribution

x - µ 2
( )
1
?
2
y
e
Formula
? 2 p
10
5-2

• The Standard Normal Distribution

11
Definition

• Standard Normal Distribution
• a normal probability distribution that has a
• mean of 0 and a standard deviation of 1, and the
  total area under its density curve is equal to 1.

Area found in z - table or calculator
1
z2
2
y
e
2 p
0.4429
z 1.58
0
Note Typically z denotes a standard normal
random variable
12
Notation

• P(a lt z lt b)
• denotes the probability that the z score is
  between a and b
• P(z gt a)
• denotes the probability that the z score is
  greater than a
• P (z lt a)
• denotes the probability that the z score is
  less than a

13
Z - Table

• Uses calculus to find the area under density
  function curve associated with a particular
  z-value
• Table available on website
• Calculator is easier

14
POSITIVE Z SCORES
15
NEGATIVE Z SCORES
16
To findz Score

• the distance along horizontal scale of the
  standard normal distribution refer to the
  leftmost column and top row of Table
• Area
• the region under the curve refer to the values
  in the body of Table

17
TI-83 Calculator

• Finding Areas of Standard Normal Distribution
• Press 2nd Distr
• Choose normcdf
• Enter normcdf(low z value, high z value)
• Press enter
• Note if low z value is negative infinity use
  E99
• if high z value is positive infinity use E99

18
Example If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of
1 degree for freezing water and if one
thermometer is randomly selected, find the
probability that, at the freezing point of water,
the reading is less than 1.58 degrees.
P (z lt 1.58) 0.9429

• The probability that the chosen thermometer will
  measure freezing water less than 1.58 degrees is
  0.9429.

19
Example If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of
1 degree for freezing water and if one
thermometer is randomly selected, find the
probability that, at the freezing point of water,
the reading is less than 1.58 degrees.
P (z lt 1.58) 0.9429

• 94.29 of the thermometers have readings less
  than 1.58 degrees.

20
Example If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of
1 degree for freezing water, and if one
thermometer is randomly selected, find the
probability that it reads (at the freezing point
of water) above 1.23 degrees.
P (z gt 1.23) 0.8907

• The probability that the chosen thermometer with
  a reading above 1.23 degrees is 0.8907.

21
Example If thermometers have an average (mean)
reading of 0 degrees and a standard deviation of
1 degree for freezing water, and if one
thermometer is randomly selected, find the
probability that it reads (at the freezing point
of water) above 1.23 degrees.
P (z gt 1.23) 0.8907

• 89.07 of the thermometers have readings above
  1.23 degrees.

22
Example A thermometer is randomly selected.
Find the probability that it reads (at the
freezing point of water) between 2.00 and 1.50
degrees.
P (z lt 2.00) 0.0228 P (z lt 1.50) 0.9332 P
(2.00 lt z lt 1.50) 0.9332 0.0228 0.9104

• The probability that the chosen thermometer has
  a reading between 2.00 and 1.50 degrees is
  0.9104.

23
Example A thermometer is randomly selected.
Find the probability that it reads (at the
freezing point of water) between 2.00 and 1.50
degrees.
P (2.00 lt z lt 1.50) 0.9104

• If many thermometers are selected and tested at
  the freezing point of water, then 91.04 of them
  will read between 2.00 and 1.50 degrees.

24
Lets try several examples examples (sketch a
graph)

1. P(Z lt 1.45)
2. P(Z lt -1.45)
3. P(Z gt 1.45)
4. P(Z gt -1.45)
5. P(0 lt Z lt 1.45)
6. P(-1.45 lt Z lt 0)
7. P(-1.45 lt Z lt 1.45)

