Download Example: Mean of a Random Variable 5/90) The probability

Examples: Mathematical Expectation
Example: Mean of a Random Variable
5/90) The probability distribution of X, the number of imperfections
per 10 meters of a synthetic fabric in continuous rolls of uniform
width, was given as
x
0
1
f(x) 0.41 0.37
2
3
4
0.16 0.05 0.01
Find the average number of imperfections per 10 meters of this fabric.
Solution:
µ = E(X) = Σ x f(x)
= (0)(0.41) + (1)(0.37) + (2)(0.16) + (3)(0.05) + (4)(0.01)
= 0.88
11/90) A private pilot wishes to ensure his airplane for P50,000. The
insurance company estimate that a total loss may occur with
probability 0.002, a 50% loss with probability 0.01, and a 25% loss
with probability 0.1. Ignoring all other partial losses, what premium
should the insurance company charge each year to realize an average
profit of P500.
Solution:
X
F(x)
500,000
0.002
25,000
0.01
12,500
0.10
Let Y - the premium the insurance company has to charge the pilot.
Y - E(X) = P500
µ = E(X) = (50,000)(0.002) + (25,000)(0.01) + (12,500)(0.10)
= P1,600
Therefore, Y - 1,600 = 500; Y = P2,100
ENGSTAT Notes of AM Fillone
Examples: Mathematical Expectation
19/91) A large industrial firm purchases several new type writers at the end of
each year, the exact number depending on the frequency of repairs in the
previous year. Suppose that the number of typewriters, X, that are purchased
each year has the following probability distribution:
x
f(x)
0
1/10
1
3/10
2
2/5
3
1/5
If the cost of the desired model will remain fixed at P1,200 throughout this year
and a discount of 50X2 pesos is credited toward any purchase, how much can
this firm expect to spend on new typewriters at the end of this year?
Solution:
g(x) = cost of the desired model = 1,200X - 50X2
µg(X) = E[g(X)] = ∑ g(x) f(x)
= E[1,200X - 50X2] = ∑ (1,200x - 50x2)f(x)
= [1,200(0) - 50(0)2](1/10) + [1,200(1) - 50(1)2](3/10) +
[1,200(2) - 50(2)2](2/5) + [1,200(3) - 50(3)2](1/5)
= 0 + 3450/10 + 4400/5 + 3150/5
= P1,855
ENGSTAT Notes of AM Fillone
Examples: Mathematical Expectation
23/91) Suppose that X and Y have the following joint probability
function:
X
f(x,y)
2
4
1
0.10 0.15
y
3
0.20 0.30
5
0.10 0.15
2
(a) Find the expected value of g(X, Y) = XY .
(b) Find µX and µY.
(a)
E(XY) = ∑ ∑ XY2 f(x,y)
x
y
= 2(1)2f(2,1) + 4(1)2f(4,1) + 2(3)2f(2,2) + 4(3)2f(4,2)
+ 2(5)2f(2,3) + 4(5)2f(4,3)
= 2(0.1) + 4(0.15) + 18(0.2) + 36(0.3) + 50(0.1)
+ 100(0.15)
= 35.2
(b)
µX = ∑ ∑ x f(x,y)
x
y
= ∑ x g(x)
x
= 2(0.1) + 2(0.2) + 2(0.1) + 4(0.15) + 4(0.3) + 4(0.15)
= 3.20
µY = ∑ ∑ y f(x,y) = ∑ y h(y)
x
y
y
= 1(0.1) + 1(0.15) + 3(0.2) + 3(0.3) + 5(0.1) + 5(0.15)
= 3.00
ENGSTAT Notes of AM Fillone
Examples: Mathematical Expectation
7/99) The total no. of hours, in units of 100 hrs. that an engineering
firm runs its generator set over a period of one year is a random
variable X having the density function
 x,
0<x<1

f(x) =
 2 - x,
1≤ x ≤ 2

 0,
elsewhere,
Find the variance of X.
