Download STAT 330 Spring 08 Assignment 3

STAT 330 Spring 08 Assignment 3
Dr. Paula Smith
Due Thurs July 10
1. (Adapted from Larson, Intro. to Probability Theory, 1969) A nursery specializes in the
installation of circular flower beds. When laying out the circle, a workman puts a peg in
the center and cuts a length of rope (already tied in a loose loop to the stake) equal to the
radius of the desired circle, and uses this to mark out the bed on the ground. Assume the
desired radius is r meters. Also assume the workman is a little sloppy, and is equally
likely to cut the rope to any length within the interval (r – 0.1, r + 0.1). Let X be the
length of the rope from stake to cut end.
a. Determine the probability density function for X.
b. Let Y be the surface area of the circle. State Y in terms of X.
c. Determine the probability density function for Y.
d. Find E(Y)
e. Find P(Y > r2).
Solution
a) Since it is assumed that the workman is a little sloppy, and is equally likely to cut the rope to
any length within the interval (r – 0.1, r + 0.1), the random variable X, the length of the rope
from stake to cut end follows a Uniform distribution over the range (r – 0.1, r + 0.1). So the
probability density function of X is given by
1
f X ( x) 
,
r  0.1  x  r  0.1
0.2
= 0 otherwise
[Note that the probability density function of a Uniform distribution over the range (a ,b) is
1
f X ( x) 
,
a xb]
ba
b) Let Y be the surface area of the circle. Then Y can be written as Y   X 2
[Note that the surface area of a circle with radius r is  r 2 ]
c) Since X varies from (r – 0.1) to (r + 0.1), Y varies from  (r  0.1)2 to  (r  0.1)2
Now the probability density function of Y is given by
dx
gY ( y )  f X ( x)
dy
From the transformation y   x 2 we get x 
y

and hence
dx
1
.

dy 2  y
Thus the probability density function of Y is given by
1
1
gY ( y) 
0.2 2  y
1
,  (r  0.1)2  y   (r  0.1)2 ,

0.4  y
= 0 otherwise
[Note that the above method to find the p.d.f. is valid because the transformation from X to Y is
one to one as X takes only positive values since r >0.1 always.]
d) we have E (Y )   ygY ( y )dy
y
 ( r  0.1)2


y
 ( r  0.1)2
1

0.4 
1

0.4 
1

0.6 



1
0.4  y
dy
 ( r  0.1)2

ydy
 ( r  0.1)2
 ( r  0.1)2
 2 3/ 2 
 3 y 
 ( r  0.1)2
 (r  0.1)2 3/ 2   (r  0.1) 2 3/ 2 


 
(r  0.1)3  (r  0.1)3 
0.6 

3
2
(r  0.3r  0.03r  0.001)  (r 3  0.3r 2  0.03r  0.001) 
0.6 

0.6
(0.6r 2  0.002)
e) We have P[Y   r ] 
 ( r  0.1)2

2
r
gY ( y )dy
2
 ( r  0.1)2


r
2
1

0.4 
1

0.4 
1

0.2 


1
0.4  y
 ( r  0.1)2

 r2
dy
1
dy
y
 ( r  0.1)2
2 y  2

 r
  (r  0.1)2   r 2 


(r  0.1)  r 
0.2 
0.1

0.2
1
= = 0.50
2
2. The Weibull density function is given by
f(y) = (1/) m ym–1 exp(–ym /) if 0< y <; and f(y) = 0 otherwise,
where  and m are positive constants. This density function is often used as a model for
the length of life of physical systems. Suppose Y has the Weibull density just given.
a. Find the density function of U = Ym.
b. Find E(Yk) for any positive integer k.
c. Suppose W has an exponential distribution with mean . Prove that Y = W has
a Weibull density with  =  and m = 2.
d. Find E(Wk/2)
Solution
Here it is given that the p.d.f. of Y is
 ym 
1
f ( y)  m   y m1 exp    , 0  y  
 
  
a) Let U  Y m . Let g (u ) be the probability density function of U.
Then we have g (u )  f ( y )
dy
du
Since u  y m , we have y  u1/ m and hence
dy u (1/ m )1

du
m
Thus the probability density function of U is given by
1
g (u)  m   u
 
m 1
m
(1/ m ) 1
 uu
exp   
  m
1
 u
   exp    , 0  u  
 
 
Hence U  Y m follows an Exponential distribution with mean  .

b) E (Y k )   y k f ( y )dy
0

 ym 
 1  k m1
 m    y y exp   dy
  0
  

 ym 
1
 m    y k m1 exp   dy
  0
  
Put u  y m , then y  u1/ m and hence dy 
u (1/ m )1
du
m
Now the above integral becomes

1
E (Y k )  m    u
  0
k  m 1
m
(1/ m ) 1
 u u
exp   
du
  m

k
1
 u
    u m exp   du
  0
 

k

 1   11
 u
    u  m  exp   du
  0
 
k

  1
k

k

 1  m 
, where   1  Gamma   1
 
k

1
m 
m 
    1 m
 
 
k

  k / m   1
m 
c) Suppose W has an exponential distribution with mean . Then the density function of W is
given by
 w
1
h( w)  exp    , 0  w  