1. P(-1.35 lt Z lt 1.45)
2. P(.56 lt Z lt 1.45)

Test Problems
25
Frequently used Z Scores

1. Z 1.28
2. Z 1.645
3. Z 1.96
4. Z 2.33
5. Z 2.58

To illustrate why, find

1. P(Z gt 1.28)
2. P(Z gt 1.645)
3. P(Z gt 1.96)
4. P(Z gt 2.33)
5. P(Z gt 2.58)

26
The Empirical Rule Standard Normal Distribution
µ 0 and ? 1
99.7 of data are within 3 standard deviations of
the mean
95 within 2 standard deviations
68 within 1 standard deviation
34
34
2.4
2.4
0.1
13.5
13.5
x - 3s
x - 2s
x - s
x
x 2s
x 3s
x s
27
Finding a z - score when given a probability
(Area)

• 1. Draw a bell-shaped curve, draw the
  centerline, and identify the region under the
  curve that corresponds to the given probability.
• 2. Use the invNorm calculator function to
  determine the corresponding z score.

28
Finding z Scores when Given Probabilities
Finding the 95th Percentile
95
5
5 or 0.05
0.45
0.50
1.645
0
invNorm(.95)
(z score will be positive)
29
Finding z Scores when Given Probabilities
Finding the 10th Percentile
90
10
Bottom 10
0.40
0.10
z
0
(z score will be negative)
30
Finding z Scores when Given Probabilities
Finding the 10th Percentile
90
10
Bottom 10
0.40
0.10
-1.28
0
invNorm(.10)
(z score will be negative)
31
Examples

• Find the Z score that corresponds the following
  percentiles
• 50th
• 25th
• 75th
• Find the Z score(s) for the following areas from
  the mean
• 0.4495
• 0.3186

32
Find the Z score when given the probability
(sketch each)

• P(0 lt z lt a) .34
• P(-b lt z lt b) .8442
• P(z gt c) .056
• P(z gt d) .8844
• P(z lt e) .3400
• These are similar to HW 4 (Extra Credit)

33
5-3 Applications of Normal Distributions

• If ? ? 0 or ???? 1 (or both), we will convert
  values to standard scores using the z-score
  formula we learned in Chapter 2, then procedures
  for working with all normal distributions are the
  same as those for the standard normal
  distribution.

Note z scores are real numbers that have no
units
34
Converting to Standard Normal Distribution

35
Probability of Sitting Heights Less Than 38.8
Inches
36
Probability of Sitting Heights Less Than 38.8
Inches
P ( x lt 38.8 in.) P(z lt 2) normalcdf (-E99,2)
0.9772
37
Probability of Weight between 150 pounds and 200
pounds
200 - 150
µ 150
z
2.00
s ??25
25
x
Weight
150 200
z
0 2.00
38
Probability of Weight between 150 pounds and 200
pounds
normalcdf (0,2) .4772
µ 150
s ??25
x
Weight
150 200
z
0 2.00
39
Probability of Weight between 150 pounds and 200
pounds
There is a 0.4772 probability of randomly
selecting a woman with a weight between 150 and
200 lbs.
µ 150
s ??25
x
Weight
150 200
z
0 2.00
40
Probability of Weight between 150 pounds and 200
pounds
OR - 47.72 of women have weights between 150 lb
and 200 lb.
µ 150
s ??25
x
Weight
150 200
z
0 2.00
41

• Normal Distribution Finding Values
• From Known Areas

42
Cautions to keep in mind

• Dont confuse z scores and areas.
• Choose the correct (right/left) side of the
  graph.
• A z score must be negative whenever it is located
  to the left of the centerline of 0.

43
Procedure for Finding Values Using TI-83
Calculator

• 1. Sketch a normal distribution curve, enter the
  given probability or percentage in the
  appropriate region of the graph, and identify the
  x value(s) being sought.
• Find the z score
• invNorm(area)
• Make the z score negative if it is located to
  the left of the centerline.
• 3. Use x µ (z ?) to find the value for x
  or invNorm(area,µ,s)
• 4. Refer to the sketch of the curve to verify
  that the solution makes sense.

