The Normal Probability Distribution

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The Normal Probability Distribution
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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,

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There are many Normal distributions depending on
by m and s
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The Standard Normal Distributionm 0, s 1
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• 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.

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

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• Perform the z-transformation

Area under the Normal distribution with mean m
and standard deviation s
then
Area under the standard normal distribution
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Area under the Normal distribution with mean m
and standard deviation s
s
m
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Area under the standard normal distribution
1
0
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Using the tables for the Standard Normal
distribution
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Example

• Find the area under the standard normal curve
  between z -? and z 1.45

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Example

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

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Example

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

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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

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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

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

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

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

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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.

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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
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Graphical Illustration
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Example, Part 2)
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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?
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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.)

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• 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.

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Graphs
X (Mathematics) m 90, s 3.
Z (Social Studies) m 70 , s 7.
Y (English Literature) m 60, s 10.
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• 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?
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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
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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)
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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)
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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
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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?
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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
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The distribution of averages (the mean)

• Because the mean is a linear combination

and
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• 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
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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

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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

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Solution

• Find the probability distribution of

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Graphs
The probability distribution of the mean
The probability distribution of individual
observations
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Normal approximation to the Binomial distribution

• Using the Normal distribution to calculate
  Binomial probabilities

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Binomial distribution n 20, p 0.70
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Normal Approximation to the Binomial distribution

• X has a Binomial distribution with parameters n
  and p

• Y has a Normal distribution

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Example

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

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• Using the Normal approximation to the Binomial
  distribution

Where Y has a Normal distribution with
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• Hence

0.4052 - 0.2327 0.1725
Compare with 0.1643
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Normal Approximation to the Binomial distribution

• X has a Binomial distribution with parameters n
  and p

• Y has a Normal distribution

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Example

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

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• Using the Normal approximation to the Binomial
  distribution

Where Y has a Normal distribution with
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• Hence

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

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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.

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Solution

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

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