MAT 470, Worksheet 5
Name:_____________________
1. One of the continuous random variables that we have not discussed is the Laplace random
variable Y whose moment generating function is given by
a 2 ebt
m(t )  2 2 , where a and b are constants with a > 0.
a t
Use this moment generating function to help you determine the mean of Y in terms
of the constants, a and/or b.
Simply use the quotient rule to compute the derivative and you find that E(Y) = m’(0) = b.
2. A continuous random variable Y is uniformly distributed on the interval [2, 8].
Hence the density function is given by
1/ 6, 2  y  8;
f ( y)  
 0, otherwise.
a). Determine the probability, P( Y < 6 ) =
“It’s a rectangle!” (6 – 2)(1/6) = 4/6
b). Determine the cumulative distribution function F(y).
0, y  2;


F ( y )  ( y  2)(1/ 6), 2  y  8;

1, y  8.

c). Determine the 4th moment of Y, E( Y4 ) =

8
2
y 4 (1/ 6)dy  1091.2
d). Define a random variable W = 3Y + 2. Determine the mean and variance of W.
Note E(Y) = (8+2)/2 = 5 and V(Y) = (8 – 2)2/12 = 3.
Thus, E(W) = 3E(Y) + 2 = 17 and V(W) = 9V(Y) = 27.
e). Define a random variable W = 3Y2 + 2. Determine the mean and variance of W.
8
Note that E (Y 2 )   y 2 (1/ 6)dy  28 and so E(W) = 3(28) + 2 = 86.
2
Also, E( W2 ) = E( 9Y4 + 12Y2 + 4 ) = 9(1091.2) + 12(86) + 4 = 10160.8
and so we have V(W) = 10160.8 – (86)2 = 2764.8
3. Consider the random variables whose familiar moment generating functions are given below.
e 4 t  et
a). For a random variable Y, suppose mY (t ) 
.
3t
i). Determine E(Y) and V(Y).
Note that Y is uniform on (1, 4). So we know E(Y) = (1 + 4)/2 = 2.5
and V(Y) = (4 – 1)2 / 12 = 9/12 = 0.75
ii). Determine the expected value E (eY / 2 )  mY (1/ 2) 
b). For a random variable X, suppose mX (t ) 
e4/ 2  e1/ 2
 3.827
3(1/ 2)
5
1

.
5  12t 1  2.4t
i). Determine E(Y) and V(Y).
Note that Y is exponential with  = 2.4 and so E(Y) = 2.4 and V(Y) = (2.4)2 = 5.76
ii). Define W = 3X. Give the moment generating function for W, mW (t ) .
5
mW (t )  mX (3t ) 
5  36t
4. Suppose the height Y (in inches) of a crop of wheat is normally distributed with standard
deviation 2.5 inches.
a). Determine the average height of the hay if 67% of the crop measures less than 31 inches
in height.
Note 33% in the upper tail corresponds to z = 0.44.
Thus, 0.44 = (31 – )/2.5 and so  = 29.9 inches.
b). Now, suppose the average height is 30 inches and standard deviation is 2.5 inches.
Compute the following probabilities.
i). P( 32 < Y < 35 ) = 0.1891
ii). P( Y > 35 ) = 0.02275
5. The amount of fill dispensed by a bottling machine is normally distributed with
standard deviation  = 0.6 ounces.
a). If the dispenser fills 16 ounce cups such that the cups overflow with probability 0.005,
what is the mean  for the amount dispensed by the machine?
16  
 2.5758, and so   14.455
0.6
b). If the machine is adjusted so the amount dispensed is normally distributed with  = 16
and  = 0.6 ounces, what is the probability that a 17-ounce cup is filled to overflowing
(i.e., more than 17 ounces is dispensed)?
P(Y  17)  0.50  P(16  Y  17)  0.0478
6. Define a continuous random variable Y with distribution function given by
y0
 0,
F ( y)  
 0.2 y
,y0
1  e
a). Determine the density function for Y.
y0
 0,
f ( y)  
0.2 y
,y0
0.2e
b). Give the mean. E(Y) = 1/ 0.2 = 5, since this is an exponential distribution.
c). Determine the probability P( 4 < Y < 8 ) = = F(8) – F(4) = 0.2474
d). Determine the conditional probability P( Y > 8 | Y > 6 ) = P( Y > 2 ) = e -0.4 = 0.6703
using the fact that this is a memoryless distribution!
7. Consider a continuous random variable Y whose moment generating function is
3e 3t
given by m(t ) 
. Use this moment generating function to help you determine the
3t
mean, E(Y).
(3  t )9e 3t  3e 3t (1)
(3  t )2
10
E (Y )  m(0) 
3
m(t ) 
8. Let Y be an exponentially distributed continuous random variable with moment generating
function m(t ) 
2
1
3

,  .
3
2  3t 1  2 t
2
a). Give the mean and variance of Y.
E(Y) = 3/2 and V(Y) = 9/4
b). Recall the moments of Y are given by E( Yk ) = (k!)k . For W = Y2 + 3 ,
determine the second moment of W.
E( W2 ) = E( Y4 ) +6E( Y2 ) + 9 = 157.5