3. Expected Value
μ
f(x
x
E(X)
x
range
)
f(x
x
)
E(X
x
range
2
2
)
f(x
x
)
E(X
x
range
K
K
)
2
x
range
2
f(x
)
-
(x
V(X)
)
(x
V(X) 2
2
2
)]
(
[
) x
E
E
Continuous RV
3
Dr. Yehya Mesalam
f(x)
thing)
(any
thing)
E(any
4. Expected Value
b
x
E
a
b
ax
E
)
(
.
)
(
2
2
2
2
2
.
)
(
a
a
σ
b
ax
V x
b
ax
If a and b are constants, then
4
Dr. Yehya Mesalam
2
2
2
2
2
.
)
(
a
a
σ
ax
V x
ax
If X and Y are independent random variables
b
a
σ
by
ax
V y
x
by
ax
2
2
2
2
2
.
.
)
(
)
(
)
(
)
( y
E
x
E
y
x
E
)
(
).
(
)
.
( y
E
x
E
y
x
E
5. Example
• A lot containing 7 components is sampled by a
quality inspector; the lot contains 4 good
components and 3 defective components. A
sample of 3 is taken by the inspector. Find the
expected value of the number of good
components in this sample, Then calculate the
variance
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5
6. Solution
• Let X represent the number of good components in
the sample
• The probability distribution of X is
3
7
3
3
4
)
(
x
x
x
f
D
G
3
4
G
3
2
1
0
{ } .
D
0
1
2
3
X =0, 1, 2, 3
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6
7. X 0 1 2 3 Sum
f(x) 1/35 12/35 18/35 4/35 1
x.f(x) 0 12/35 36/35 12/35 60/35
x2.f(x) 0 12/35 72/36 36/35 120/35
7
Solution
71
.
1
7
12
35
60
μ
f(x)
x
E(X)
x
3647
.
0
)
35
60
(
35
120
)]
(
[
) 2
2
x
E
E 2
2
(x
V(X)
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8. Example
• Let X be the random variable that denotes the
life in hours of a certain electronic device. The
probability density function is
Find the expected life of this type of device
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8
10. Example
• Suppose that the number of cars X that pass through a
car wash between 4:00 P.M. and 5:00 P.M. on any
sunny Friday has the following probability
distribution:
10
Let g(X) = 2X−1 represent the amount of money, in
dollars, paid to the attendant by the manager. Find the
attendant’s expected earnings for this particular time
period.
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12. Summary Measures
1. Expected Value (Mean of probability distribution)
• Weighted average of all possible values
• = E(x) = x p(x)
2. Variance
• Weighted average of squared deviation about
mean
• 2 = E[(x 2(x 2p(x)
3. Standard Deviation
2
●
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12
13. Summary Calculation Table
x p(x) x p(x) x –
Total (x 2p(x)
(x – 2 (x – 2p(x)
xp(x)
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13
14. Expected Value & Variance
Solution
0 .25 –1.00 1.00
1 .50 0 0
2 .25 1.00 1.00
0
.50
.50
= 1.0
x p(x) x p(x) x – (x – 2 (x – 2p(x)
.25
0
.25
2.50
.71
2 E[(x 2
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14
15. Expected Value & Variance
Solution
0 .25 –1.00 1.00
1 .50 0 0
2 .25 1.00 1.00
0
.50
.50
= 1.0
x p(x) x p(x) x – (x – 2 (x 2p(x)
0
.5
1
E(x)21.50
.71
21.51.00.5
2 E(x 2) [E(x2
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15
16. Joint Probability function
16
Let X and Y be random variables with joint probability
distribution f(x, y). The mean, or expected value, of the
random variable g(X, Y ) is
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17. Example
17
Let X and Y be the random variables with joint probability
distribution indicated in Table on slid 25from chapter 3.
Find the expected value of g(X, Y ) = XY . The table is
reprinted here for convenience.
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22. Remarks
b
x
E
a
b
ax
E
)
(
.
)
(
2
2
2
2
2
.
)
(
a
a
σ
b
ax
V x
b
ax
If a and b are constants, then
22
Dr. Yehya Mesalam
2
2
2
2
2
.
)
(
a
a
σ
ax
V x
ax
If X and Y are independent random variables
b
a
σ
by
ax
V y
x
by
ax
2
2
2
2
2
.
.
)
(
)
(
)
(
)
( y
E
x
E
y
x
E
)
(
).
(
)
.
( y
E
x
E
y
x
E
![Expected Value
μ
P(x)
x
E(X)
x
x
2
2
P(x)
x
)
E(X
x
K
K
P(x)
x
)
E(X
f(x)
)
-
(x
V(X) 2
x
2
(x
V(X) 2
2
2
)]
(
[
) x
E
E
Discrete RV
2
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x
P(x)
thing)
(any
thing)
E(any](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)