Probability and statistics for engineering

1. Probability and Statistics
Chapter 6
Some Continuous Probability
Distributions
Dr. Yehya Mesalam 1

2. 2
























b
x
for
b
x
a
for
a
b
a
x
a
x
for
x
F
otherwise
b
x
a
for
a
b
x
f
b
a
U
X
1
0
)
(
0
1
)
(
]
,
[
a b
f(x)
a b
F(x)
Continuous Uniform Distribution
a
b 
1
)
(x
F
)
(x
f
a
b
a
x


2
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3. Mean:
Probability density function:
Variance:
P a  x  b
  b  a
  d  c
 , c  a  b  d
b
x
a
a
b
x
f 



1
)
(
2
b
a 


12
)
( 2
2 a
b 


Continuous Uniform Distribution
3
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4. Example
You’re production manager of a
soft drink bottling company. You
believe that when a machine is set
to dispense 12 oz., it really
dispenses between 11.5 and 12.5
oz. inclusive. Suppose the
amount dispensed has a uniform
distribution. What is the
probability that less than 11.8 oz.
is dispensed?
SODA
4
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5. Solution
P(11.5  x  11.8) = (Base)*(Height)
= (11.8 – 11.5)*(1) = .30
11.5 12.5
f (x)
x
11.8
1.0
1
5
.
11
5
.
12
1


a
b 
1
5
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6. 6



















0
1
0
0
)
(
0
0
0
1
)
(
x
for
e
x
for
x
F
x
for
x
for
e
x
f
x
x



0 
f(x)
1
F(x)
1/
Exponential Distributiom
)
(x
f )
(x
F
Density function
Cumulative function
6
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7. Exponential Distribution
Mean:
Probability density function:
Standard Deviation:

 

 
0
1
)
( 


x
for
e
x
f
x


7
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8. Example
An electrical firm manufactures light bulbs that have a
length of life is approximately negative exponential
distribution with mean 1500 hr and density function is
What is the probability that a
given bulbs will last;
a. More than 2000 hours;
b. b. Less than 300 hours
c. Find the value of true mean to give the probability of
a given bulbs will last less than 10 hours at most 0.001.
x
-
e
)
( 


x
f
8
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9. Solution
Approximately negative exponential distribution with
mean 1500 hr and
x
-
e
)
( 


x
f


1
 2
2 1

 
Mean: Variance:
dx


2000
X
-
e 
 1500
2000

 e
a- More than 2000 hours; P(X>2000) =
=0.264
Or F(x) =
X
X
X
e
dx
dx
x
f 

 


 
 1
e
)
( x
-
0
0
P(X>2000)=1-P(X<2000)=1-F(2000)=1-(
X
e 


1 ) =0.264
9
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10. Solution
dx

300
0
X
-
e 

5
/
1

e
b. Less than 300 hours P(X<300) =
= 1- = 0.181
X
e 


1 1500
/
300
1 
e
Or P(X<300) = F(300) = P(X<300) = = = 0.181
c. Find the value of true mean to give the probability of a given bulbs will last
less than 10 hours at most 0.001.
001
.
0
)
10
( 
F

  X
e 
1 
10
1 
 e 001
.
0

999
.
0
10

 
e 999
.
0
ln
10 
 



 

 )
1(10
10
-
ln0.999/
10
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11. Normal Distribution
1. ‘Bell-shaped’ &
symmetrical
2. Mean, median, mode
are equal x
f(x )
Mean
Median
Mode
f (x) 
1
 2
e

1
2






x







2
where
µ = Mean of the normal random variable x
 = Standard deviation
11
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12. Normal Distribution
c d
x
f (x)
Probability is
area under
curve!
P(c  x  d)  f (x)
c
d
 dx?
f (x) 
1
 2
e

1
2






x







2
12
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13. Standard Normal Distribution
The standard normal distribution is a normal
distribution with µ = 0 and  = 1. A random variable
with a standard normal distribution, denoted by the
symbol z, is called a standard normal random variable.
z 
x  µ

13
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14. x
f(x)
Non-standard Normal
Distribution
Normal distributions differ by
mean & standard deviation.
Each distribution would
require its own table.
That’s an infinite
number of tables!
14
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15. Standardize the
Normal Distribution
Normal
Distribution
x


