Transcript Document 7211722
Chapter
1
The Normal Probability Distribution
THIS CHAPTER’S GOALS
2
TO LIST THE CHARACTERISTICS OF THE NORMAL DISTRIBUTION.
TO DEFINE AND CALCULATE Z VALUES.
TO DETERMINE PROBABILITIES ASSOCIATED WITH THE STANDARD NORMAL DISTRIBUTION.
TO USE THE NORMAL DISTRIBUTION TO APPROXIMATE THE BINOMIAL DISTRIBUTION.
CHARACTERISTICS OF A NORMAL PROBABILITY DISTRIBUTION
3
The normal curve is
bell-shaped
and has a single peak at the exact center of the distribution.
The arithmetic mean, median, and mode of the distribution are equal and located at the peak.
Half the area under the curve is above this center point, and the other half is below it.
The normal probability distribution is
symmetrical
about its mean.
It is
asymptotic -
the curve gets closer and closer to the x-axis but never actually touches it.
CHARACTERISTICS OF A NORMAL DISTRIBUTION 4
Normal curve is symmetrical two halves identical Tail Theoretically, curve extends to - infinity Tail
0.5
0.5
Mean, median, and mode are equal Theoretically, curve extends to + infinity
Normal Distributions with Equal Means but Different Standard Deviations.
s = 3.1
s =
3.9
s
= 5.0
5 m = 20
Normal Probability Distributions with Different Means and Standard Deviations.
6 m
= 5,
s m m
= 9,
s
= 14,
s
= 3 = 6 = 10
THE STANDARD NORMAL PROBABILITY DISTRIBUTION
7
A normal distribution with a mean of 0 and a standard deviation of 1 is called the standard normal distribution .
z value:
The distance between a selected value, designated
X
, and the population mean
m
, divided by the population standard deviation,
s .
Z
s m
The z-value is the number of standard deviations
X
is from the mean.
EXAMPLE
8
The monthly incomes of recent MBA graduates in a large corporation are normally distributed with a mean of $2,000 and a standard deviation of $200. What is the z value for an income X of $2,200? $1,700?
For X = $2,200 and since z = (X -
m)/s ,
then
z
= (2,200 - 2,000)/200 = 1.
A z value of 1 indicates that the value of $2,200 is 1 standard deviation
above
the mean of $2,000.
EXAMPLE (continued)
9
For X = $1,700 and since z = (X -
m)/s ,
then
z
= (1,700 - 2,000)/200 = -1.5.
A z value of -1.5 indicates that the value of $2,200 is 1.5 standard deviation
below
the mean of $2,000.
How might a corporation use this type of information?
AREAS UNDER THE NORMAL CURVE
10
About
68 percent
of the area under the normal curve is within plus one and minus one standard deviation of the mean. This can be written as
m
± 1
s
.
About
95 percent
of the area under the normal curve is within plus and minus two standard deviations of the mean, written
m
± 2
s
.
Practically all
(99.74 percent)
of the area under the normal curve is within three standard deviations of the mean, written
m
± 3
s
.
Between:
1. 68.26% 2. 95.44% 3. 99.97%
m+3s m2s m1s m m+1s m+2s m+3s 11
P(z)=?
12
A typical need is to determine the probability of a z-value being greater than or less than some value.
Tabular Lookup (Appendix D, page 474)
EXCEL Function =NORMSDIST(z)
EXAMPLE
13
The daily water usage per person in Toledo, Ohio is normally distributed with a mean of 20 gallons and a standard deviation of 5 gallons.
About 68% of the daily water usage per person in Toledo lies between what two values?
EXAMPLE
14
The daily water usage per person in Toledo (X), Ohio is normally distributed with a mean of 20 gallons and a standard deviation of 5 gallons.
What is the probability that a person selected at random will use
less than
20 gallons per day?
What is the probability that a person selected at random will use
more than
20 gallons per day?
EXAMPLE (continued)
15
What percent uses between 20 and 24 gallons?
The z value associated with X = 20 is z = 0 and with X = 24, z = (24 - 20)/5 = 0.8 P(20 < X < 24) = P(0 < z < 0.8) = 0.2881=28.81%
What percent uses between 16 and 20 gallons?
0.8
P(0 < z < 0.8) = 0.2881
16
EXAMPLE (continued)
17
What is the probability that a person selected at random uses more than 28 gallons?