Solution:
2
∞
2
σ = E[(X - µ) ] = ∫ (x - µ)2 f(x) dx
-∞
1
-∞
2
µx = E(X) = ∫ x f(x) dx = ∫ x (x) dx + ∫ x (2 - x) dx
∞
0
1
3
2
3
1
2
= x /3 ] + x - x /3 ]
0
1
= (1/3 - 0) + [(22 - 23/3) - (12 - 13/3)] = 1
2
∞
2
1
2
2
σ = E[(X - µ) ] = ∫ (x-µ) f(x) dx =∫ (x-1) x dx + ∫ (x - 1)2 (2 - x) dx
-∞
2
0
1
1
2
2
= ∫ (x - 2x +1) x dx + ∫ (x2 - 2x + 1)(2 - x) dx
0
1
3
1
2
2
= ∫ (x - 2x + x) dx + ∫ (- x3 + 4x2 - 5x + 2) dx
0
1
4
3
2
1
4
3
2
= x /4 - 2x /3 + x /2] + [ -x /4 + 4x /3 - 5x /2 + 2x]
0
2
1
= [ 1/4 - 2/3 + 1/2 ] + {[-24/4 + 4(2)3/3 - 5(2)2/2 + 2(2)] [-14/4 + 4(1)3/3 - 5(1)2/2 + 2(1)]}
= 1/6
ENGSTAT Notes of AM Fillone
Examples: Mathematical Expectation
Example: Variance
3/99) The random variable X, representing the number of errors per
100 lines of software code, has the following probability distribution
X
f(x)
2
0.01
3
0.25
4
0.40
5
6
0.30 0.04
Using Theorem 4.2, find the variance of X.
Solution:
µX = E(X) = 2(.01) + 3(.25) + 4(.40) + 5(.30) + 6(.04) = 4.11
E(X2) = 4(.01) + 9(.25) + 16(.40) + 25(.30) + 36(.04) = 17.63
σ2 = E(X2) - µ2
= 17.63 - (4.11)2
= 0.7379
4/99) Suppose that the probabilities are 0.4, 0.3, 0.2, and 0.1,
respectively, that 0, 1, 2, or 3 power failures will hit a certain
subdivision in any given year. Find the mean and variance of the
random variable X representing the number of power failures hitting
this subdivision.
Solution:
x
f(x)
0
0.4
1
0.3
2
0.2
3
0.1
µ = E(X) = 0(0.4) + 1(0.3) + 2(0.2) + 3(0.1) = 1.0
σ2 = E[(X - µ)2] = ∑ (x - µ)2f(x)
x
= (0 - 1)2(0.4) + (1 - 1)2(0.3) + (2 - 1)2(0.2)+ (3 - 1)2(0.1)
= 0.4 + 0 + 0.2 + 0.4 = 1.0
ENGSTAT Notes of AM Fillone
Examples: Mathematical Expectation
Example: Variance
5/99) The dealer’s profit, in units of $1000, on a new automobile is a random
variable X having the density function given by
2(1 - x),
0<x<1
f(x) =

 0,
elsewhere,
Find the variance of X
a) using Definition 4.3, and
b) using Theorem 4.2.
Sol’n:
∞
2
2
a. σ = E[(X - µ) ] = ∫ (x - µ)2 f(x) dx
-∞
∞
1
1
2
3
µ = E(X) = ∫ x f(x) dx = ∫ x[2(1-x)]dx = 2 [x /2 – x /3]
-∞
0
0
= 2(1/2-1/3) = 2(1/6) = 1/3
1
1
2
2
σ = ∫ (x – 1/3) [2(1-x)] dx = 2 ∫ (x2 – 2x/3 + 1/9)(1-x)dx
0
0
1
= 2 ∫ (x2 –2x/3 + 1/9 –x3 +2x2/3 –x/9) dx
0
1
= 2 ∫ ( -x3 + 5x2/3 – 7x/9 + 1/9)dx
0
4
3
1
2
= 2[-x /4 + 5x /9 – 7x /18 + x/9] = 2[-1/4 + 5/9 – 7/18 + 1/9] = 2/36 = 1/18
0
b. Using Theorem 4.2
σ2 = E(X2) - µ2
µ = 1/3
1
2
E(X ) = ∫ x2 [2(1-x)] dx
0
1
= 2 [x3/3 – x4/4 ] = 2 (1/3 – 1/4) = 2/12 = 1/6
0
σ2 = E(X2) - µ2 = 1/6 – (1/3)2 = 1/6 – 1/9 = 1/18
ENGSTAT Notes of AM Fillone
Same.
Examples: Mathematical Expectation
Example: Covariance
13/100) Find the covariance of the random variables X and Y of
Exercise 13 on page 79.
