Assignment 4
ENGR371
Fall 2010
Problem 1:
An engineering statistics class has 40 students and 60% are electrical engineering majors, 10%
are industrial engineering majors, and 30% are mechanical engineering majors. A sample of four
students is selected randomly, without replacement, for a project team. Let X and Y denote the
number of industrial engineering and mechanical engineering majors, respectively. Determine
the following:
(a) f XY ( x, y ) (b) f X (x) (c) E ( X ) (d) f Y 3 ( y) (e) E (Y X = 3) (f) V (Y X = 3)
(g) Are X and Y independent?
Solution 1:
1
Applied Statistics and Probability for Engineers, 5th edition
18 January 2010
Assignment 4
ENGR371
Fall 2010
0.222
Problem 2:
Determine the value of c that makes the function f ( x, y ) = cxy a joint probability density
function over the range 0 < x < 3 and 0 < y < x .
Determine the following:
a)
b)
c)
d)
e)
f)
g)
h)
P ( X < 1, Y < 2)
P (1 < X < 2)
P (Y > 1)
P ( X < 2, Y < 2)
E( X )
E (Y )
Marginal probability distribution of X.
Conditional probability distribution of Y given X=1
2
Applied Statistics and Probability for Engineers, 5th edition
18 January 2010
Assignment 4
ENGR371
i)
E (Y X = 1)
j)
P(Y > 2 X = 1)
Fall 2010
Solution 2:
3 x
x
3
3
y2
x3
x 4 81
c ∫ ∫ xydyd x = c ∫ x
dx = c ∫
dx
= c. Therefore, c = 8/81
2 0
2
8
8
0 0
0
0
1 x
1
8
8 x3
8 ⎛1⎞ 1
xydyd x = ∫ dx = ⎜ ⎟ = .
a) P(X<1,Y<2)=
∫
∫
81 0 0
81 0 2
81 ⎝ 8 ⎠ 81
2 x
2
4
8
8
x2
⎛ 8 ⎞x
b) P(1<X<2) =
xydyd x = ∫ x dx = ⎜ ⎟
81 ∫1 ∫0
81 1 2
⎝ 81 ⎠ 8
2
4
⎛ 8 ⎞ (2 − 1) 5
=⎜ ⎟
=
.
27
⎝ 81 ⎠ 8
1
c)
3
3
3 x
3
3
8
8 ⎛ x2 −1⎞
8 x
8 ⎛ x4 x2 ⎞
x
⎟⎟d x = ∫
− dx = ⎜⎜ − ⎟⎟
P (Y > 1) = ∫ ∫ xydyd x = ∫ x⎜⎜
81 1 1
81 1 ⎝ 2 ⎠
81 1 2
2
81 ⎝ 8
4 ⎠1
⎞ ⎛ 14 12 ⎞⎤ 64
⎟⎟ − ⎜⎜ − ⎟⎟⎥ =
= 0.7901
⎠ ⎝ 8 4 ⎠⎦ 81
2 x
2
8
8 x3
8 ⎛ 2 4 ⎞ 16
⎜ ⎟= .
xydyd
x
=
dx
=
d) P(X<2,Y<2) =
81 ∫0 ∫0
81 ∫0 2
81 ⎜⎝ 8 ⎟⎠ 81
=
8 ⎡⎛ 3 4 3 2
⎢⎜ −
81 ⎣⎜⎝ 8
4
e)
3 x
3 x
3
3 x
3
3
8
8
8 x2 2
8 x4
2
(
)
x
xy
dyd
x
x
ydyd
x
x
d
x
dx
=
=
=
81 ∫0 ∫0
81 ∫0 ∫0
81 ∫0 2
81 ∫0 2
E( X ) =
5
⎛ 8 ⎞⎛ 3 ⎞ 12
= ⎜ ⎟⎜⎜ ⎟⎟ =
⎝ 81 ⎠⎝ 10 ⎠ 5
f)
3 x
E (Y ) =
x3
8
8
8
2
y
xy
dyd
x
xy
dyd
x
x
dx
=
=
(
)
81 ∫0 ∫0
81 ∫0 3
81 ∫0 ∫0
3
5
8 x4
⎛ 8 ⎞⎛ 3 ⎞ 8
⎜
dx =⎜ ⎟⎜ ⎟⎟ =
= ∫
81 0 3
⎝ 81 ⎠⎝ 15 ⎠ 5
8
4x 3
xydy
=
81 ∫0
81
x
g)
f ( x) =
0< x<3
3
Applied Statistics and Probability for Engineers, 5th edition
18 January 2010
Assignment 4
ENGR371
8
(1) y
f (1, y ) 81
h) f Y | x =1 ( y ) =
=
= 2y
f (1)
4(1) 3
81
Fall 2010
0 < y <1
1
1
2 3
2
i) E (Y | X = 1) = ∫ y * 2 ydy = y
=
3 0 3
0
j) P (Y > 2 | X = 1) = 0 this isn’t possible since the values of y are 0< y < x.