44
Finding P10 for Weights of Women
x ? (z ? ?) x 150 (-1.28 25) 118.75
µ 150
s ??25
0.10
0.40
0.50
x
Weight
150
x ?
z
0
-1.28
45
Finding P10 for Weights of Women
The weight of 119 lb (rounded) separates the
lowest 10 from the highest 90.
0.10
0.40
0.50
x
Weight
150
x 119
z
0
-1.28
46
In Class Example
A tire company finds the lifespan for one brand
of its tires is normally distributed with a mean
of 48,500 miles and a standard deviation of 5000
miles. a) What is the probability that a
randomly selected tire will be replaced in
40,000 miles or less? b) If the manufacturer is
willing to replace no more than 10 of the
tires, what should be the approximate number of
miles for a warranty?
Similar to hw 9 Test problem
47
REMEMBER!
Make the z score negative if the value is located
to the left (below) the mean. Otherwise, the z
score will be positive.
48
Extra Credit
A teacher informs her statistics class that a
test if very difficult, but the grades will be
curved. Scores are normally distributed with a
mean of 25 and standard deviation of 5. The
grades will be curved according to the following
scheme. As Top 10 Bs Between the top
10 and top 30. Cs Between the bottom 30 and
top 30 Ds Between the bottom 10 and bottom
30 Fs Bottom 10 Find the numerical
limits for each grade. Include sketch.
49
5-4

• Determining Normality

50
Determining Normality

• Use Descriptive Methods
• Histogram
• Dot Plot
• Stem-leaf

51
Determining Normality

• Use Normal Probability Plots
• Plot the observed data versus expected z-score of
  the data.
• Expected z-scores are based on the position of
  the data in ascending order similar to
  percentiles.
• If the data are from a normal distribution then
  the Normal Probability Plot will be approximately
  linear. Note X µ Zs (linear relationship
  between X and Z)
• Due to the tedious nature of this process, use
  technology.

52
TI-83 Calculator

• Normal Probability Plot
• Enter data in L1
• Choose statplot (2nd Y)
• Press Enter
• Plot on should be ON
• Cursor to Type and choose the last plot type
• Press Zoom
• Choose 9

53
5-5

• The Central Limit Theorem
• (sampling distribution of the mean)

54
Distribution of 200 digits from Social Security
Numbers (Last 4 digits from 50 students)
20
Frequency
10

• 0

0 1 2 3 4 5 6 7
8 9
Distribution of 200 digits
55
SSN digits x
56
Distribution of 50 Sample Means for 50 Students
15
Frequency
10
5

• 0

0 1 2 3 4 5 6 7
8 9
57
Definition
Sampling Distribution of the mean the
probability distribution of sample means, with
all samples having the same sample size n. See
Excel
58
Sampling Distributions

http//onlinestatbook.com/stat_sim/sampling_dist/i
ndex.html
59
Central Limit Theorem
Given

• 1. The random variable x has a distribution
  (which may or may not be normal) with mean µ and
  standard deviation ?.
• 2. Samples all of the same size n are randomly
  selected from the population of x values.

60
Central Limit Theorem
Conclusions
1. The distribution of sample x will, as the
sample size increases, approach a normal
distribution regardless of the original
population distribution. 2. The mean of the
sample means will be the population mean µ. 3.
The standard deviation of the sample means will
approach ??????????????
n
61
Notation

• the mean of the sample means
• the standard deviation of sample mean
• ???
• (often called standard error of the mean)

µx µ
?
?x
n
62
Central Limit Theorem
Conditions for Application

• Original population is not normal or unknown
• Sample must be sufficiently large (n gt 30)
• 2. The original population is normal.
• Sample may be of any size

Test Question
63
Describing the Sampling Distribution of the Mean
As the sample size increases, the sampling
distribution of sample means approaches a normal
distribution with
µx µ
?
?x
n
Test Question
64
Example The mean age of students at a community
is 25 with a standard deviation of 8. A sample
of 64 students is drawn. Describe the sampling
distribution of the mean.

Solution XN(25,1)
Test Question
65
Example Given the population of women has
normally distributed weights with a mean of 143
lb and a standard deviation of 29 lb, a.) if
one woman is randomly selected, find the
probability that her weight is greater than 150
lb.b.) if 36 different women are randomly
selected, find the probability that their mean
weight is greater than 150 lb.

