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## The Normal Probability Distribution

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### Approximating. Normal distribution. Binomial distribution n = 20, p = 0.70 ... Using the Normal approximation to the Binomial distribution ... – PowerPoint PPT presentation

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Title: The Normal Probability Distribution

1
The Normal Probability Distribution
2
Main characteristics of the Normal Distribution
• Bell Shaped, symmetric
• Points of inflection on the bell shaped curve are
at m s and m s. That is one standard
deviation from the mean
• Area under the bell shaped curve between m s
and m s is approximately 2/3.
• Area under the bell shaped curve between m 2s
and m 2s is approximately 95.
• Close to 100 of the area under the bell shaped
curve between m 3s and m 3s,

3
There are many Normal distributions depending on
by m and s
4
The Standard Normal Distributionm 0, s 1
5
• There are infinitely many normal probability
distributions (differing in m and s)
• Area under the Normal distribution with mean m
and standard deviation s can be converted to area
under the standard normal distribution
• If X has a Normal distribution with mean m and
standard deviation s than has a standard normal
distribution

has a standard normal distribution.
• z is called the standard score (z-score) of X.

6
• Converting Area
• under the Normal distribution with mean m and
standard deviation s
• to
• Area under the standard normal distribution

7
• Perform the z-transformation

Area under the Normal distribution with mean m
and standard deviation s
then
Area under the standard normal distribution
8
Area under the Normal distribution with mean m
and standard deviation s
s
m
9
Area under the standard normal distribution
1
0
10
Using the tables for the Standard Normal
distribution
11
Example
• Find the area under the standard normal curve
between z -? and z 1.45

12
Example
• Find the area to the left of -0.98 P(z lt -0.98)

13
Example
• Find the area under the normal curve to the
right of z 1.45 P(z gt 1.45)

14
Example
• Find the area to the between z 0 and of z
1.45 P(0 lt z lt 1.45)
• Area between two points differences in two
tabled areas

15
Notes
• Use the fact that the area above zero and the
area below zero is 0.5000
• the area above zero is 0.5000
• When finding normal distribution probabilities, a
sketch is always helpful

16
• Example
• Find the area between the mean (z 0) and z
-1.26

17
• Example Find the area between z -2.30 and z
1.80

18
• Example Find the area between z -1.40 and z
-0.50

19
Computing Areas under the general Normal
Distributions(mean m, standard deviation s)
Approach
• Convert the random variable, X, to its z-score.
• Convert the limits on random variable, X, to
their z-scores.
• Convert area under the distribution of X to area
under the standard normal distribution.

20
Example
• Example A bottling machine is adjusted to fill
bottles with a mean of 32.0 oz of soda and
standard deviation of 0.02. Assume the amount
of fill is normally distributed and a bottle is
selected at random

1) Find the probability the bottle contains
between 32.00 oz and 32.025 oz 2) Find the
probability the bottle contains more than 31.97 oz
21
Graphical Illustration
22
Example, Part 2)
23
Combining Random Variables
• Quite often we have two or more random variables
• X, Y, Z etc

We combine these random variables using a
mathematical expression. Important question What
is the distribution of the new random variable?
24
An Example
• Suppose that a student will take three tests in
the next three days
• Mathematics (X is the score he will receive on
this test.)
• English Literature (Y is the score he will
receive on this test.)
• Social Studies (Z is the score he will receive on
this test.)

25
• Assume that
• X (Mathematics) has a Normal distribution with
mean m 90 and standard deviation s 3.
• Y (English Literature) has a Normal distribution
with mean m 60 and standard deviation s 10.
• Z (Social Studies) has a Normal distribution
with mean m 70 and standard deviation s 7.

