Second Midterm Solution - Introduction to the Theory of Probability | MATH 431, Exams of Probability and Statistics

Material Type: Exam; Professor: Valko; Class: Introduction to the Theory of Probability; Subject: MATHEMATICS; University: University of Wisconsin - Madison; Term: Spring 2009;

Typology: Exams

2010/2011

Uploaded on 07/25/2011

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UW-Madison 431 - Introduction to Probability Theory Spring 2009
Second Midterm Solutions
1. (12 points) In each of the following cases identify the name of the distribution of the random
variable Xand determine the probability mass function.
(a) We have an urn with 5 red and 18 blue balls and we pick 4 balls without replacement. We
denote the number of red balls in the sample by X.
This is a hypergeometric random variable with parameters n= 4, N = 23 and m= 5. the
probability mass function is
P(X=k) = ¡5
k¢¡18
5k¢
¡23
4¢,0k4
(b) We have an urn with 5 red and 18 blue balls and we pick 4 balls with replacement. We
denote the number of red balls in the sample by X.
This is a binomial random variable with parameters n= 4, p =5
23 . The probability mass
function is
P(X=k) = µ4
kµ5
23kµ18
234k
,0k4
(c) We roll a fair die until we get a three or a four. Xdenotes the number of rolls needed.
This is a geometric random variable with parameter p=2
3. The probability mass function
is
P(X=k) = 2
3·1
3k1k1
2. (18 points) The discrete random variable Xcan take on the possible values 1,0,1,2 and it
has a probability mass function p(a) = 1
10 (a2+ 1).
(a) What is the value of the cumulative distribution function of Xat 0.3?
P(X0.3) = P(X=1) + P(X= 0) = 1
5+1
10 =3
10
(b) Compute the expectation of X.
E[X] =
2
X
i=1
p(a)a= (1)1
5+ 0 ·1
10 + 1 ·1
5+ 2 ·1
2= 1
(c) Compute the variance of X.
We first compute the second moment:
E[X2] =
2
X
i=1
a2p(a) = 1 ·1
5+ 0 ·1
10 + 1 ·1
5+ 4 ·1
2=12
5
which gives
V ar[X] = E[X2]E[X]2=7
5
1
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Second Midterm Solutions

  1. (12 points) In each of the following cases identify the name of the distribution of the random variable X and determine the probability mass function.

(a) We have an urn with 5 red and 18 blue balls and we pick 4 balls without replacement. We denote the number of red balls in the sample by X. This is a hypergeometric random variable with parameters n = 4, N = 23 and m = 5. the probability mass function is

P(X = k) =

k

5 −k

4

) , 0 ≤ k ≤ 4

(b) We have an urn with 5 red and 18 blue balls and we pick 4 balls with replacement. We denote the number of red balls in the sample by X. This is a binomial random variable with parameters n = 4, p = 235. The probability mass function is P(X = k) =

k

)k ( 18 23

) 4 −k , 0 ≤ k ≤ 4

(c) We roll a fair die until we get a three or a four. X denotes the number of rolls needed. This is a geometric random variable with parameter p = 23. The probability mass function is P(X = k) =^2 3

3 k−^1

k ≥ 1

  1. (18 points) The discrete random variable X can take on the possible values − 1 , 0 , 1 , 2 and it has a probability mass function p(a) = 101 (a^2 + 1).

(a) What is the value of the cumulative distribution function of X at 0.3?

P(X ≤ 0 .3) = P(X = −1) + P(X = 0) =

10 =^

(b) Compute the expectation of X.

E[X] =

∑^2

i=− 1

p(a)a = (−1)^1 5

(c) Compute the variance of X. We first compute the second moment:

E[X^2 ] =

∑^2

i=− 1

a^2 p(a) = 1 · 1 5

=^12

which gives V ar[X] = E[X^2 ] − E[X]^2 =^7 5

  1. (16 points) In Florida there are 2.4 shark attacks per year on average. Estimate the probability that there will be exactly one attack next year. (State your assumptions!) We count rare events which are essentially independent. That means we can use a Poisson random variable X to approximate the number of shark attacks in a year, and the parameter is λ = 2.4. Thus P(exactly one attack) = P(X = 1) = 2. 4 e−^2.^4
  2. (18 points) The random variable X has a probability density f (x) given by

f (x) =

{ C

x if 1^ ≤^ x^ ≤^3 0 otherwise Compute the expectation and the variance of X. We first compute C using the fact that

−∞ f^ (x)dx^ = 1. ∫ (^) ∞

−∞

f (x)dx =

1

C

x dx^ =^ C^ log 3 which gives C = (^) log 3^1. The expectation is E[X] =

−∞

xf (x)dx =

1

x C x

dx = 2C = 2 log 3

The second moment is

E[X^2 ] =

−∞

x^2 f (x)dx =

1

x^2

C

x dx^ =^

Cx^2 2

x= x=1 =^

log 3. Thus the variance is V ar[X] = E[X^2 ] − E[X]^2 = 4 log 3

(log 3)^2

  1. (18 points) We flip two fair coins 4800 times and count the number of times we got two heads. Use normal approximation to estimate the probability that this number is between 1176 and
    (Use the table on the next page for the distribution function of the standard normal random variable. The numbers are set up so that you do not need a calculator to compute the numerical answer. Do not worry about the continuity correction.) Let X be the number of times we got two heads. This is a binomial random variable with parameters n = 4800 and p = 12 × 12 = 14. To estimate P(1176 ≤ X ≤ 1236) we standardize X and use the normal approximation. Note that np = 1200 and np(1 − p) = 900 = 30^2.

P(1176 ≤ X ≤ 1236) = P( √^1176 −^ np np(1 − p)

√X^ −^ np np(1 − p)

√^1236 −^ np np(1 − p)

= P(

30 ≤^

√X^ −^ np np(1 − p)

= P(− 0. 8 ≤

√X^ −^ np np(1 − p)