Unformatted text preview: Homework 7.• Homework must be answered in the order shown here (else please make a note telling the reader where it is). • Write your answers neatly • Work must be shown for full credit • No late homework accepted under any circumstances • Staple the homework 1.
The joint density function of X and Y is given by Are X and Y independent? What if f(x, y) were given by 2.
Let X and Y have a joint density function f(x,y) = k xy 0 ≤x ≤1 0 ≤ y ≤1 The correlation between X and Y is (a) 1/4 (b) 0 (c) 1.4 (d) 9.64 Answer: (b) . Show why. 3.
The joint probability density function of X and Y is given by f ( x, y ) = e − ( x + y ) 0 ≤ x < ∞, 0 ≤ y < ∞ 1 Find (a) P{X < Y} (b) P{X < a}, where a is a constant. Your answer will depend on a. P( X < Y ) = ∫ [e
∞ (a) = 0 = 1− −y ∫ [∫
∞ y − x− y 0 o e − e−2 y ] dy = ∫ dx dy = ∞ −y
0 e dy − ∫ [e
∞ 0 1
2 ∫ −y ∞ −2 y
0 (−e− x ) y dy
0 e d2 y 11
=
22 (b) .
4.A soft drink machine has a random supply Y2 at the beginning of a given day and dispenses a
random amount Y1 during the day (with measurements in gallons). It is not resupplied during the
day, hence Y1 ≤ Y2. It has been observed that Y1 and Y2 have joint density ⎧y
f ( y1, y 2 ) = ⎨ 2
⎩0 0 ≤ y1 ≤ y 2 , 0 ≤ y 2 ≤ 2
elsewhere (i) The conditional expectation of the amount of sales of Y1 given that Y2 = 1 is:
(a) 0.7 (b) 1.2 (c) ½ (d) ¾ (ii) Find the conditional expectation of the amount of sales of Y1 given that Y2 = 1.6. Show work. ( ) E y1 y 2 = ∫ ( y2 0 ) 1 y12
y1 (1 / y 2 ) dy1 =
y2 2 y2 =
0 2
y 21 y 2
=
2 y2 2 E y1 y 2 = 1.6 = 1.6 / 2 = 0.8 2 5.Let X1 and X2 denote the proportion of time, out of one workweek, that employees I and II,
respectively ,actually spend performing their assigned tasks. The join relative frequency behavior
of X1 and X2 is modeled by the p.d.f.: ⎧ x + x2
0 ≤ x1 ≤ 1, 0 ≤ x 2 ≤ 1
f ( x1, x 2 ) = ⎨ 1
elsewhere
⎩0
a)Find the probability that the total proportion of time spent by both employees working during a
week is less than or equal to 1. That is, fin P[(X1 + X2) ≤ 1]
b) Find the Expected value of X1 1 1 1⎞
1
⎛
E ( x1 ) = ∫ x1 x1 + dx1 = ∫ ( x12 + x1 ) dx1 = 7 /12
⎝
2⎠
2
0
0
6.
A radioactive particle is randomly located in a square area with sides that are 1 unit length. Let X and Y denote the coordinates of the particle. Since the particle is equally likely to fall in any subarea of fixed size, a reasonable model for (X, Y) is given by f(x, y) = 1 0 ≤ x ≤1 0 ≤ y ≤ 1 Find the conditional probability that X is less that 0.3 given that Y is less than 0.5 Show work. 1 f ( x 2 ) = ∫ 1dx1 = 1 0 ≤ x2 ≤ 1 0 f ( x1 ) = 1 0 ≤ x1 ≤ 1
f ( x1, x 2 ) 1
f ( x1  x 2 ) =
= = 1 = f(x1 )
f ( x2 )
1 0 ≤ x1 ≤ 1
0.3 P ( X1 < 0.3  Y < 0.5) = P ( X1 < 0.3) = ∫ 1dx1 = 0.3
0 7.
The proportion X and Y of two chemicals found in samples of an insecticide have the joint probability density function f(x, y) = 2 0 ≤ x ≤ 1 0 ≤ y ≤ 1 0 ≤ x+ y ≤ 1 3 Find P(X≤ 1/2  Y ≤ 1/2) . Show work. ⎛
1
1⎞
P⎜ X ≤ ,Y ≤ ⎟
⎛
1
1⎞
1/ 2
⎝
2
2⎠
P⎜ X ≤  Y ≤ ⎟ =
= 1 / 21− y
= 2 / 3 ⎛
1⎞
⎝
2
2⎠
P⎜Y ≤ ⎟
∫ ∫ 2dxdy
⎝
2⎠
00 8.