66
Example Given the population of women has
normally distributed weights with a mean of 143
lb and a standard deviation of 29 lb, a.) if
one woman is randomly selected, find the
probability that her weight is greater than 150
lb. (same approach as section 5.3.
z 150-143 0.24 29
P (X gt 150) P(z gt .24) Normalcdf (.24,E99)
0.4052
0.5948
x
? 143
150
z
??? 29
0
0.24
67
Example Given the population of women has
normally distributed weights with a mean of 143
lb and a standard deviation of 29 lb, a.) if
one woman is randomly selected, the probability
that her weight is greater than 150 lb. is
0.4052.
P (X gt 150) P(z gt .24) Normalcdf (.24,E99)
0.4052
0.5948
x
? 143
150
z
??? 29
0
0.24
68
Example Given the population of women has
normally distributed weights with a mean of 143
lb and a standard deviation of 29 lb, b.) if 36
different women are randomly selected, find the
probability that their mean weight is greater
than 150 lb.

P (x gt 150)
x
?x 143
150
?x? 29 4.83333
36
69
Example Given the population of women has
normally distributed weights with a mean of 143
lb and a standard deviation of 29 lb, b.) if 36
different women are randomly selected, the
probability that their mean weight is greater
than 150 lb is 0.0735.
z 150-143 1.45 29
P (x gt 150) P(z gt 1.45) Normalcdf (1.45,E99)
0.0735
36
0.9265
x
?x 143
150
z
?x? 4.83333
0
1.45
70
Example Given the population of women has
normally distributed weights with a mean of 143
lb and a standard deviation of 29 lb,

• a.) if one woman is randomly selected, find the
  probability that her weight is greater than 150
  lbs. P(x gt 150) 0.4052
• b.) if 36 different women are randomly selected,
  find the probability that their mean weight is
  greater than 150 lbs. P(x gt 150) 0.0735It
  is much easier for an individual to deviate from
  the mean than it is for a group of 36 to deviate
  from the mean.

71
5-6

• Normal Distribution as Approximation to Binomial
  Distribution

72
Approximate a Binomial Distributionwith a Normal
Distribution if

• np ? 5
• nq ? 5

Basically if n is large your ok. Typically
this approximation is only used when n is large
anyway.
73
Approximate a Binomial Distributionwith a Normal
Distribution if

• np ? 5
• nq ? 5

then µ np and ? npq and the random
variable has
a
distribution.
(normal)
74
Approximate a Binomial Distributionwith a Normal
Distribution
http//bcs.whfreeman.com/ips4e/cat_010/applets/CLT
-Binomial.html
75
Definition

• Continuity Correction
• Needed when we use the normal distribution
  (continuous) as an approximation to the binomial
  distribution (discrete)
• Made to a discrete whole number x in the binomial
  distribution by representing the single value x
  by the interval from
• x - 0.5 to x 0.5.

76
Procedure for Using a Normal Distribution to
Approximate a Binomial Distribution

• 1. Verify np ? 5 and nq ? 5.
• 2. Calculate µ np and ? npq.
• 3. Identify the continuity correction x - 0.5 to
  x 0.5.
• 4. Draw a normal curve and enter the values of µ
• 4. Change x by replacing it with x - 0.5 or x
  0.5, as appropriate.
• 5. Find the area corresponding to the desired
  probability.

77
Finding the Probability of At Least 520 Men
Among 1000 Accepted Applicants
78
P(x ? 520) 520, 521, 522, . Use P(x ? 519.5)
.
520
519.5
P(x gt 520) 521, 522, 523, . . . Use P(x gt 520.5)
521
520.5
P(x 520) 0, 1, . . . 518, 519, 520 Use P(x
520.5)
520
520.5
P(x lt 520) 0, 1, . . . 518, 519 Use P(x lt 519.5)
519
519.5
79
P(x520)
520
519.5
520.5
Interval represents discrete number 520