σXY = E(XY) - µxµy
µX = E(X) = Σ x g(x) = Σ Σ x f(x,y)
x
x y
3
Σ
=
x g(x) = 1(0.1) + 2(0.35) + 3(0.55) = 2.45
x =1
µY = E(Y) = Σ y h(y) = Σ Σ y f(x,y)
y
x y
3
= Σ y h(y) = 1(0.2) + 2(0.5) + 3(0.3) = 2.10
y=1
3
3
E(XY) = Σ Σ xy f(x,y)
x=1 y=1
= 1(1)(.05) + 1(2)(.05) + 1(3)(.1)
+ 2(1)(.05) + 2(2)(.1) + 2(3)(.35)
+ 3(1)(0) + 3(2)(.2) + 3(3)(.1)
= .05 + .1 + .3 + .1 + .4 + 2.1 + 0 + 1.2 + .9 = 5.15
σXY = E(XY) - µXµY = 5.15 – 2.45(2.1)
= 0.005
ENGSTAT Notes of AM Fillone
Examples: Mathematical Expectation
14/100) Find the covariance of the random variables X and Y of Exercise
8 on page 79.
σXY = E(XY) - µXµY
µX = E(X) = ∫ x g(x) dx
g(x) = ∫ f(x,y) dy
50
= 7.653x10 ∫ (x2 + y2) dy
30
-7
50
-7
2
3
= 7.653x10 [x y + y /3]
30
g(x) = 7.653x10-7[ (50x2 + 503/3) – (30x2 + 303/3)]
g(x) = 7.653x10-7(20x2 + 32,666.667)
µX = E(X) = ∫ x g(x) dx
50
= ∫ x [7.653x10-7(20x2 + 32,666.667)] dx
30
50
-7
= 7.653x10 ∫ (20x3 + 32,666.667x) dx
30
50
-7
4
2
= 7.653x10 [5x + 16,333.333x ]
30
-7
4
= 7.653x10 [(5(50) + 16,333.33(50)2) – (5(30)4 + 16,333.33(30)2)]
µX = 40.816
Also, µY = 40.816
ENGSTAT Notes of AM Fillone
Examples: Mathematical Expectation
50 50
E(XY) = ∫ ∫ xy(7.653x10-7)(x2+y2) dx dy
30 30
50
50
4
2 3
= 7.653x10-7 ∫ (x y/4 + x y /2) ] dy
30
30
50
= 7.653x10-7 ∫ [(504y/4 + 502y3/2) – (304y/4 + 302y3/2)] dy
30
50
= 7.653x10-7 ∫ [(504y/4 + 502y3/2) – (304y/4 + 302y3/2)] dy
30
50
= 7.653x10-7 ∫ [1,360,000y + 800y3]dy
30
50
2
4
= 7.653x10-7 [680,000y + 200y ]
30
= 7.653x10-7[(680,000(50)2 + 200(50)4) – (680,000(30)2 + 200(30)4]
= 1665.2928
Therefore, σXY = E(XY) - µXµY
= 1665.2928 – 40.816(40.816)
= -0.6531
ENGSTAT Notes of AM Fillone
Examples: Mathematical Expectation
8/110) Suppose that X and Y are independent random variables
having the joint probability distribution
x
f(x,y)
1
3
5
y
2
0.10
0.20
0.10
4
0.15
0.30
0.15
Find
(a) E(2X – 3Y);
(b) E(XY).
Sol’n:
a. E(2X – 3Y) = 2E(X) – 3E(Y)
E(X) = Σ x g(x)
x
= 2(0.4) + 4(0.6) = 0.8 + 2.4 = 3.2
E(Y) = Σ y h(y)
y
= 1(0.25) + 3(0.5) + 5(0.25) = 0.25 + 1.5 + 1.25
= 3.0
therefore, E(2X – 3Y) = 2 E(X) – 3 E(Y)
= 2(3.2) – 3(3)
= -2.6
b. E(XY) = Σ Σ xy f(x,y)
x
y
= 1(2)(0.1) + 1(4)(0.15) + 3(2)(0.2) + 3(4)(0.3) + 5(2)(0.10)
+ 5(4)(0.15)
= 9.60
ENGSTAT Notes of AM Fillone
Examples: Mathematical Expectation
14/111) If X and Y are independent random variables with variances
σ2X = 5 and σ2Y = 3, find the variance of the random variable Z = -2X
+ 4Y –3.
σ2Z = σ2-2X + 4Y – 3
σ2Z = σ2-2X + 4Y
(by Theorem 4.9, Corollary 1)
= 4σ2X + 16σ2Y (by Theorem 4.10, Corollary 2)
σ2Z = 4(5) + 16(3)
= 68
15/110) Repeat Exercise 14 if X and Y are not independent and σXY =
1.
