1. A random variable X has the following cumulative distribution function
F(x) = { 1 - e^(-(x+1)) -1</ x < oo
0 elsewhere.
a) 25% of the time, X exceeds what value?
Solution. Since the cumulative distribution function is
1  e ( x 1) ,1  x  
F ( x)  P( X  x)  
0, otherwise
e ( x 1) ,1  x  
So, the density function is f ( x)  F ' ( x)  
0, otherwise
Now, we want to find x such that P( X  x)  1  F ( x)  0.25 , so F ( x)  0.75 .
Let F ( x)  0.75 = 1  e  ( x 1) , we have e  ( x 1)  0.25 . So
x   ln( 0.25)  1  0.3863
b) Find the moment generating function of X, or Mx(t)
Solution. By the definition, we have

M X (t )  E (e tX ) 

tx
tx  ( x 1)
t
 e f ( x)dx   e e dx  11t e , t  1

1
tX
Note: when t>=1 , the expectation E (e ) is not finite
c) Using your result in (b), find the E(X^2)
Solution. It is easy to know that E ( X 2 ) = M X
( 2)
(0) . Now we compute M X
( 2)
( 0)
M X ' (t )  e  t [ (11t ) 2  11 t ] and M X " (t )  e t [ (12t )3  (12t ) 2  11 t ]
So, M X
( 2)
(0) =1,
Thus, E ( X 2 ) = M X
( 2)
(0) =1.
2. The lifetime of a certain brand of tire, in 10's of thousands of kilometers (or 10,000
kms), is a random variable having the following probability density function.
f(x) = { 20/(3x^2) 4</ x </ 10
0 elsewhere.
a) find the cumulative distribution function of X, or
F(x)
Solution. By the definition, we have
10 20
  3t 2 dt , x  10;
4
x
 x
F ( x)  P( X  x)   f (t )dt   320t 2 dt ,4  x  10;

4
0, x  4


1, x  10;

=  53  320x ,4  x  10;
0, x  4.

b) How long would you expect one of these tires to last, in kilometers?
Solution. We can compute the expected value of X as follows.

E( X ) 
10
 xf ( x)dx   x

10
20
3x2
dx 

4
20
3x
dx 
20
3
ln 52  6.109 (10,000kms)
4
c) If a tire lasts for at least 65,000 kilometers, what is the probability it will last at
most 85,000 kilometers?
Solution. The probability that a tire lasts for at least 65,000 kilometers, denoted by

P(X>=65000)=
10
 f ( x)dx  
6.5
20
3x2
dx 0.359
6.5
The probability that a tire lasts for between 65000 and 85,000 kilometers,
denoted by P(65000<=X<=85000). We have
8.5
P(65000<=X<=85000)=
8.5
 f ( x)dx  
6.5
20
3x2
dx 0.2413
6.5
So, if a tire lasts for at least 65,000 kilometers, the probability it will last at most
85,000 kilometers is
P(X<=85000|X>=65000)=P(65000<=X<=85000)/ P(X>=65000)
=0.2413/0.359=0.6721
3. (refer to question 2). X1, X2, ...,Xk represent a random sample of k-tires
a) What is the probability that the 28th tire selected and the 5th tire to have a lifetime
exceeding 90,000 kms?
Solution. First, we need to compute that any given a tire , the probability p that it has
a lifetime exceeding 90,000kms.

p= P(X>=90000)=
10
 f ( x)dx  
9.5
9.5
20
3x2
dx  0.0351
So, if X1, X2, ...,Xk represent a random sample of k-tires, then the probability that
the 28th tire selected and the 5th tire to have a lifetime exceeding 90,000 kms is
p 2 (1  p) k 2  0.03512 * 0.9649 k 2
b) To ensure that there is a probability of 0.90 that at least one tire in the k sampled
will have a lifetime exceeding 90,000 kms, how many tires should be randomly
selected? (k=?)
Solution. Assume that the number of tires having a lifetime exceeding 90,000kms in
the k sampled is W. Then W is a random variable and follows binomial distribution,
that is, W~b(k,p), i.e, W~b(k,0.0351)
Now, to ensure that there is a probability of 0.90 that at least one tire in the k sampled
will have a lifetime exceeding 90,000 kms, ie., P(W  1)  0.9
We have P(W  0)  0.1
k 
But, P(W  0)   0.03510 * 0.9649 k  0.9649 k  0.1
0
So, k 
ln(0.1)
ln(0.9649)

2.3025809
0.0357308
 64.443
So, k min  65
4. The number of mice in a grain bin, at any given time, is a random variable which
follows a Poisson distribution with a mean of 4.6.
a) What is the probability that there is exactly 12 mice in two randomly selected grain
bins?
Solution. Denote the number of mice in a grain bin A, at any given time by X, then
X follows a Poisson distribution with a mean of 4.6. Denote the number of mice in a
grain bin B, at any given time by Y, then Y follows a Poisson distribution with a
mean of 4.6 So
4.6
P(X=k)= e k ! 4.6 k , k=0,1,2,…
4.6
P(Y=k)= e k ! 4.6 k , k=0,1,2,…
So,
P(X+Y=12)=P(X=0)P(Y=12)+P(X=1)P(Y=11)+P(X=2)P(Y=10)+P(X=3)P(Y=9)+
P(X=4)P(Y=8)+P(X=5)P(Y=7)+P(X=6)P(Y=6)+P(X=7)P(Y=5)+P(X=8)P(Y=4)+
P(X=9)P(Y=3)+P(X=10)P(Y=2)+P(X=11)P(Y=1)+P(X=12)P(Y=0)
= e 4.6 * e 4.6 4.612 / 12!(e 4.6 4.6)(e 4.6 4.611 / 11!)  ...  (e 4.6 4.612 / 12!)(e 4.6 )
= e 9.2 4.612 [2 / 12!2 / 11!1 / 10!2 /(3!9!)  2 /( 4!8!)  2 /(5!7!)  1 /(6!6!)]
b) What is the probability that 10 of the 50 randomly selected grain bins have exactly
six mice?
Solution. Denote the number of the grain bins in each of which there is exactly six
mice by Z. Then Z is a random variable which follows binomial distribution, that is,
Z~b(50,p). Where p is the probability that there is six mice in a grain bin.
We can compute p as follows. Since the number of mice in a grain bin, at any given
time, is a random variable which follows a Poisson distribution with a mean of 4.6
So, p= 4.6 6 e 4.6 / 6! 0.1323
So, the probability that 10 of the 50 randomly selected grain bins have exactly six
mice is
 50 
P(Z=10)=  0.132310 (1  0.1323) 40 =
10 
 50 
 0.132310 0.8677 40
10 