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Stats 241 Assignment 5 – Solutions
1. A random variable, X, has probability density function
0 x3
 ax

f ( x )  6a  ax
3 x  6
 0
otherwise

a) Find a.
Solution:

1
3
6
0
3
 f  x  dx    ax  dx    6a  ax  dx

3
6
 x2 

x2 
 a  xdx  a   6  x  dx  a    a  6 x  
2 3
 2 0

0
3
2
2

6  
3  


9
 a    a  6  6  
   6  3 

2  
2  
2


1
9
9
 a   36  18  18    9a Thus a 
9
2
2
b) Find P(X  4).
3
6

6

x2 
P  X  4   f  x  dx    6a  ax  dx  a   6  x  dx  a 6 x  
2 4

4
4
4
2
2

6  
4 


 a  6  6  
   6  4 

2  
2  


16 

1 2
 a 36  18  24    2a  2   
2

9 9
6
6
2. Suppose X has the distribution function
0
x0


1 2
F ( x )  ax  4 x
0 x2

1
x2

a) Find a.
Solution: Since X is a continuous random variable (It has a density function), Then
F(x) is continuous. This will be true if ax  14 x 2  1 at x  2.
Thus a2  14 4  1 or a  1.
b) Find P(X  1)
 1 1
Solution: PX  1  1  PX  1  1  F (1)  1  1   
 4 4
c) Find f(x) the density function of X.
1  x 0  x  0
Solution: f  x   F   x    2
 0 otherwise
Page 1
Stats 241 Assignment 5 – Solutions
3. In preparation for a long weekend, a hospital in an National park is purchasing
antidotes for rattle snake bites. Past experience has shown that the number of rattle
snake bites occurring during this long weekend has a Poisson distribution with mean
= 0.80. How many antidotes should the hospital have on hand so that there is at
least a 99.99% chance that an antidote will be available to all individuals who suffer a
rattle snake bite. How should this be altered if the mean number of rattlesnake bites
could be as high as =3.1.
Solution: Let X = # of snake bites during the long weekend.
P  X  x  p  x  
x
x!
e

 0.8

x
x!
e0.8 x  0,1, 2,3,
  0.8 x 0.8 
We need to find c such that P  X  c    
e   F  c   0.9999
x 0 
 x!

Here is a table of x, p(x) and F(x):
p (x)
F (x)
x
c
0
1
2
3
4
5
6
7
8
9
10
11
0.44933
0.35946
0.14379
0.03834
0.00767
0.00123
0.00016
0.00002
0.00000
0.00000
0.00000
0.00000
0.44933
0.80879
0.95258
0.99092
0.99859
0.99982
0.99998
1.00000
1.00000
1.00000
1.00000
1.00000
Thus c = 6
4. The expected number of suicides in a large metropolitan city in a month has is known
to be = 4.5. The number of suicides is known and to have a Poisson distribution.
a) Compute the probability of at least 8 suicides in a month
Solution:
P  X  x  p  x  
x

 4.5

x
e
e4.5 x  0,1, 2,3,
x!
x!
P  X  8  1  P  X  7   1   p  0   p 1  p  7  
  4.50 4.5  4.51 4.5  4.5 2 4.5  4.5 3 4.5  4.5 4 4.5  4.5 5 4.5  4.5 6 4.5  4.5 7 4.5 
 1 
e 
e 
e 
e 
e 
e 
e 
e 
1!
2!
3!
4!
5!
6!
7!
 0!

=0.08659
b) Compute the probability of at most 10 suicides in a month, if it is known that
there were at least 3 suicides. (Assume that the number of suicides for a month in
the city follows a Poisson distribution)
Page 2
Stats 241 Assignment 5 – Solutions
.
Solution:
P  X  10 X  3 

P  X  10   X  3
P  X  3

P 3  X  10
1  P  X  2

p  3  p  4  
 p 10 
1   p  0   p 1  p  2  
0.81975
 0.99193
0.82642
5. Suppose that a random variable X, has probability density function.
2 xe  x for x  0
f x   
for x  0
 0
2
determine E(X) and E(X ).
Solution

EX  




0
0
xf  x  dx   x  2 xe   x  dx   2   x 2e   x  dx
 2
  3

3


3
   3  x e
2  x
0
  3   3 2! 2
 dx   2 3 
 


 
The gamma dist’n with  = 3
EX2 



 2


0
0
x 2 f  x  dx   x 2  2 xe   x  dx   2   x 3e   x  dx   2
  4

4

  4

2

3!