σ2-2X + 4Y –3 = σ2-2X + 4Y
= 4σ2X + 16σ2Y + 2(-2)(4)σXY
= 4(5) + 16(3) –16(1)
= 52
ENGSTAT Notes of AM Fillone
(by Theorem 4.10)
Examples: Mathematical Expectation
Example: Independent random variables
20/110) Suppose that X and Y are independent random variables with
probability densities
g(x) =
 8/x3,

0,
x>2
elsewhere
and
h(y) =
2y,

0,
0< y < 1
elsewhere.
Find the expected value of Z = XY.
Sol’n:
E(Z) = E(XY) = E(X)(E(Y)
∞
E(X) =
∫ x g(x) dx
-∞
∞
∞
3
=
∫ x(8/x ) dx = ∫ (8/x2)dx
2
2
∞
= 8[ -1/x] = - 8[1/∞ - 1/2] = 4
2
0
∞
1
E(Y) =
∫ y h(y) dy
=
∫ y(2y) dy
-∞
0
1
3
= 2[y /3] = 2[1/3 - 0] = 2/3
0
Therefore E(Z) = E(X)E(Y) = 4(2/3) = 8/3
ENGSTAT Notes of AM Fillone
Examples: Mathematical Expectation
Examples: Mean of a random variable
xx./89) A shipment of 7 television sets contains 2 defectives. A hotel makes a random
purchase of 3 of the sets. If X is the number of defective sets purchased by the hotel, find the
mean of X.
Soln:
2  5 
 x   3-x 
f(x) = --------------- ,
7
3
x = 0, 1, 2
2 5
0 3
10
2
f(0) = --------------- = ------ = -----7
35
7
3
2 5
1 2
20
4
f(1) = --------------- = ------ = -----7
35
7
3
2 5
2 1
5
1
f(x) = --------------- = ------ = -----7
35
7
3
µ = E(X) = (0)(2/7) + (1)(4/7) + (2)(1/7) = 6/7
2/89) The probability distribution of the discrete random variable x is
f(x) =  3   1 x 3 3-x
x4 4 ,
x = 0, 1, 2.
Find the mean of X.
Solution:
 3  1 0 3 3
f(0) =  0  4   4  = 1(1)(27/64) = 27/64
 3  1 1 3 2
f(1) =  1  4   4  = 3(1/4)(9/16) = 27/64
ENGSTAT Notes of AM Fillone
Examples: Mathematical Expectation
 3  1 2 3 1
f(2) =  2  4   4  = 3(1/16)(3/4) = 9/64
 3  1 3 3 0
f(3) =  3  4   4  = 1(1/64)(1) = 1/64
µ = E(X) = (0)(27/64) + (1)(27/64) + (2)(9/64) + (3)(1/64) = 48/64 = 3/4
5/90) The probability distribution of X, the number of imperfections per 10 meters of a
synthetic fabric in continuous rolls of uniform width, was given as
x
f(x)
0
0.41
1
0.37
2
0.16
3
0.05
4
0.01
Find the average number of imperfections per 10 meters of this fabric.
Solution:
µ = E(X) = (0)(0.41) + (1)(0.37) + (2)(0.16) + (3)(0.05) + (4)(0.01) = 0.88
11/90) A private pilot wishes to ensure his airplane for P50,000. The insurance company
estimate that a total loss may occur with probability 0.002, a 50% loss with probability 0.01,
and a 25% loss with probability 0.1. Ignoring all other partial losses, what premium should
the insurance company charge each year to realize an average profit of P500.
Solution:
x
f(x)
500,000
0.002
25,000
0.01
12,500
0.10
Let Y - the premium the insurance company has to charge the pilot.
Y - E(X) = P500
µ = E(X) = (50,000)(0.002) + (25,000)(0.01) + (12,500)(0.10) = P1,600
Therefore, Y - 1,600 = 500 , Y = P2,100
12/90) If a dealer’s profit, in units of P1,000, on a new automobile can be looked upon as a
random variable X having the density function
2(1 - x),
f(x) = 
 0,
0<x<1
elsewhere,
find the average profit per automobile.
ENGSTAT Notes of AM Fillone
Examples: Mathematical Expectation
Solution:
Average profit per automobile = E(X)(P1,000)
1
1
1
2
3
µ = E(X) = ∫ x f(x) dx = ∫ x[2(1 - x)] dx = x - 2x /3] = 1 - 2/3 = 1/3
0
0
0
Therefore, average profit per automobile = (1/3)(P1,000) = P 333.33
xx/86) Let X represent the outcome when a balanced die is tossed. Find µg(X) = 3X2 + 4.