18
P(z > 1.6) = 0.5 - 0.4452 = 0.0048
Area =
0.4452
1.6
z
EXAMPLE (continued)
What percent uses between 18 and 26 gallons?
19
Area = 0.1554
Total area = 0.1554 + 0.3849 = 0.5403
20
Area = 0.3849
.4
1.2
z
EXAMPLE (continued)
21
How many gallons or more do the top 10% of the population use?
Let
X Y
be the least amount. Then we need to find such that P(X
Y
) = 0.1 To find the corresponding z value look in the body of the table for (0.5 - 0.1) = 0.4. The corresponding z value is 1.28 Thus we have (
Y
- 20)/5 = 1.28, from which
Y
= 26.4. That is, 10% of the population will be using
at least
26.4 gallons daily.
(
Y
- 20)/5 = 1.28
22
Y
= 26.4
0.4
1.28
0.1
z
EXAMPLE
23
A professor has determined that the final averages in his statistics course is normally distributed with a mean of 72 and a standard deviation of 5. He decides to assign his grades for his current course such that the top 15% of the students receive an A. What is the lowest average a student must receive to earn an A?
(
Y
- 72)/5 = 1.04
24
Y
= 77.2
0.35
1.04
0.15
z
EXAMPLE
25
The amount of tip the waiters in an exclusive restaurant receive per shift is normally distributed with a mean of $80 and a standard deviation of $10. A waiter feels he has provided
poor
service if his total tip for the shift is less than $65. Based on his theory, what is the probability that he has provided
poor
service?
26
Area = 0.5 - 0.4332 = 0.0668
- 1.5
Area =
0.4332
z
THE NORMAL APPROXIMATION TO THE BINOMIAL
27
Using the normal distribution (a continuous distribution) as a substitute for a binomial distribution (a discrete distribution) for large values of
n
seems reasonable because as
n
increases, a binomial distribution gets closer and closer to a normal distribution.
When to use the normal approximation?
The normal probability distribution is generally deemed a good approximation to the binomial probability distribution when
np
and n(1 - p)
are both greater than 5.
P(r)
0.5
Binomial Distribution with n = 3 and p = 0.5.
28
0.4
0.3
0.25
0 1 2
r
P(r)
Binomial Distribution with n = 5 and p = 0.5.
29
r
P(r)
Binomial Distribution with n = 20 and p = 0.5.
30
Observe the Normal shape.
r
THE NORMAL APPROXIMATION (continued)
31
Recall for the binomial experiment:
There are only two mutually exclusive outcomes (success or failure) on each trial.
A binomial distribution results from counting the number of successes.
Each trial is independent.
The probability p is fixed from trial to trial, and the number of trials n is also fixed.
CONTINUITY CORRECTION FACTOR
32
The value
0.5
subtracted or added, depending on the problem, to a selected value when a binomial probability distribution, which is a discrete probability distribution, is being approximated by a continuous probability distribution--the normal distribution.
The basic concept is that a slice of the normal curve from x-0.5 to x+0.5 is approximately equal to P(x).
EXAMPLE
33
A recent study by a marketing research firm showed that 15% of the homes had a video recorder for recording TV programs. A sample of 200 homes is obtained. (Let X be the number of homes).
Of the 200 homes sampled how many would you expect to have video recorders?
m
= np (0.15)(200) = 30 & n(1 - p) = 170
What is the variance?
s
2 = np(1 - p) = (30)(1- 0.15) = 25.5
EXAMPLE (continued)
34
What is the standard deviation?
s
=
(25.5) = 5.0498.
What is the probability that
less than
40 homes in the sample have video recorders?
We need P(X < 40) = P(X
39). So, using the normal approximation, P(X
39.5)
P[z
(39.5 - 30)/5.0498] = P(z
P(z
1.8812) 1.88) = 0.5 + 0.4699 = 0.9699
Why did I use 39.5 ? ...
How would you calculate P(X=39) ?
35
P(z
1.88) = 0.5 + 0.4699 = 0.9699
0.5
0.4699
1.88
z
EXAMPLE (continued)
36
What is the probability that
more than
24 homes in the sample have video recorders?
37
P(z
-1.09) = 0.5 + 0.3621 = 0.8621.
-1.09
0.3621
0.5
z
EXAMPLE (continued)
38
What is the probability that
exactly
40 homes in the sample have video recorders?
P(1.88
z
2.08) = 0.4812 - 0.4699 = 0.0113
39
0.4699
0.4812
1.88
2.08
z