Solutions for Examples in Review II
1. Suppose that 25% of students at a particular college have SAT scores over 1200. If
I pick 5 students at random from this student body, what is the probability that at
least two of those picked have SAT scores over 1200?
Sol) Let X be the number of student who have SAT scores over 1200, then
X~Binomial(5, 0.25)
P( X  2)  1  P ( X  2)  1  P ( X  0)  P( X  1)
   
   
5
0
1
5
 5
 1    * 1 * 3
   * 1 * 3
4
4
4
4
0
1
 376
1024
4
 4   5 * 1 4 3 4 
 1 3
5
4
2. Find the expected value and variance of the number of times one must throw a die
until the outcome 1 has occurred 4 times.
Sol) Let X be the number of throwing a die until 1 has occurred 4 times, then
X~Neg.Bin (p=1/6, r=4). Therefore,
E( X )  r
p

5
4
r (1  p) 4 * 6
 24 and Var ( X ) 

 120
2
1
p2
1
6
6
 
 
3. Suppose that the moment generating function of a random variable X is given
t
by M (t )  e 3(e 1) . What is P(X>1)?
Sol) The m.g.f of a r.v. X given is the that of Poisson (3), so X ~ Poisson (3).
 e 3 30   e 3 31 
  
  1  e 3  3e 3
0
!
1
!

 

Therefore, P( X  1)  1  P( X  1)  1  
4. Suppose that the moment generating function of a random variable X is given
1
2
by M (t ) 
, for t  1 . If Y  2 X  4 , what is the mean of Y?
6
2
(1  2t )
Sol) The m.g.f of a r.v. X given is the that of Gamma (α=6, θ=2), so X~Gamma
(α=6, θ=2). Using the thm. of gamma distribution and rules for expectation,




E (2 X 2  4)  2E ( X 2 )  4  2 Var( X )  E ( X ) 2  4  2  2  ( ) 2  4


 2 2 2 * 6  (2 * 6) 2  4  2(168)  4  340
Or, as German suggested, use m.g.f of X to get E(X2) since it is easy to
calculate M”(t) unlike I expected it would be:
E ( X 2 )  M " (t ) |t 0  168 * (1  2t ) 8 |t 0  M " (0)  168
Then, simply, E (2 X 2  4)  2 E ( X 2 )  4  2 *168  4  340
5. Suppose that the length of a phone cell in minutes is an exponential random
variable with parameter θ=10. If someone arrives immediately ahead of you at a
public telephone booth, find the probability that you will have to wait.
Sol) Let X be the waiting time until phone is available, then X~exp (10).
Therefore,
a) between 10 and 20
20
P(10  X  20)   e 10   e 1  e  2

 10
b) more than 20 minutes given you have already waited 10minutes.
x

P( X  20 | X  10) 
 e  x10 

 20
P( X  20)
e 2


 e 1


1
P( X  10)   x10 
e
e

 10
6. Suppose that the cumulative density function of a random variable X is
F ( x)  x 2 c for 0  x  2 .
a) Find c such that f(x) satisfies the conditions of a p.d.f.
Sol) Since

0
2

2
0
f ( x)dx  F ( x)0  1 from the condition of p.d.f,
2
 c
2
2
f ( x)dx  F ( x)0  x
0
2
 2  1  c  2
c
b) Find the median or 50th percentile of X (e.g.,  0.5 )
Sol) F ( 0.5 ) 
 0.5
 0.5   0.5  1
2
c) Find the moment generating function M(t) of X.
2
 f ( x)  x ,
Sol) Since F ( x)  x
2
2
2
21
2 t
1

M (t )  E (e tx )   e tx xdx   xetx    e tx dx 
e
0
0
t
t
t
0
2

1 t
e
t2
2

1
t2
7. Suppose that the random variable X has the p.d.f f ( x)  3x 2 8 , 0  x  2. Consider a
new random variable Y, where Y=X3. Find the p.d.f of Y.
1
1 2
Sol) Since inverse function is X   ( y )  y 3 ,  ' ( y )  y 3 , and 0 < y < 8,
3
2
3 1
1 2
f Y ( y )  f X ( ( y )) *  ' ( y )   y 3  * y 3  1
8
 3
8
Therefore, Y~Uniform (0, 8)