Download 2015 Lesson 4: The Normal Probability

Lesson 4 Probability & Normal Probability Distribution
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P(A)= probability of event A = (number of outcomes of event A)/(all of the outcomes)
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a
P A.  1 – P(A) = the complement of event A
P(A or B) = P(A)+P(B) - P(A and B) = the union of events A and B
P(A and B)= P(A) x P(B) = the intersection of independent event A and independent event B
P(an impossible event) = 0
P( a certain event) = 1
P(at least one) = 1 – P(none)
P(B|A)= Conditional probability of event B, given event A has an outcome = P(A and B) / P(A)
A ball is drawn at random from a box containing 6 red balls, 4 white balls and 5 blue balls.
Find the probability that a) it is red b) white c) blue d) not red e) red or white f)red and
white
a) 6 / ( 6 + 4 + 5) = 6 / 15 = 2 / 5 = 0.40
b) 4 / ( 6 + 4 + 5) = 4/ 15 = 0.267
c) 5 / (6 + 4 + 5) = 5/15 = 0.33
d) P(not red) = 1 – P(red) = 1 - 0.40 = 0.60
e) P(red union white) = (6 + 4) / (6 + 4 + 5)= .667 or 1- P(blue)= 1- .0.333 = .667
f) P(red intersect white)= P(red) x P(white)= 0.40 x 0.267 = 0.11
CONTINGENCY TABLE
Wireless System Preference by Gender
Male
Female
Type A
200
35
Type B
125
300
Type C
19
21
Total
344
356
Type A
Type B
Type C
Total
Male
0.29
0.18
0.02
0.48
Female
0.05
0.43
0.03
0.51
Total
235
425
40
700
Total
0.34
0.61
0.05
1
P(M)=
P(F)=
P(M) or Type A=
P((F) or Type B=
P(F) or Type C=
P(M) and Type A=
P(F) and Type C=
0.48+ 0.34-0.29
0.51+0.61-0.43
0.51+0.05-0.03
0.48 x 0.29
0.51 x 0.03
0.48
0.51
0.53
0.69
0.53
0.14
0.02
P(M)+P(A)-P(M and A)
P(F)+P(B)-P(F and B)
P(F)+P©-P(F and C)
P(M)x P(A|M)
P(F)xP(C|F)
fffff
Using z-scores to find Normal Probabilities: An SAT Exam had X bar= 21.0 and s= 4.7.
When a nearby high school student took the exam, he scored 32. His z-score is
Z= (32-21)/4.7 = 2.34 . This may also be obtained using Excel:
=STANDARDIZE(32, 21, 4.7)=2.34 . Looking up the probability in a Normal
Distribution table, for z=2.34 p= .9904. This means that about 1 student in 100
scored better than the nearby high school student.
Assume the commute time to travel from Barnegat to Toms River is normally
distributed with a mean of 65 minutes and a standard deviation of 5 minutes.
a) What is the z-score for a commute time of 53 minutes?
b) About what percent of commutes will be faster than 60 minutes?
a) =STANDARDIZE(53, 65, 5) =z = -2.4
b) =STANDARDIZE(60, 65, 5) =z =-1.0 , approximately 16% are faster.
Assume that men’s weights are normally distributed with a mean of 172 lb and a
standard deviation of 29 lb. Find the probability that if an individual man is
selected, his weight will be greater than 200 lb.
=STANDARDIZE(200, 172, 29)= z = .9655,
83 % of men will have a weight > 200 lb. 201 lb= (201-172)=29 lb = within one
standard deviation above the mean.
When designing a door, how high should it be if 95% of men will fit through without bumping
their head? Heights of men are normally distributed with a mean of 69.0 inches, and a
standard deviation of 2.8 inches. Using Excel,
NORMINV(.95, 69, 2.8)= 73.6 inches = 6.13 feet high.
To find the z-scores for any value of X, use =STANDARDIZE(X, Mean, Standard deviation).
To find the area to the left of a z-score in the standard normal dist., use =NORMSDIST(z).
To find the area to the right of a z-score, use =1 – NORMSDIST(z).
To find the area between two z-scores, when z1 < z2, use =NORMSDIST(z2) - NORMSDIST(z1).
By the Empirical Rule, if X is normally distributed, approximately 68% of observations will be
within 1 standard deviation of x bar. 95% within 2 standard deviations of x bar, and
99.6 % within 3 standard deviations of x bar.
Age
Salary
zAge
22
35,000 -1.4863
23
36,000 -1.15601
24
37,000 -0.82572
25
38,000 -0.49543
26
39,000 -0.16514
27
40,000 0.165145
28
41,000 0.495434
29
42,000 0.825723
30
43,000 1.156012
31
44,000 1.486301
You see in the Excel sheet, there are 6 Ages that have z-scores
between -1 and 1. 6 Ages/ 10 total Ages = 60 % Empirical
Probability, compared with the Theoretical Probability for a normal
distribution of 68% for z values between -1 and 1. For a large
amount of data, using Excel to sort and filter data is easier, faster
and more accurate. (We did not use the Salary column).
TOTAL
26.5 Mean
3.02765 Standard dev.
http://prod.campuscruiser.com/cruiser/occ/aaikin/NormalDistribution.xls
allows you to do calculations using the Normal Distribution, knowing the mean and the
standard deviation. (Right-click on the link and click “Open hyperlink.” You may view
and/or save the file - it is very useful.
Normal Heights & z-scores
Height
62
66
61
70
72
71
73
70
68
71
69
72
66
65
68
67
69
TOTAL
Mean=
St. Dev.=
z
-1.81287
-0.64327
-2.10526
0.526316
1.111111
0.818713
1.403509
0.526316
-0.05848
0.818713
0.233918
1.111111
-0.64327
-0.93567
-0.05848
-0.35088
0.233918
12 of 17
68.2
3.42
Given 17 women’s heights, x bar
and s were calculated; then z-scores were calculated
using the Excel “fill handle.” Then using Filter, 12 of
17 z-scores were within the between -1 and 1 criteria
which equals 70.59 % probability for this Normal
distribution.
(Minitab 16 graph)
70%