 
Consider the transformation Y  W .
Now the p.d.f. of Y is given by
f ( y )  h( w)
dw
dy
Since y  w , we have w  y 2 and hence
dw
 2y
dy
Thus the p.d.f. of Y is given by
 y2 
f ( y)  exp    2 y

  
 y2 
1
 2   y 21exp    , 0  y   , which is the p.d.f. of a Weibull density with  = 
 
  
and m = 2.
1
d) Since Y = W has a Weibull density with  =  and m = 2, using the result (b) we have
E (W k / 2 )  E (Y k )
k 
  k / 2   1
2 
3. A small orchard contains 5 Winesap apple trees and 3 Mutsu apple trees, all at peak
bearing age. The yearly harvests for Winesap apple trees at that stage are normally
distributed with mean 1 and variance 2, while the yearly harvest for Mutsu apple trees
at that stage are normally distributed with mean 2 and variance 22.
a. What is the expected yearly harvest for this orchard?
b. Assume you have lists of yearly harvest amounts per tree and you can randomly
select n of the Winesap harvest amounts and m of the Mutsu harvest amounts.
Suppose 2 is known. Construct a 95% confidence interval for the expected
yearly harvest for this orchard.
c. Under the same conditions as in part b, but with 2 unknown, construct a 95%
confidence interval for the expected yearly harvest for this orchard.
Solution
Let X denotes the yearly harvests for Winesap apple trees and Y denotes the yearly
harvest for Mutsu apple trees in the orchard. Here it is given that X follows a Normal distribution
with mean 1 and variance 2 and Y follows a Normal distribution with mean 2 and variance
22.
a) Since the orchard contains 5 Winesap apple trees and 3 Mutsu apple trees, the expected yearly
harvest for this orchard is E(5X + 3Y) = 5E(X) + 3E(Y) = 51  32 .
b) Let U= 5X + 3Y. Then U represents the yearly harvest for the orchard. Since X follows a
Normal distribution with mean 1 and variance 2 and Y follows a Normal distribution with
mean 2 and variance 22, assuming that X and Y are independent U= 5X + 3Y follows a
Normal distribution with mean 51  32 and variance 432.
[Note that Var(U) = Var(5X + 3Y)=25Var(X)+9Var(Y)=252 + 9*22 = 432]
Let x1 , x2 ,..., xn denote the random sample of the Winesap harvest amounts and y1 , y2 ,..., ym
denote the random sample of the Mutsu harvest amounts. Suppose x 
y
1 n
 xi and
n i 1
1 m
 yi are the sample means of the two random samples respectively.
m i 1
Since x1 , x2 ,..., xn are random samples from a Normal distribution with mean 1 and
variance 2, x follows a normal distribution with mean 1 and variance
x
2
. That is
n
N ( 1 ,  2 n) . Similarly since y1 , y2 ,..., ym are random samples from a Normal distribution
with mean 2 and variance 22, y follows a normal distribution with mean 2 and variance
2 2
. That is y N ( 2 , 2 2 m) .
m
Thus 5 x  3 y follows a normal distribution with mean 51  32 and variance
25
2
n
9
2 2  25 18  2
    .
m  n m
 25 18 
N (51  32 ,     2 )
 n m
5 x  3 y  (51  32 )
Hence Z 
N (0,1) , a Standard Normal distribution.
 25 18 
   
 n m
For a Standard Normal random variable Z, we know that
That is 5 x  3 y
P[1.96  Z  1.96]  0.95




5 x  3 y  (51  32 )

That is P 1.96 
 1.96   0.95


 25 18 
   




 n m

 25 18 
 25 18  
i.e., P  1.96     5 x  3 y  (51  32 )  1.96      0.95
 n m
 n m  


 25 18 
 25 18  
i.e., P (5 x  3 y )  1.96     (51  32 )  (5 x  3 y )  1.96      0.95
 n m
 n m  

Thus a 95% confidence interval for the expected yearly harvest for this orchard is

 25 18 
 25 18  
 (5 x  3 y )  1.96    , (5 x  3 y )  1.96    
 n m
 n m 

c) Let S X2 
1 m
1 n
2
2
(
x

x
)
S

( yi  y ) 2 denote the sample variances of the two
and


Y
i
m i 1
n i 1
samples. Then we have
Also
2
Y
2
mS
2
nS X2

 (2n 1) , a Chi-Square distribution with n 1 degrees of freedom.
2
mSY2


 2 2 2
5 x  3 y  (51  32 )
2
( m 1)
. Thus
nS X2
 (2n m2)
 25 18 
Hence t 

   
 n m
 nS X2 mSY2 
 2 
 (n  m  2)
2 2 
 
5 x  3 y  (51  32 )
 25 18 
 mSY2 2  (n  m  2)   
 n m
5 x  3 y  (51  32 )

t( n m2) , a Student’s t distribution with (n  m  2) degrees of
 25 18 
SP   
 n m
 nS
2
X
freedom, where SP 
[Note that if X
 nS
2
X
 mSY2 2 (n  m  2) is the pooled standard deviation.
N (0,1) and Y
(2n) , then t 
X
Y n
t( n ) ]
From Student’s t –table with (n  m  2) degrees of freedom, we can find a t0.025 such that
P[t0.025  t  t0.025 ]  0.95




5 x  3 y  (51  32 )

That is P t0.025 
 t0.025   0.95


 25 18 
SP   


 n m



 25 18 
 25 18  
i.e., P  t0.025 S P     5 x  3 y  (51  32 )  t0.025 S P      0.95
 n m
 n m  


 25 18 
 25 18  
i.e., P (5 x  3 y )  t0.025 S P     (51  32 )  (5 x  3 y )  t0.025 S P      0.95
 n m
 n m  

Thus a 95% confidence interval for the expected yearly harvest for this orchard is

 25 18 
 25 18  
 (5 x  3 y )  t0.025 S P    , (5 x  3 y )  t0.025 S P    
 n m
 n m 
