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The Normal Probability Distribution

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,

There are many Normal distributions depending on

by m and s

The Standard Normal Distributionm 0, s 1

- 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.

- Converting Area
- under the Normal distribution with mean m and

standard deviation s - to
- Area under the standard normal distribution

- Perform the z-transformation

Area under the Normal distribution with mean m

and standard deviation s

then

Area under the standard normal distribution

Area under the Normal distribution with mean m

and standard deviation s

s

m

Area under the standard normal distribution

1

0

Using the tables for the Standard Normal

distribution

Example

- Find the area under the standard normal curve

between z -? and z 1.45

Example

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

Example

- Find the area under the normal curve to the

right of z 1.45 P(z gt 1.45)

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

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

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

-1.26

- Example Find the area between z -2.30 and z

1.80

- Example Find the area between z -1.40 and z

-0.50

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.

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

Graphical Illustration

Example, Part 2)

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?

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.)

- 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.

Graphs

X (Mathematics) m 90, s 3.

Z (Social Studies) m 70 , s 7.

Y (English Literature) m 60, s 10.

- 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?

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

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)

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)

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

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?

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

The distribution of averages (the mean)

- Because the mean is a linear combination

and

- 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

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

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

Solution

- Find the probability distribution of

Graphs

The probability distribution of the mean

The probability distribution of individual

observations

Normal approximation to the Binomial distribution

- Using the Normal distribution to calculate

Binomial probabilities

Binomial distribution n 20, p 0.70

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

- Using the Normal approximation to the Binomial

distribution

Where Y has a Normal distribution with

- Hence

0.4052 - 0.2327 0.1725

Compare with 0.1643

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

- Using the Normal approximation to the Binomial

distribution

Where Y has a Normal distribution with

- Hence

0.5948 - 0.0436 0.5512

Compare with 0.5357

- Comment
- The accuracy of the normal appoximation to the

binomial increases with increasing values of n

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.

Solution

- X has a Binomial distribution with parameters n

2000 and p 0.85

where Y has a Normal distribution with

0.9004 - 0.0436 0.8008

Solution part 2.

1.000