26
Graphs
X (Mathematics) m 90, s 3.
Z (Social Studies) m 70 , s 7.
Y (English Literature) m 60, s 10.
27
• Suppose that after the tests have been written an
overall score, S, will be computed as follows

S (Overall score) 0.50 X (Mathematics) 0.30
Y (English Literature) 0.20 Z (Social Studies)
10 (Bonus marks)
What is the distribution of the overall score, S?
28
Sums, Differences, Linear Combinations of R.V.s
A linear combination of random variables, X, Y, .
. . is a combination of the form L aX
bY where a, b, etc. are numbers
positive or negative. Most common Sum X
Y Difference X Y Others Averages 1/3 X
1/3 Y 1/3 Z Weighted averages 0.40 X 0.25
Y 0.35 Z
29
Means of Linear Combinations
If L aX bY
The mean of L is Mean(L) a Mean(X) b
Mean(Y) mL a mX b mY Most
common Mean( X Y) Mean(X) Mean(Y)
Mean(X Y) Mean(X) Mean(Y)
30
Variances of Linear Combinations
If X, Y, . . . are independent random variables
and L aX bY then Variance(L) a2
Variance(X) b2 Variance(Y) Most
common Variance( X Y) Variance(X)
Variance(Y) Variance(X Y) Variance(X)
Variance(Y)
31
Combining Independent Normal Random Variables
If X, Y, . . . are independent normal random
variables, then L aX bY is normally
distributed. In particular X Y is normal
with X Y is normal with
32
Example Suppose that one performs two
independent tasks (A and B)
X time to perform task A (normal with mean 25
minutes and standard deviation of 3 minutes.) Y
time to perform task B (normal with mean 15
minutes and std dev 2 minutes.) X and Y
independent so T X Y total time is normal
with
What is the probability that the two tasks take
more than 45 minutes to perform?
33
The distribution of averages (the mean)
• Let x1, x2, , xn denote n independent random
variables each coming from the same Normal
distribution with mean m and standard deviation
s.
• Let

What is the distribution of
34
The distribution of averages (the mean)
• Because the mean is a linear combination

and
35
• Thus if x1, x2, , xn denote n independent
random variables each coming from the same Normal
distribution with mean m and standard deviation
s.
• Then

has Normal distribution with
36
Example
• Suppose we are measuring the cholesterol level of
men age 60-65
• This measurement has a Normal distribution with
mean m 220 and standard deviation s 17.
• A sample of n 10 males age 60-65 are selected
and the cholesterol level is measured for those
10 males.
• x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, are
those 10 measurements
• Find the probability distribution of
• Compute the probability that is between 215
and 225

37
Example
• Suppose we are measuring the cholesterol level of
men age 60-65
• This measurement has a Normal distribution with
mean m 220 and standard deviation s 17.
• A sample of n 10 males age 60-65 are selected
and the cholesterol level is measured for those
10 males.
• x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, are
those 10 measurements
• Find the probability distribution of
• Compute the probability that is between 215
and 225

38
Solution
• Find the probability distribution of

39
Graphs
The probability distribution of the mean
The probability distribution of individual
observations
40
Normal approximation to the Binomial distribution
• Using the Normal distribution to calculate
Binomial probabilities

41
Binomial distribution n 20, p 0.70
42
Normal Approximation to the Binomial distribution
• X has a Binomial distribution with parameters n
and p
• Y has a Normal distribution

43
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46
Example
• X has a Binomial distribution with parameters n
20 and p 0.70

47
• Using the Normal approximation to the Binomial
distribution

Where Y has a Normal distribution with
48
• Hence

0.4052 - 0.2327 0.1725
Compare with 0.1643
49
Normal Approximation to the Binomial distribution
• X has a Binomial distribution with parameters n
and p
• Y has a Normal distribution

50
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52
Example
• X has a Binomial distribution with parameters n
20 and p 0.70

53
• Using the Normal approximation to the Binomial
distribution

Where Y has a Normal distribution with
54
• Hence

0.5948 - 0.0436 0.5512
Compare with 0.5357
55
• Comment
• The accuracy of the normal appoximation to the
binomial increases with increasing values of n

56
Example
• The success rate for an Eye operation is 85
• The operation is performed n 2000 times
• Find
• The number of successful operations is between
1650 and 1750.
• The number of successful operations is at most
1800.

57
Solution
• X has a Binomial distribution with parameters n
2000 and p 0.85

where Y has a Normal distribution with
58
0.9004 - 0.0436 0.8008
59
Solution part 2.
1.000