2

  4   4
 x3e   x  dx
 4 0   4  
6
2
6. Suppose that X is a Poisson random variable with parameter .
Find  if P(X = 2) = P(X = 3).
Solution
p  x 
Then

x
x!
2
2!
e  and suppose that p(2) = p(3)
e 
3
3!
e   or
3!
   3.
2!
7. Show, if X is a Poisson random variable with parameter  where  is an integer, that
some 2 consecutive values of X have equal probabilities.
Solution
 x 
 x 1  
p  x 
e  p  x  1 
e if
x!
 x  1!
x
x!

 x 1
 x  1!
or
 x  1!   x 1
x!
x
and   x  1 . Thus p(x) = p(x + 1) if x =  – 1.
Page 3
Stats 241 Assignment 5 – Solutions
8. Let X be Poisson with parameter .
a) Find a recursion for P(X = x + 1) in terms of P(X = x).
Solution
 x 
 x 1  
p  x 
e and p  x  1 
e
x!
 x  1!
Thus p  x  1 
 x 1
 x  1!
e

 x

x  1 x!
Hence the recursion formula is:

p  x  1 
e 

x 1
p  x
p  x  with p  0   e  
x 1
b) Use the recursion in part a) to find  and 2.
Solution


Note:   1  E  X    xp  x  and 2  E  X 2    x 2 p  x 
x 0
x 0
n
n
x 0
x 0
Let mn   xp  x  and vn   x 2 p  x  then m0  0 and v0  0.
Also mn1  mn   n  1 p  n  1  mn   p  n 
And vn 1  vn   n  1 p  n  1  vn    n  1 p  n 
2


Hence 1   xp  x   lim  mn  and 2   x 2 p  x   lim  vn  and
x 0
n 
x 0
n 
Finally  2  2  12
9. Ten people are wearing tags numbered 1, 2,..., 10. Three people are chosen at
random. Let X = the smallest badge number among the three. Find the probability
distribution of X.
Solution
Note the possible values of X are x = 1, 2, 3, 4, 5, 6, 7 and 8.
In order for x to be the smallest number it has to be chosen and the remaining 2
numbers have to be chosen from the 10 – x numbers larger than x.
The number of ways of choosing 3 numbers from 10 is:
10  10  9  8
 10  3  4  120
 
 3  3  2 1
The number of ways of choosing 3 numbers from 10 so that x is the smallest is:
10  x  10  x 10  x  1 10  x  9  x  90  19 x  x 2
1 



2 1
2
2
 2 
90  19 x  x 2
2
Thus p  x   P  X  x  
x
p(x)
1
2
3
72
240
56
240
42
240
90  19 x  x 2
for x = 1 to 8.
120
240
4
5
6
7

30
240
Page 4
20
240
12
240
6
240
8
2
240
Stats 241 Assignment 5 – Solutions
10. Let X be a random variable with probability density function
1 x  2
2x3
otherwise
 k

f ( x )  k ( 3  x )
 0

a) Find k.

Solution: 1 


3
 
x2 
f  x  dx   kdx   k  3  x  dx   kx 1  k  3x   
2  2
1
2
 
2
3
2
 
2
9 
4  
3
 k  2  1  k  9    6     k   , hence k 
3
2 
2  
2
 
b) Calculate E(X).
Solution:
3
 2  2  x 2 x 3  
2
2
2 x
E  X      xf  x  dx  3  xdx  3  x  3  x  dx  3    3  3   
 2 1

1
2
 2 3   2
 23  421   32  3 92  273    3 24  83    1  32 152  193   1 97

2
3
2
c) Find the cumulative distribution function, F(x).
Solution:
F  x 
x


0
x 1

0


x

x
2
2

1 x  2 
3  u 1
3  1du


x
1

f  u  du   2

u 2 
2
x
2
2
 2 1du  2 3  u du 2  x  3  3 u 1  3  3u   

2  2
3 


3 
1
2


1

1
3 x

0
x 1


2
1 x  2
3  x  1

 2 1
2
 3  3  6 x  x  8 2  x  3

1
3 x

Page 5
x 1
1 x  2
2 x3
3 x
Stats 241 Assignment 5 – Solutions
11. The percentage X of antiknock additive in a particular gasoline is a random variable
with probability density function
kx 3 1  x  0  x  1
f ( x)  
0
otherwise

a) Find k.

Solution: 1 


1
 x 4 x5 
k
f  x  dx   kx 1  x  dx  k     k  14  15  
20
 4 5 0
0
1
3
hence k  20
b) Evaluate P[X < E(X)].
Solution:

1
 x5 x 6 
20 2
E  X    xf  x  dx   20 x 1  x  dx  20     20  15  16  

30 3
 5 6 0

0
2/ 3
2/ 3
  23 4  23 5 
 x 4 x5 
3

P  X  E  X     20 x 1  x  dx  20     20 

 4

5
 4 5 0
0


4
2
 
7  2 112
41
4
 20  23    3    32   5  83   5 
3
243
4 5 
c) Find F(x).
1
4
Solution: F  x  
x


0
x0

x

f  u  du    20u 3 1  u  du 0  x  1
0

1
x 1
0


x
  u 4 u5 
 20   
  4 5 0

1

0

 4
 5 x  4 x5

1

Page 6
x0
0  x 1
x 1
x0
0  x 1
x 1