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ECE302 Spring 2006 Exam 1 February 27, 2006     1

Name:   Solution   Score: /100

You must show ALL of your work for full credit.This exam is closed-book. One 8.5” by 11” page of notes is permitted. It should be turned in

with the exam.Calculators may be used.

1. The CDF of random variable  R   is

F R(r) =

0   r

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ECE302 Spring 2006 Exam 1 February 27, 2006     2

2. Let the random variable  C   represent the amount the telephone company charges for callsto directory assistance.   C   is a function of the number   R  of phone numbers requested.Given that

C  = $0.50   R ∈ {1, 2}$0.75   R = 3$1.00   R = 4

(6)

and

P R(r) =

0.5   R = 10.3   R = 20.1   R = 30.1   R = 40 otherwise

(7)

answer the following questions:

(a) What is the expected number of requests E [R]?

E [R] = 0.5(1) + 0.3(2) + 0.1(3) + 0.1(4) = 1.8 (8)

(b) What is the variance Var[R] of the number of requests?

Var [R] =   E [R2] − (E [R])2 (9)= 0.5(1)2 + 0.3(2)2 + 0.1(3)2 + 0.1(4)2 − (1.8)2 (10)= 4.20

−3.24 = 0.96 (11)

(c) For calls that request more than one number, what is the conditional PMF   P R|R>1of the number of numbers requested?

P [R|R > 1](r) =

3/5   r = 21/5   r = 31/5   r = 4

0 otherwise

(12)

(d) Given that a call requests more than one number, what is the expected number of requests E [R|R > 1]?

E [R|R > 1] = 35

(2) + 1

5(3) +

 1

5(4) = 2.6 (13)

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(e) What is the expected cost of a call to directory assistace?

E [C ] = 0.5(0.50) + 0.3(0.50) + 0.1(0.75) + 0.1(1) = 0.575 (14)

so the expected cost of a call to directory assistance is 57.5 cents.

(f) What is the standard deviation of a call to directory assistace?

σC  = 

Var [C ] = 

E [C 2] − (E [C ])2 (15)

E [C 2] = 0.5(0.50)2 + 0.3(0.50)2 + 0.1(0.75)2 + 0.1(1)2 (16)

= 1/8 + 3/40 + 9/160 + 1/10 = 57/160 (17)

(E [C ])2 = (1/4 + 3/20 + 3/40 + 1/10)2 = (23/40)2 = 529/1600 (18)

so

σC  =

 570/1600 − 529/1600 =

 41/1600 =

√ 41/40 (19)

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3. The random variable  N  has PMF

P N (n) =

  c (|n| + 1)   n ∈ {−2, −1, 0, 1, 2}0 otherwise

  (20)

(a) What is the value of the constant  c?

Over the range  S N  = {−2, −1, 0, 1, 2}   of the random variable  N ,  theprobabilities  P N (n)  must sum to 1, so

c[3 + 2 + 1 + 2 + 3] = 1 =⇒   c =   111

  (21)

(b) What is the probability that N  = 0?

P [N  = 0] =  1

11 (|0| + 1) =   1

11  (22)

(c) What is the probability that |N |   is less than 2?

P [|N | 

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ECE302 Spring 2006 Exam 1 February 27, 2006     5

4. You and a friend find a penny and a nickel. To decide who gets to keep them you flip thecoins. If a coin comes up heads you win it; tails your friend wins it.

(a) What is the sample space of the experiment?

Where   H   represents the nickel coming up heads,   h   representsthe penny coming up heads,   T   represents the nickel coming uptails, and   t   represents the nickel coming up tails, the samplespace is

{Hh,Ht,Th,Tt} .   (24)

(b) Let  S  be the amount of money you win. What is the PMF of  S ?

Under the default assumption that the coins are fair, the PMF is

P S (s) =

1/4   s = 01/4   s = 0.01

1/4   s = 0.051/4   s = 0.06

(25)

(c) What is the expected value of  S ?