Solution:
The probability distribution when a die is tossed is
x
f(x)
1
1/6
2
1/6
3
1/6
4
1/6
5
1/6
6
1/6
µg(X) = E[g(X)] = ∑ g(X)f(X) = [3(1)2 + 4](1/6) + [3(2)2 + 4](1/6) + [3(3)2+ 4](1/6)
x
+ [3(4)2 + 4](1/6) + [3(5)2 + 4](1/6) + [3(6)2 + 4](1/6)
2
σ
g(X)
= 7/6 + 16/6 + 31/6 + 52/6 + 79/6 + 112/6 = 49.5
= E{[g(X) - µg(X)] } = ∑[g(X) - µg(X)]2 f(x)
2
x
2
= [7 - 49.5] (1/6) + [16 - 49.5]2(1/6) + [31 - 49.5]2(1/6) + [52 - 49.5]2(1/6)
+ [79 - 29/5]2(1/6) + [112 - 49.5]2(1/6)
= 1,342.5
17/91) Let X be a random variable with the following probability distribution:
x
f(x)
-3
1/6
6
1/2
9
1/3
Find µg(X), where g(X) = (2X + 1)2.
Solution:
µg(X) = E[g(X)] = ∑ g(x) f(x) = [2(-3) + 1]2(1/6) + [2(6) + 1]2(1/2) + [2(9) + 1]2(1/3)
= 25/6 + 169/2 + 361/3
= 209
19/91) A large industrial firm purchases several new type writers at the end of each year, the
exact number depending on the frequency of repairs in the previous year. Suppose that the
number of typewriters, X, that are purchased each year has the following probability
distribution:
ENGSTAT Notes of AM Fillone
Examples: Mathematical Expectation
x
f(x)
0
1/10
1
3/10
2
2/5
3
1/5
If the cost of the desired model will remain fixed at P1,200 throughout this year and a
discount of 50X2 pesos is credited toward any purchase, how much can this firm expect to
spend on new typewriters at the end of this year?
Solution:
g(x) = cost of the desired model = 1,200X - 50X2
µg(X) = E[g(X)] = ∑ g(x) f(x)
= E[1,200X - 50X2] = ∑ (1,200x - 50x2)f(x)
= [1,200(0) - 50(0)2](1/10) + [1,200(1) - 50(1)2](3/10) + [1,200(2) - 50(2)2](2/5) +
[1,200(3) - 50(3)2](1/5)
= 0 + 3450/10 + 4400/5 + 3150/5
= P1,855
22/91) The hospital period, in days, for patients following treatment for a certain type of
kidney disorder is a random variable Y = X + 4 where X has the density function
32/(x + 4)3, x > 0
f(x) = 
0,
elsewhere,
Find the average number of days that a person is hospitalized following treatment for this
disorder.
Solution:
∞
µg(X) = E[g(X)] = ∫ g(x) f(x) dx
-∞
∞
∞
0
0
= ∫ (x + 4)[32/(x + 4)3] dx = ∫ [32/(x + 4)2] dx = - [32/(x + 4)]
∞
0
= -32/(∞ + 4) - (-32/8) = 0 + 8 = 8
23/91) Suppose that X and Y have the following joint probability function:
x
f(x,y)
y
ENGSTAT Notes of AM Fillone
1
3
5
2
0.10
0.20
0.10
4
0.15
0.30
0.15
Examples: Mathematical Expectation
(a) Find the expected value of g(X, Y) = XY2.
(c) Find µX and µY.
(a) E(XY) = ∑ ∑ XY2 f(x,y) = 2(1)2f(2,1) + 4(1)2f(4,1) + 2(3)2f(2,2) + 4(3)2f(4,2)
x
y
+ 2(5)2f(2,3) + 4(5)2f(4,3)
= 2(0.1) + 4(0.15) + 18(0.2) + 36(0.3) + 50(0.1)
+ 100(0.15)
= 35.2
(c) µX = ∑ ∑ x f(x,y)
x
= ∑ x g(x)
y
x
= 2(0.1) + 2(0.2) + 2(0.1) + 4(0.15) + 4(0.3) + 4(0.15)
= 3.20
µY = ∑ ∑ y f(x,y) = ∑ y h(y)
x
y
y
= 1(0.1) + 1(0.15) + 3(0.2) + 3(0.3) + 5(0.1) + 5(0.15)
= 3.00
Examples: Variance and covariance
2/99) Let X be a random variable with the following probability distribution:
x
f(x)
-2
0.3
3
0.3
5
0.5
Find the standard deviation of X.