A particular fast food restaurant is interested in the joint behavior of the random variable X, defined as the total time between a customer’s arrival at the store and his or her leaving the service window, and Y, the time that the customer waits in line before reaching the service window. Because X includes the time a customer waits in line, we must have X ≥ Y. The relative frequency distribution of observed values of X and Y can be modeled by the probability density function f ( x, y ) = e− x 0 ≤ y ≤ x ≤ ∞ The random variable X and Y represent the time spent at the service window. (a) Find the E(X
Y) f ( x) = x ∫e −x dy = xe− x 0 ∞ ∫xe E[ X 2 ] = 3 −x dx = 3!= 6 0 ∞ ∫xe E[ X ] = 2 −x dx = 2 0 Var( X ) = 6 − 2 2 = 2 f ( y) = ∞ ∫e −x dx = e− y y ∞ ∫ye E [Y 2 ] = 2 −y dy = 2 0 E [Y ] = ∞ ∫ ye −y dy = 1 0 Var(Y ) = 2 − 12 = 1
4 E(X
Y)=E(X)
E(Y)= 2 – 1 = 1 (b) Find the Var(X
Y) where Var means variance. 5 ( a) E [ X − Y ] = ∞∞ ∫ ∫ ( x − y )e −x dx dy 0y = ∞∞ ∫ ∫ xe −x − ye− x dx dy 0y = ∞ ∫ ye −y + e− y − ye− y dy 0 = ∞ ∫e −y 0 ∞ dy = −e− y 0 = 1 (b) Var( X − Y ) = Var( X ) + Var(Y ) − 2Cov ( X ,Y )
f ( x) = x ∫e −x dy = xe− x 0 ∞ ∫xe E[ X 2 ] = 3 −x dx = 3!= 6 0 ∞ ∫xe E[ X ] = 2 −x dx = 2 0 Var( X ) = 6 − 2 2 = 2
f ( y) = ∞ ∫e −x dx = e
−y y ∞ ∫ye E [Y ] =
2 2 −y dy = 2 0 E [Y ] = ∞ ∫ ye −y dy = 1 0 Var(Y ) = 2 − 12 = 1
E [ XY ] = ∞∞ ∫ ∫ xye −x dx dy 0y = ∞ ∫ye 2 −y + ye− y dy = 3 0 Cov ( X ,Y ) = E [ XY ] − E [ X ] E [Y ]
= 3 − (2)(1) = 1
⇒ Var( X − Y ) = 2 + 1 + 2(1)(−1)(1) = 1
6 9.Consider students at a university. Let X be their combined SAT scores (Math and Verbal), and Y their freshman GPA (out of 4). The random variables X and Y are bivariate
normally distributed. Suppose a study reveals that Show work. 10.
The joint density of X and Y is given by e−x / ye− y
f ( x, y ) =
, 0 < x < ∞, 0 < y < ∞ y Compute E[X2  Y=y]. x 1 −y − y
ee
x
1 −y
y
f X Y ( x  y ) =
=e
e− y
y
E ( X 2  Y = y) = ∫ ∞
0 ∞⎛ x ⎞
1
x
x 2 e dx = y 2 ∫ ⎜ ⎟ e d = 2 y 2
0y
y
y
⎝⎠
x
−
y 2 x
−
y 11.
The number of bacteria per cubic centimeter is Poisson distributed with parameter lambda equal to 5. Consider 50 independent cubic centimeters and denote Xi the number of bacteria in cubic centimeter i. (a) Write down the joint distribution of the 50 Xi’s. (b) Compute the probability that it will take 20 cubic centimeters to find 5 cubic centimeters with less than 2 bacteria each. 12.
Adam, Beth, Carlos, and Donna are running a onemile run.
Their completion times in minutes are uniformly distributed in the following intervals:
Adam’s time is A ~ Uniform( 8.3; 10.2)
Beth’s time is B~ U(8.9; 10.5)
Carlo’s time is C ~ U(8.1; 10.2)
Donna’s time is D ~ U(8.4; 10.7)
7 (a) Find the probability that the earliest completion time is less than 10 minutes
(b) Find the probability that the latest completion time is less than 10 minutes 8 ...
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This note was uploaded on 03/16/2012 for the course STAT 110a taught by Professor J.sanchez during the Winter '12 term at UCLA.
 Winter '12
 j.sanchez

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