E [S ] = 1

4(0) +

 1

4(0.01) +

 1

4(0.05) +

 1

4(0.06) =

 1

4(0.12) = 0.03 (26)

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ECE302 Spring 2006 Exam 1 February 27, 2006     6

5. Short Questions

(a) Suppose we are given that the PDF of the random variable X   is f X (x) = e−bx + c for

x ∈ [0, 1] and zero elsewhere. What two properties of PDF’s would you use in orderto determine the acceptable values for the parameters   b  and   c. (DON’T DO THE

MATH. Just tell me the two facts.)

  ∞−∞

f X (x) = 1 and   f X (x) ≥ 0   ∀x   (27)(b) A source wishes to transmit data packets to a receiver over a radio link. The receiver

uses error detection to identify packets that have been corrupted by radio noise. Whena packet is received error-free, the receiver sends an acknowledgment (ACK) back tothe source. When the receiver gets a packet with errors, a negative acknowledgment(NAK) message is sent back to the source. Each time the source receives a NAK, thepacket is retransmitted. We assume that each packet transmission is independentlycorrupted by errors with probability   q . Would you model the distribution of the

number of times the packet is transmitted by the source as Poisson (P X (x) below),Exponential (f X (x) below), or (c) Neither? (Circle the correct answer below andindicate the values of   x   for which the first option holds if you choose Poisson orExponential. If you choose neither, suggest an alternative.)

P X (x) =

  αxe−α

x!  x ∈

0 otherwise

f X (x) =

  λe−λx x ∈0 otherwise

NeitherThe point here was to realize that a random variable representing

the number of times something happens must be a discrete random

variable. I accepted either the Poisson or any other discrete

probability mass distribution that made sense. The ranges of the

values of   x   are   x ∈ {0, 1, 2, . . .}   for the Poisson distribution and   x ∈[0, ∞)   for the Exponential.

(c) What are the mean and the variance of a standard Gaussian (Normal) random vari-able?

A standard normal random variable has mean 0 and variance 1.

(d) How do you obtain a standard Gaussian (Normal) random variable from a nonstan-

dard one?Let   X   be the nonstandard Gaussian random variable,   µX    be its mean,and   σ2X    its variance. Then the random variable   Z  = (X −µX )/σX    isa standard normal random variable.

(e) The graph of the PDF of a Gaussian random variable   X   is symmetric about whatvalue?

the mean  µX 

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ECE302 Spring 2006 Exam 1 February 27, 2006     7

6. A cellular phone call may be placed by a pedestrian or a driver. As the phone changeslocation along with its owner, a call may requires no handoffs (event   H 0), one handoff (H 1) or more than one handoff (H 2) from base station to base station. We are given thatthe probability that a call comes from someone in a vehicle is 0.8. The probability that a

call from a vehicle will not require a handoff is 0.5; the probability that it will require onehandoff is 0.2. The probability that a call from a pedestrian will not require a handoff is0.4; the probability that it will require one handoff is 0.1. (Note that this is not a veryrealistic model.)

(a) What are the probabilities that calls from pedestrians, respectively drivers, will re-quire two handoffs?

I will use   P   to designate the pedestrian and   V    to denote the driverof the vehicle.

P P [H 2] = 1 − P P [H 0] − P P [H 1] = 1 − 0.4 − 0.1 = 0.5 (28)P V  [H 2] = 1 − P V   [H 0] − P V   [H 1] = 1 − 0.5 − 0.2 = 0.3 (29)

(b) What is the probability that a call (of unknown origin) will not require a handoff?

P [H 0] = P P [H 0|P ]P [P ] + P V  [H 0|V ]P [V ] = 0.5(0.8) + 0.4(0.2) = 0.48 (30)

(c) When there is no handoff, what is the probability that the caller is in a vehicle?

P [V |H 0] =  P [H 0|V ]P [V ]P [H 0]

  = 0.5(0.8)

0.48  =

 5

6  (31)

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7. Suppose that a digit is selected at random from the set  S   = {1, 2, 3, 4, 5, 6, 7}  where theprobability of selecting  n  is

P [n] =   0.2   n ∈ {1, 2, 3}

0.1

  n ∈ {4

,5

,6

,7}

  (32)

Consider the following events

A1   =   {odd numbers in  S }   (33)A2   =   {even numbers in  S }   (34)A3   =   {n > 2}   (35)A4   =   {2, 3, 5}   (36)A5   =   {n