Solution;
µ = E(X) = -2(0.3) + 3(0.2) + 5(0.5) = 2.5
σ2 = E[(X - µ)2] = ∑ (x - µ)2f(x) = (-2 - 2.5)2(0.3) + (3 - 2.5)2(0.2) + (5 - 2.5)2(0.5)
x
= 9.25
σ = 3.041
3/99) The random variable X, representing the number of errors per 100 lines of software
code, has the following probability distribution
X
f(x)
2
0.01
3
0.25
4
0.40
Using Theorem 4.2, find the variance of X.
Solution:
ENGSTAT Notes of AM Fillone
5
0.30
6
0.04
Examples: Mathematical Expectation
µX = E(X) = 2(.01) + 3(.25) + 4(.40) + 5(.30) + 6(.04) = 4.11
E(X2) = 4(.01) + 9(.25) + 16(.40) + 25(.30) + 36(.04) = 17.63
σ2 = E(X2) - µ2
= 17.63 - (4.11)2
= 0.7379
4/99) Suppose that the probabilities are 0.4, 0.3, 0.2, and 0.1, respectively, that 0, 1, 2, or 3
power failures will hit a certain subdivision in any given year. Find the mean and variance of
the random variable X representing the number of power failures hitting this subdivision.
Solution:
x
f(x)
0
0.4
1
0.3
2
0.2
3
0.1
µ = E(X) = 0(0.4) + 1(0.3) + 2(0.2) + 3(0.1) = 1.0
σ2 = E[(X - µ)2] = ∑ (x - µ)2f(x) = (0 - 1)2(0.4) + (1 - 1)2(0.3) + (2 - 1)2(0.2)
x
+ (3 - 1)2(0.1)
= 0.4 + 0 + 0.2 + 0.4 = 1.0
5/99) The dealer’s profit, in units of $1000, on a new automobile is a random variable X
having the density function given by
2(1 - x),
f(x) = 
 0,
0<x<1
elsewhere,
Find the variance of X
c) using Definition 4.3, and
d) using Theorem 4.2.
Sol’n:
∞
a. σ2 = E[(X - µ)2] = ∫ (x - µ)2 f(x) dx
-∞
∞
1
1
µ = E(X) = ∫ x f(x) dx = ∫ x[2(1-x)]dx = 2 [x2/2 – x3/3]
-∞
0
0
= 2(1/2-1/3) = 2(1/6) = 1/3
ENGSTAT Notes of AM Fillone
Examples: Mathematical Expectation
1
1
σ2 = ∫ (x – 1/3)2[2(1-x)] dx = 2 ∫ (x2 – 2x/3 + 1/9)(1-x)dx
0
0
1
= 2 ∫ (x2 –2x/3 + 1/9 –x3 +2x2/3 –x/9) dx
0
1
= 2 ∫ ( -x3 + 5x2/3 – 7x/9 + 1/9)dx
0
1
4
3
2
= 2[-x /4 + 5x /9 – 7x /18 + x/9]
0
= 2[-1/4 + 5/9 – 7/18 + 1/9] = 2/36
= 1/18
b. Using Theorem 4.2
σ2 = E(X2) - µ2
µ = 1/3
1
E(X2) = ∫ x2 [2(1-x)] dx
0
1
= 2 [x3/3 – x4/4 ] = 2 (1/3 – 1/4) = 2/12 = 1/6
0
σ2 = E(X2) - µ2
= 1/6 – (1/3)2 = 1/6 – 1/9 = 1/18
Same.
7/99) The total no. of hours, in units of 100 hrs. that an engineering firm runs its generator set
over a period of one year is a random variable X having the density function
 x,

f(x) =  2 - x,

 0,
Find the variance of X.
Solution:
ENGSTAT Notes of AM Fillone
0<x<1
1≤ x ≤ 2
elsewhere,
Examples: Mathematical Expectation
∞
σ2 = E[(X - µ)2] = ∫ (x - µ)2 f(x) dx
-∞
-∞
1
2
1
3
2
2
3
µx = E(X) = ∫ x f(x) dx = ∫ x (x) dx + ∫ x (2 - x) dx = x /3 ] + x - x /3 ]
∞
0
2
1
3
0
2
1
3
= (1/3 - 0) + [(2 - 2 /3) - (1 - 1 /3)] = 1
2
∞
2
1
2
2
σ = E[(X - µ) ] = ∫ (x - µ) f(x) dx = = ∫ (x - 1) x dx + ∫ (x - 1)2 (2 - x) dx
-∞
0
1
2
2
1
= ∫ (x2 - 2x +1) x dx + ∫ (x2 - 2x + 1)(2 - x) dx
0
1
1
2
= ∫ (x3 - 2x2 + x) dx + ∫ (- x3 + 4x2 - 5x + 2) dx
0
1
1
2
0
1
= x4/4 - 2x3/3 + x2/2] + [ -x4/4 + 4x3/3 - 5x2/2 + 2x]
= [ 1/4 - 2/3 + 1/2 ] + {[-24/4 + 4(2)3/3 - 5(2)2/2 + 2(2)] [-14/4 + 4(1)3/3 - 5(1)2/2 + 2(1)]}
= 1/6
11/99) The length of time, in minutes, for an airplane to wait for clearance to take off at a
certain airport is a random variable Y = 3X – 2, where X has the density function
 (1/4)e-x/4,
f(x) = 
 0,
x>0
elsewhere.