Problem 3:
Determine the covariance and correlation for the following joint probability distribution:
− 1 − 0.5 0.5
x
−2
y
f XY ( x, y ) 1 / 8
−1
1/ 4
1
1
2
1/ 2 1/ 8
Solution 3:
E ( X ) = −1(1 / 8) + ( −0.5)(1 / 4) + 0.5(1 / 2) + 1(1 / 8) = 0.125
E (Y ) = −2(1 / 8) + (−1)(1 / 4) + 1(1 / 2) + 2(1 / 8) = 0.25
E(XY) = [-1× −2 × (1/8)] + [-0.5 × -1× (1/4)] + [0.5 × 1 × (1/2)] + [1 × 2 × (1/8)] = 0.875
V(X) = 0.4219
V(Y) = 1.6875
σ XY = 0.875 − (0.125)(0.25) = 0.8438
ρ XY = σσ σ =
0.8438
XY
X
Y
0.4219 1.6875
=1
Problem 4:
Detrmine the covariance and correlation for the joint probability density function
f XY ( x, y ) = e − x − y over the range 0 < x and 0 < y .
Solution 4:
E(X) = 1 E(Y) = 1
∞∞
E ( XY ) = ∫ ∫ xye − x − y dxdy
0 0
∞
∞
0
0
= ∫ xe − x dx ∫ ye − y dy
= E ( X ) E (Y )
Therefore, σ XY = ρ XY = 0 .
4
Applied Statistics and Probability for Engineers, 5th edition
18 January 2010
Assignment 4
ENGR371
Fall 2010
Problem 5:
Test results from an electronic circuit board indicate that 50% of board failures are caused by
assembly defects, 30% are due to electrical components, and 20% are due to mechanical defects.
Suppose that 10 boards fail independently. Let the random variable X, Y, and Z denote the
number of assembly, electrical, and mechanical defects among the 10 boards.
Calculate following:
a) P ( X = 5, Y = 3, Z = 2)
b) P ( X = 8)
c) P( X = 8 Y = 1)
d) P( X ≥ 8 Y = 1)
e) P( X = 7, Y = 1 Z = 2)
Solution 5:
a) board failures caused by assembly defects = p1 = 0.5
board failures caused by electrical components = p2 = 0.3
board failures caused by mechanical defects = p3 = 0.2
P ( X = 5, Y = 3, Z = 2) =
10!
0.550.330.2 2 = 0.0851
5!3!2!
8
2
P( X = 8) = (10
8 )0.5 0.5 = 0.0439
P( X = 8, Y = 1)
c) P ( X = 8 | Y = 1) =
. Now, because x+y+z = 10,
P(Y = 1)
10!
P(X=8, Y=1) = P(X=8, Y=1, Z=1) =
0.58 0.310.21 = 0.0211
8!1!1!
10
1
9
P(Y = 1) = 1 0.3 0.7 = 0.1211
P ( X = 8, Y = 1) 0.0211
P ( X = 8 | Y = 1) =
=
= 0.1742
P (Y = 1)
0.1211
P( X = 8, Y = 1) P( X = 9, Y = 1)
d) P ( X ≥ 8 | Y = 1) =
+
. Now, because x+y+z = 10,
P(Y = 1)
P(Y = 1)
10!
P(X=8, Y=1) = P(X=8, Y=1, Z=1) =
0.58 0.310.21 = 0.0211
8!1!1!
10!
P(X=9, Y=1) = P(X=9, Y=1, Z=0) =
0.59 0.310.2 0 = 0.0059
9!1!0!
