### Department of Mathematics & Statistics STATISTICS   3083 (with solutions)

Final Examination 1997 April 17

### Do questions I and II and any TWO of III through VI

 name student #
QUESTION I   ((15))   (Identify distributions)
Supply the names of the distributions specified.
1. At a party attended by 20 people (10 women, 10 men), a randomly chosen three (different) people get a door prize. What is the distribution of the number of prizes won by women?
__ HYPERGEOMETRIC __

2. Five bowls each hold two tokens, one token marked 0 and one token marked 1. One token is drawn at random from each bowl and the five numbers added. What is the distribution of the sum?
__ BINOMIAL __

3. Your club meets once a week. Each week all 20 members attend. A randomly chosen member gets a door prize. Last week you got the door prize. What is the distribution of the number of weeks until you win again?
__ GEOMETRIC (or NEG BINOMIAL)__

4. What is the distribution usually applied to the number of people logging into a large computer network during a five second period?
__ POISSON __

5. A multiple choice test has 20 questions; each question has 5 choices. What is the distribution of the number of correct answers obtained by a person guessing at random?
__ BINOMIAL __

6. What is the distribution of D, a single digit produced by a random digit generator?
__ UNIFORM __

7. In a list of 100 randomly generated digits, 45 happen to be even. What is the (conditional) distribution of the number of even digits in the first 50 digits of the entire list?
__ HYPERGEOMETRIC __

8. In triad taste test, the subject gets three specimens (of cheese, wine, tea, soft drink, ...). Two specimens are alike, one is different. The subject is to identify the different one. The subject is tested five times. If the subject really cannot tell and is randomly guessing what is the distribution of the number of correct identifications?
__ BINOMIAL __

QUESTION II   ((35))   (Two dimensions)

There are three bowls, each with three numbered tokens:
bowl 1 contains tokens numbered:   2, 2, and 3
bowl 2 contains tokens numbered:   2, 2, and 3
bowl 3 contains tokens numbered:   2, 3, and 3
A random token is taken from bowl 1, call its value X.
If X=2 a second token is drawn at random from bowl 2.
If X=3 a second token is drawn at random from bowl 3.
Call the value of the second token, Y.
1. ```((10)) Complete this table of the joint distribution of X and Y.

Y =        2              3            sum
X =  2   |      4/9       |                |   2/3             at the
+----------------+----------------+----------         bottom
3   |                |                |   1/3             of the
+----------------+----------------+----------         page
sum   |                |                |```
2. ```((5)) Calculate Pr[ X=2 | Y=2 ]                          (4/9)/(5/9) = 4/5
((5)) Evaluate the expectation of the product, E{ X * Y }            52/9
((5)) Evaluate the expectation of the sum, E{ X + Y }                43/9
((3)) Evaluate the probability distribution of Y-2 Y-2         0         1
P[Y-2]     5/9       4/9
((2)) What is the name of the distribution of Y-2?              Bernoulli
or  Binomial
((1)) Are X and Y statistically independent?                           NO
((4)) Explain
Various answers are possible, among them
(a) The body of the table is not the product of the
margins, for example,(2/3)×(5/9) does not equal 16/9
(b) The description of the process makes it clear that the
conditional distribution of Y given X depends on X.```

 Y = 2 Solution for (1) and worksheet for expectations 4/9 XY=4     Prob×(XY)   = 16/9 X+Y=4   Prob×(X+Y) = 16/9 2/9 XY=6     Prob×(XY)   = 12/9 X+Y=5   Prob×(X+Y) = 10/9 1/9 XY=6     Prob×(XY)   = 6/9 X+Y=5   Prob×(X+Y) = 5/9 2/9 XY=9     Prob×(XY)   = 18/9 X+Y=6   Prob×(X+Y) = 12/9

### Do any TWO of questions III through VI

QUESTION III   ((15))   (Simple Expecation)
A random digit generator generates four (decimal) digits. Let X be the number of these digits which are even other than zero (i.e., one of 2, 4, 6, or 8).
Evaluate   E{ X (4-X) } __________________

 ```The distribution of X is binomial, n=4, p=4/10 First Solution X 0 1 2 3 4 Sum   P[X] 0.1296 0.3456 0.3456 0.1536 0.0256 1 X*(4-X) 0 3 4 3 0 P[X]*X*(4-X) 0 1.0368 1.3824 0.4608 0 2.88 Second Solution E{ X*(4-X) } = 4E{X} - E{X2} = 4E{X} - (V{X} + E{X}2) = 6.4 - (0.96 + 2.56)```

QUESTION IV   ((15))   (Continuous distribution)

```1)    f(x)  =  C (1-x*x)      for -1 < x < 1
=  0              elsewhere

Calculate the value of C which makes f(x) a
continuous probability density function.                   C    =  3/4
Calculate the expectation.                                 E{X} =    0

Comment: The distribution being symmetric, it's not
necessary to take time to evaluate the integral.

2) Calculate expectation and median of the distribution whose
probability density is f(x-5),  f as in part (1).
E{X}  =    5
Median =   5
Comment: f(x-5) = (3/4)[ 1 - (x-5)2].  The distribution
has been moved 5 units to the right.  Of course the
expectation and median move 5 units with it.```

QUESTION V   ((15))   (Text table 3.8 and variance)

```Suppose X, Y, and Z are independently distributed:
X has a binomial distribution,    N=12,   p=1/2
Y has a binomial distribution,    N=18,   p=1/3
Z has a geometric distribution,   p=1/2
Define S  as  S = X+Y+Z.
Evaluate these variances,
(Hint: Use text table 3.8 as an aid.)
V{S} =   _____________________
V{S-Z} = _____________________
V{S+Z} = _____________________```
 ``` V{X} = 12 (1/2) (1/2) = 3 V{Y} = 18 (1/3) (2/3) = 4 V{Z} = (1/2) / (1/2)2 = 2 V{S} = V{X}+V{Y}+V{Z} = 3 + 4 + 2 V{S-Z} = V{X+Y} = V{X}+V{Y} = 3 + 4 V{S+Z} = V{X+Y+2Z} = V{X}+V{Y}+4V{Z} = 3 + 4 + 8 Comment: Don't forget that S and Z are not independent```

QUESTION VI   ((15))   (Conditional probability)

Suppose in a mass medical screening --
A fraction, F, of the population has the ailment under test,
When applied to those with the ailment, the test is negative with probability 0.10.
When applied to those without the ailment, the test is positive with probability 0.10
Express the conditional probability a person testing positive has the ailment, i.e   Pr[sick | positive test],   in terms of   F.

(Note: A solution for F in general is expected, that is, an algebraic expression containing F), NOT a numeric answer for a particular value of F such as 0.10 or 0.03, etc. Half Credit will be awarded for evaluating the probability for a particular value of F such as F=0.1.)
 ``` sick not sick | sum test + 0.9 F 0.1 (1-F) | 0.1+0.8F - 0.1 F 0.9 (1-F) | 0.9-0.8F -------------------------------+----------- sum F 1-F | 1``` The conditional probability is (0.9F)/(0.1+0.8F) Comment: Check answers when possible: In this case, from the posted solution of the second assignment we know that if F=0.1 the conditional probability is one half.   Substituting F=0.1 into the formula indeed results in 0.09/0.18 = 1/2