5/109) Let X be a random variable with the following probability distribution:
x
f(x)
-3
1/6
6
1/2
9
1/3
Find E(x) and E(x2) and then, using these values, evaluate E{(2x + 1)2].
E(x) = Σ xf(x) = -2(1/6) + 6(1/2) + 9(1/3) = -1/2 + 3 + 3
= 11/2
E(x2) = Σx2f(x) = (-3)2(1/6) + 62(1/2) + 92(1/3)
= 9/6 + 18 + 27
= 93/2
ENGSTAT Notes of AM Fillone
Examples: Mathematical Expectation
E[(2x+1)2] = E[4x2 + 4x + 1] = E[4x2] + E[4x] + E[1]
= 4E[x2] + 4E[x] + E[1]
= 4(93/2) + 4(11/2) + 1
= 186 + 22 +1 = 209
13/100) Find the covariance of the random variables X and Y of Exercise 13 on page 79.
σxy = E(XY) - µxµy
3 3
µx = E(X) = Σ x f(x) = Σ Σ x f(x,y)
x
x=1 y=1
3
= Σ x g(x) = 1(0.1) + 2(0.35) + 3(0.55) = 2.45
x =1
3
3
µy = E(Y) = Σ y h(y) = Σ
Σ y f(x,y)
y
x=1 y=1
3
= Σ y h(y) = 1(0.2) + 2(0.5) + 3(0.3) = 2.10
y=1
3
3
E(XY) = Σ
Σ xy f(x,y)
x=1 y=1
= 1(1)(.05) + 1(2)(.05) + 1(3)(.1)
+ 2(1)(.05) + 2(2)(.1) + 2(3)(.35)
+ 3(1)(0) + 3(2)(.2) + 3(3)(.1)
= .05 + .1 + .3 + .1 + .4 + 2.1 + 0 + 1.2 + .9 = 5.15
σxy = E(XY) - µxµy = 5.15 – 2.45(2.1)
= 0.005
14/100) Find the covariance of the random variables X and Y of Exercise 8 on page 79.
σXY = E(XY) - µXµY
µX = E(X) = ∫ x g(x) dx
g(x) = ∫ f(x,y) dy
50
= 7.653x10-7 ∫ (x2 + y2) dy
30
ENGSTAT Notes of AM Fillone
Examples: Mathematical Expectation
= 7.653x10-7 [x2y + y3/3]
50
30
g(x) = 7.653x10-7[ (50x2 + 503/3) – (30x2 + 303/3)]
g(x) = 7.653x10-7(20x2 + 32,666.667)
µX = E(X) = ∫ x g(x) dx
50
= ∫ x [7.653x10-7(20x2 + 32,666.667)] dx
30
50
= 7.653x10-7 ∫ (20x3 + 32,666.667x) dx
30
50
= 7.653x10-7 [5x4 + 16,333.333x2]
30
= 7.653x10-7 [(5(50)4 + 16,333.33(50)2) – (5(30)4 + 16,333.33(30)2)]
= 40.816
Also, µY = 40.816
50 50
E(XY) = ∫ ∫ xy(7.653x10-7)(x2+y2) dx dy
30 30
50
50
= 7.653x10-7 ∫ (x4y/4 + x2y3/2) ] dy
30
30
50
[(504y/4 + 502y3/2) – (304y/4 + 302y3/2)] dy
= 7.653x10-7 ∫
30
50
= 7.653x10-7 ∫
[(504y/4 + 502y3/2) – (304y/4 + 302y3/2)] dy
30
50
= 7.653x10-7 ∫ [1,360,000y + 800y3]dy
30
50
= 7.653x10-7 [680,000y2 + 200y4]
30
ENGSTAT Notes of AM Fillone
Examples: Mathematical Expectation
= 7.653x10-7[(680,000(50)2 + 200(50)4) – (680,000(30)2 + 200(30)4]
= 1665.2928
σXY = E(XY) - µXµY
= 1665.2928 – 40.816(40.816)
= -0.6531
7/110) If a random variable X is defined such that E[(X – 1)2] = 10, E[(X – 2)2] = 6, find µ
and σ2.