10
1
9
P(Y = 1) = 1 0.3 0.7 = 0.1211
P( X = 8, Y = 1) P( X = 9, Y = 1) 0.0211 0.0059
P( X ≥ 8 | Y = 1) =
+
=
+
= 0.2230
P (Y = 1)
P(Y = 1)
0.1211 0.1211
b) Because X is binomial,
( )
( )
5
Applied Statistics and Probability for Engineers, 5th edition
18 January 2010
Assignment 4
ENGR371
e) P ( X = 7, Y = 1 | Z = 2) =
Fall 2010
P ( X = 7 , Y = 1, Z = 2)
P ( Z = 2)
10!
0.57 0.310.2 2 = 0.0338
7!1!2!
10
2
P( Z = 2) = 2 0.2 0.88 = 0.3020
P( X = 7, Y = 1, Z = 2) 0.0338
P ( X = 7, Y = 1 | Z = 2) =
=
= 0.1119
P ( Z = 2)
0.3020
P(X=7, Y=1, Z=2) =
( )
Problem 6:
In an acid-base titration, a base or acid is gradually added to the other until they have completely
neutralized each other. Let X and Y denote the milliliters of acid and base needed for
equivalence, respectively. Assume X and Y have a bivariate normal distribution with σ X = 5mL,
σ Y = 2mL, μ X = 120mL, μY = 100mL, and ρ = 0.6 .
Determine the following:
a) Covariance of X and Y.
b) Marginal probability distribution of X.
c) P ( X < 116)
d) Conditional probability distribution of X given that Y=102
e) P( X < 116 Y = 102)
Solution 6:
(a) ρ = cov(X,Y)/σxσy =0.6 cov(X,Y)= 0.6*2*5=6
(b) The marginal probability distribution of X is normal with mean µx , σx.
(c) P(X < 116) =P(X-120 < -4)=P((X_120)/5 < -0.8)=P(Z < -0.8) = 0.21
(d) The conditional probability distribution of X given Y=102 is bivariate normal distribution with mean
and variance
µX|y=102 = 120 – 100*0.6*(5/2) +(5/2)*0.6(102) = 123
σ2X|y=102 = 25(1-0.36) =16
(e) P(X < 116|Y=102)=P(Z < (116-123)/4)=0.040
Problem 7:
X and Y are independent, normal random variables with E ( X ) = 2, V ( X ) = 5, E (Y ) = 6, and
V (Y ) = 8. Determine the following:
(a) E (3 X + 2Y )
(b) V (3 X + 2Y )
(c) P (3 X + 2Y < 18)
(d) P(3X+2Y<28)
6
Applied Statistics and Probability for Engineers, 5th edition
18 January 2010
Assignment 4
ENGR371
Fall 2010
Solution 7:
(a) E(3X+2Y) = 3*2+2*6=18
(b) V(3X+2Y) = 9*5+4*8 =77
(c) 3X+2Y ~ N(18, 77)
P(3X+2Y < 18) = P(Z < (18-18)/770.5)=0.5
(d) P(3X+2Y < 28) = P(Z < (28-18)/770.5)=P(Z < 1.1396) =0.873
Problem 8:
The velocity of a particle in a gas is a random variable V with probability distribution
f V (v) = av 2 e − bv ,
v>0
where b is a constant that depends on the temperature of the gas and the mass of the particle.
(a) Find the value of constant a.
(b) The kinetic energy of the particle is W = mV 2 / 2 . Find the probability distribution of W.
Solution 8:
∞
a) Now,
2 −bv
∫ av e dv
∞
must equal one. Let u = bv, then 1 = a
0
definition of the gamma function the last expression is
b) If w =
∞
2 − u du
2 −u
a
∫0 ( ub ) e b = b 3 ∫0 u e du. From the
a
2a
b3
Γ
(
3
)
=
.
Therefore,
.
a
=
2
b3
b3
mv 2
, then v =
2
f W ( w) = f V
=
for
w ≥ 0.
2w
for v ≥ 0, w ≥ 0 .
m
b 3 2w −b 2mw
2 w dv
(2mw) −1 / 2
=
e
m
2m
dw
( )
b 3 m −3 / 2
2
w1 / 2 e
−b
2w
m
7
Applied Statistics and Probability for Engineers, 5th edition
18 January 2010