One table!
 = 0
 = 1
z
Standardized Normal
Distribution
z 
x  

15
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16. Standard Normal Distribution
Area from
–∞ to – a equal area from
a to +∞
X
 0
a
-a
16
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17. Standard Normal Distribution
x
 0
a
-a
Area from
– a to 0 equal area from
0 to a
17
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18. z
= 0
 = 1
1.96
Z .04 .05
1.8 .9671 .9678 .9686
.9738 .9744
2.0 .9793 .9798 .9803
2.1 .9838 .9842 .9846
Example:
.06
1.9 .9750
Standardized Normal
Probability Table (Portion)
Probabilities
.4750
Shaded area
exaggerated
Find the probability of P(0 < z < 1.96)
18
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19. Example
A certain type of storage battery lasts on the average 5
years, with a standard deviation of 10 years.
Assuming the battery lives are normally distributed,
find the Probability that a given battery will last
1. Between 5 and 6.2 years
2. Between 3.8 and 5years
3. Between 2.9 and 7.1years
4. More than 8 years
5. Between 7.1 and 8 years
6. More than 4 years
19
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20. z
= 0
 = 1
.12
Standardized Normal
Distribution
.0478
1-Between 5 and 6.2 years
P (5 < x < 6.2)
Normal
Distribution
x
= 5
 = 10
6.2
z 
x 


6.2  5
10
 .12
Solution
P( 0 < Z < 0.12) = P(Z< 0.12) –P(Z< 0)
= 0.5478 – 0.5000 = 0.0478 20
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21. 21
Standard Normal
Distribution

22. z
 = 0
 = 1
-.12
Standardized Normal
Distribution
2-Between 3.8 and 5years
P (3.8  x  5)
Normal
Distribution
x
 = 5
 = 10
3.8
.0478
z 
x 


3.8 5
10
 .12
P( -0.12 < Z < 0) = P(Z< 0.12) – P(Z< 0)
= 0.5478 – 0.5000 = 0.0478 22
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23. 0
 = 1
-.21 z
.21
Standardized Normal
Distribution
3-Between 2.9 and 7.1years
P (2.9  x  7.1)
5
 = 10
2.9 7.1 x
Normal
Distribution
.1664
.0832
.0832
z 
x 


2.9  5
10
 .21 z 
x 


7.1 5
10
 .21
P( -0.21 < Z < 0.21) = 2*[ P(Z< 0.21) – P(Z< 0)]
= 2*[ 0.5832 – 0.5000] = 0.1664 23
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24. 24
Standard Normal
Distribution

25. 4-More than 8 years
P (x  8)
x
 = 5
 = 10
8
Normal
Distribution
z
= 0 .30
Standardized Normal
Distribution

 = 1
.3821
1.0
.6179
z 
x 


8 5
10
 .30
P( Z > 0.3) = 1 -[ P(Z< 0.3)
= 1- 0.6179 = 0.3821 25
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26.  = 0
 = 1
.30 z
.21
Standardized Normal
Distribution
5-Between 7.1 and 8 years
P(7.1  X  8)
 = 5
 = 10
8
7.1 x
Normal
Distribution
.6179
.0347
.5832
z 
x 


7.1 5
10
 .21 z 
x 


8 5
10
 .30
P( 0.21 < Z < 0.30) = P(Z< 0.3) – P(Z< 0.21)
= [ 0.6179 – 0.5832] = 0.0347 26
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27. 27
Standard Normal
Distribution

28.  = 0
 = 1
z
-0.1
Standardized Normal
Distribution
6-More than 4 years
P( X  4)
 = 5
 = 10
8
7.1 x
Normal
Distribution
P( Z > -0.1) = P(Z< 0.1)
= 0.5398
1
.
0
10
5
4








x
z
.5389
28
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29. The Standard Normal Table:
P(–1.26  z  1.26)
z
 = 0
 = 1
–1.26
Standardized Normal Distribution
Shaded area exaggerated
.3962
1.26
.3962 P(–1.26 ≤ z ≤ 1.26)
= .3962 + .3962
= .7924
29
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30. 30
Standard Normal
Distribution

31. The Standard Normal Table:
P(z > 1.26)
z
 = 0
 = 1
Standardized Normal Distribution
1.26
P(z > 1.26)
= 1.0 – .8962
= .1038
.8962
1.0
31
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32. The Standard Normal Table:
P(–2.78  z  –2.00)
 = 1
 = 0
–2.78
z
–2.00
.9973
.9772
Standardized Normal Distribution
Shaded area exaggerated
P(–2.78 ≤ z ≤ –2.00)
= .9973 – .9772
= .0201
32
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2.0 2.78

33. The Standard Normal Table:
P(z > –2.13)
z
 = 0
 = 1
–2.13
Standardized Normal Distribution
Shaded area exaggerated
P(z > –2.13)
= P(z < 2.13)
= .9834
.9834
33
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2.13

