1. The Normal Probability Distribution
Points of Inflection m s + 2 3 -

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
Normal m = 100, s =20
Normal m = 100, s = 40
Normal m = 140, s =20

4. The Standard Normal Distribution m = 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. under the Normal distribution with mean m and standard deviation s
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

9. Area under the standard normal distribution1

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
A portion of Table 3: z
0.00
0.01
0.02
0.03
0.04
0.05
0.06
1.4
0.9265 .

12. Example Find the area to the left of -0.98; P(z < -0.98) P z ( 0. )= 98 .
1635

13. Example
Find the area under the normal curve to the right of z = 1.45; P(z > 1.45)

14. Example
Find the area to the between z = 0 and of z = 1.45; P(0 < z < 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
the area above zero is
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. Convert the random variable, X, to its z-score.
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 Assume the amount of fill is normally distributed and a bottle is selected at random:
1) Find the probability the bottle contains between oz and oz
2) Find the probability the bottle contains more than oz
When x z = -
32.00
32.0
0.00 ;
0.02 m s
Solutions part 1)
When x z = - 32
025
32.025
32.0 1 25 . ;
0.02 m s

21. Graphical Illustration:X z ( . ) 0.
32.0 32
025 02 1 25
3944
< = - æ è ç ö ø ÷

22. Example, Part 2) P x z ( . )
> = - æ è ç ö ø ÷ 31 97
32.0 0. 02 1
50)
0000
0668
9332

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) Y (English Literature) 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 Y 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 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 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
Approximating
Normal distribution
Binomial distribution

42. Normal Approximation to the Binomial distribution
X has a Binomial distribution with parameters n and p
Y has a Normal distribution

43. Approximating
Normal distribution
P[X = a]
Binomial distribution

45. P[X = a]

46. Example
X has a Binomial distribution with parameters n = 20 and p = 0.70

47. Where Y has a Normal distribution with:
Using the Normal approximation to the Binomial distribution
Where Y has a Normal distribution with:

48. Hence
= =
Compare with

49. Normal Approximation to the Binomial distribution
X has a Binomial distribution with parameters n and p
Y has a Normal distribution

52. Example
X has a Binomial distribution with parameters n = 20 and p = 0.70

53. Where Y has a Normal distribution with:
Using the Normal approximation to the Binomial distribution
Where Y has a Normal distribution with:

54. Hence
= =
Compare with

55. Comment:
The accuracy of the normal appoximation to the binomial increases with increasing values of n

56. The number of successful operations is between 1650 and 1750.
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. where Y has a Normal distribution with:
Solution
X has a Binomial distribution with parameters n = 2000 and p = 0.85
where Y has a Normal distribution with:

58. = =

59. Solution – part 2.
= 1.000