8/110) Suppose that X and Y are independent random variables having the joint probability
distribution
x
y
f(x,y)
1
3
5
Find
(c) E(2X – 3Y);
(d) E(XY).
Sol’n:
b. E(2X – 3Y) = 2E(X) – 3E(Y)
E(X) = Σ x g(x)
x
= 2(0.4) + 4(0.6)
= 0.8 + 2.4 = 3.2
E(Y) = Σ y h(y)
y
= 1(0.25) + 3(0.5) + 5(0.25)
= 0.25 + 1.5 + 1.25
= 3.0
ENGSTAT Notes of AM Fillone
2
0.10
0.20
0.10
4
0.15
0.30
0.15
Examples: Mathematical Expectation
therefore, E(2X – 3Y) = 2 E(X) – 3 E(Y)
= 2(3.2) – 3(3)
= -2.6
b. E(XY) = Σ Σ xy f(x,y)
x y
= 1(2)(0.1) + 1(4)(0.15) + 3(2)(0.2) + 3(4)(0.3) + 5(2)(0.10)
+ 5(4)(0.15)
= 9.60
14/111) If X and Y are independent random variables with variances σ2X = 5 and σ2Y = 3,
find the variance of the random variable Z = -2X + 4Y –3.
σ2Z = σ2-2X + 4Y – 3
σ2Z = σ2-2X + 4Y
(by Theorem 4.9, Corollary 1)
= 4σ2X + 16σ2Y (by Theorem 4.10, Corollary 2)
σ2Z = 4(5) + 16(3)
= 68
15/110) Repeat Exercise 14 if X and Y are not independent and σXY = 1.
σ2-2X + 4Y –3 = σ2-2X + 4Y
= 4σ2X + 16σ2Y + 2(-2)(4)σXY
= 4(5) + 16(3) –16(1)
= 52
4.9/121)(M/S) A panel of meteorological and civil engineers studying emergency plans for
Florida’s Gulf Coast in the event of a hurricane has estimated that it would take between 13
and 18 hours to evacuate people living in low-lying land with the probabilities shown in the
table.
Time to evacuate
(nearest hour)
13
14
15
16
17
18
ENGSTAT Notes of AM Fillone
Probability
.04
.25
.40
.18
.10
.03
Examples: Mathematical Expectation
a. Calculate the mean and standard deviation of the probability distribution of the
evacuation times.
b. Within what range would you expect the time to evacuate to fall?
c. Weather forecasters say they cannot accurately predict a hurricane landfall more than 14
hours in advance. If the Gulf Coast Civil Engineering Department waits until the 14-hour
warning before beginning evacuation, what is the probability that all residents of lowlying areas are evacuated safely (i.e. before the hurricane hits the Gulf Coast)?
a. µ = E(x) = Σ x f(x)
= 13(.04) + 14(.25) + 15(.40) + 16(.18) + 17(.10) + 18(.03)
= 15.14
σ2 = E[(x- µ)2]
= (13-15.14)2(.04) + (14-15.14)2(.25)+(15-15.14)2(.40) + (16-15.14)2(.18) +
(17-15.14)2(.10) + (18-15.14)2(.03)
= 1.2402
σ = √ σ2 = 1.1137
b. µ ± 2σ = 15.14 ± 2(1.11) ; 12.92 to 17.36
c. From table P(x =13) + P(x=14) = .04 + .25 = .29
4.10/122)(M/S) A company’s marketing and accounting departments have determined that if
the company markets its newly developed magnetron tube (the major component in
microwave ovens), the contribution of the new tube to the firm’s profit during the next 6
months is described by the probability distribution shown in the table. The company has
decided it should market the new line of magnetron tubes if the expected contribution to
profit for the next 6 months is over $10,000. Based on the probability distribution, will the
company market the new tube?
PROFIT CONTRIBUTION
-$5,000
$10,000
$5,000
µ = E(x) = Σ x f(x)
x
= -5000(.3) + 10,000(.4) + 30,000(.3)
= $11,500
ENGSTAT Notes of AM Fillone
P(Profit Contribution)
.3
.4
.3