34. Approximation
Binomial
Hypergeometric
Poisson Normal
N >> 50
P =
Np
N
__
q=1-p
=np
 =np
2 =npq 34
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35. Example
Among the 120 applicants for a job only 80 are actually qualified. If
5 of these applicants are randomly selected for an interview, find the
probability that only 2 of the 5 will be qualified for the job by using
1. The hyper geometric distribution
80 40
2 3
120
5
X=2, n=5, N=120 ,Np=80
f(2)= = 0.164
2.The binomial distribution as an approximation.
2 3
5 2 1
2 3 3
f(2) = — — = 0.165
As can be seen these results are close .
P=80/120 =2/3
35
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36. Example
If 0.8% of the fuses delivered to an arsenal are defective, what is the
probability that four fuses will defective in random sample of 400?
The binomial distribution.
4 396
400
4
p= 0.008
q=0.992
n=400
0.008 0.992
f(4) = = 0.1788
The Poisson distribution as an approximation.
 = n*p = 400*0.008 = 3.2
178
.
0
!
4
2
.
3
)
4
(
4
2
.
3



e
p
36
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37. Example
Only one personal computer per thousand is
found to be defective after assembly in a
manufacturing plant, and the defective PCs are
distributed randomly throughout the production
run.
(a) What is the probability that a shipment of 500
PCs includes no defective computer?
(b) What is the probability that a shipment of 100
PCs includes at least one defective computer?
37
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38. b- P=0.001 q=0.999 =100(0.001)=0.1
Solution
09516
.
0
!
0
)
1
.
0
(
1
)
1
.
0
1
(
0
1
.
0






e
X
P 
a- By the Poisson approximation of binomial probabilities,
P=0.001 q=0.999 =500(0.001)=0.5
6065
.
0
!
0
)
5
.
0
0
(
5
.
0





e
X
P 
38
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39. Example
A multiple-choice quiz has 200 questions, each with 4 possible
answers of which only 1 is correct. What is the probability that sheer
guesswork yields from 25 to 30 correct answers for the 80 of the 200
problems about which the student has no knowledge?
P = 0.25 q = 0.75 n = 80
P( 25<= X <= 30) = f(25)+ f(26) +f(27)+f(28)+f(29) +f(30)
=0.0433 + 0.03058 + 0.02039 + 0.0128 + 0.00769 + 0.00435
=0.11476
.0
0 25 27 29 30
80
x
26 28
24.5
30.5
24
31
30.5
24.5
39
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40. Binomial distribution
.0
.1
.2
.3
x
p(x)
Example
40
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41. Solution
1196
.
0
8770
.
0
9966
.
0
)
16
.
1
(
)
71
.
2
(
)
71
.
2
16
.
1
(








 z
P
z
P
Z
P
16
.
1
873
.
3
20
5
.
24
1 


z
71
.
2
873
.
3
20
5
.
30
2 


z
The Normal distribution as an approximation.
 = np = 80*0.25 = 20
2 = npq = 80*0.25*.75 = 15
P( 24.5< X < 305) = P( z1 < Z < z2 )
25.5
(25 + .5)
24.5
(25 – .5)
25
41
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42. Brain Storming
You work in Quality
Assurance for an investment
firm. A clerk enters 75
words per minute with 6
errors per hour. What is the
probability of 0 errors in a
255-word bond transaction?
42
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43. Solution
• 75 words/min = (75 words/min)(60 min/hr)
= 4500 words/hr
• 6 errors/hr = 6 errors/4500 words
= .00133 errors/word
• In a 255-word transaction (interval):
 = (.00133 errors/word )(255 words)
= .34 errors/255-word transaction
Finding 
43
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44. Solution
 
-
0 -.34
( )
!
.34
(0) .7118
0!
x
e
p x
x
e
p



 
44
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45. Brain Storming
You work in Quality Control for
GE. Light bulb life has a normal
distribution with  = 2000 hours
and  = 200 hours. What’s the
probability that a bulb will last
A. between 2000 and 2400
hours?
B. less than 1470 hours?
45
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46. Standardized Normal
Distribution
z
 = 0
 = 1
2.0
Solution
Normal
Distribution
x
 = 2000
 = 200
2400
.4772
z 
x 


2400  2000
200
 2.0
P(2000  x  2400)
46
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47. z
 = 0
 = 1
–2.65
Standardized Normal
Distribution
Solution
x
 = 2000
 = 200
1470
Normal
Distribution
.0040 .9960
1.0
z 
x 


1470  2000
200
 2.65
P(x  1470)
47
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2.65

48. Finding z-Values
for Known Probabilities
What is Z, given
P(z) = .1217?
Shaded area
exaggerated
z
 = 0
 = 1
?
.1217
Standardized Normal
Probability Table (Portion)
Z .00 0.2
0.0 .5000 .5040 .5080
0.1 .5398 .5438 .5478
0.2 .5793 .5832 .0871
.6179 .6255
.01
0.3 .6217
.31
48
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49. Finding x Values
for Known Probabilities
Normal Distribution
x
 = 5
 = 10
?
.1217
Standardized Normal Distribution
z
 = 0
 = 1
.31
.1217
8.1
x    z  5  .31
  10
 
49
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8.1

50. 50
50
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