Download Solution to additional exercises

Solution to additional exercises
y1

0
y2
5.108 f 1( y1 )   e  y1 dy2  y1e  y1 , y1  0. f 2( y 2 )   e  y1 dy1  e  y2 , y 2  0.
Then Y1 has a gamma distribution with α = 2 and β = 1, and Y2 has an exponential
distribution with β = 1. Thus, E(Y1 + Y2) = 2(1) + 1 = 3. Also, since
 y1
E(Y1Y2) =
 y y e
1
 y1
2
dy2 dy1  3 , Cov(Y1, Y1) = 3 – 2(1) = 1,
0 0
V(Y1 – Y2) = 2(1)2 + 12 – 2(1) = 1.
Since a value of 4 minutes is four three standard deviations above the mean of 1 minute,
this is not likely.
y2
5.133a f 2 ( y2 )   6(1  y2 )dy1  6 y2 (1  y2 ), 0  y2  1 .
0
f ( y1 | y2 )  1/ y2 , 0  y1  y2  1.
To find E(Y1 | Y2  y2 ) , note that the conditional distribution of Y1 given Y2 is
y
uniform on the interval (0, y2). So, E(Y1 | Y2  y2 ) = 2 .
2
y
6.1
6.1 The distribution function of Y is FY ( y )   2(1  t )dt  2 y  y 2 , 0 ≤ y ≤ 1.
0
a. FU1 (u)  P(U1  u)  P(2Y  1  u)  P(Y 
u1
2
)  FY ( u21 )  2( u21 )  ( u21 )2 .
Thus, fU1 (u )  FU1 (u )  12u ,  1  u  1 .
b. FU2 (u)  P(U 2  u)  P(1  2Y  u)  P(Y  12u )  FY ( 12u1 )  1  2( u21 )  ( u21 )2 .
Thus, fU 2 (u )  FU2 (u ) 
u 1
2
,  1  u  1.
c. FU3 (u )  P(U 3  u )  P(Y 2  u )  P(Y  u )  FY ( u )  2 u  u Thus,
fU3 (u)  FU3 (u) 
1
u
 1, 0  u  1.
d. E (U1 )  1 / 3, E (U 2 )  1 / 3, E (U 3 )  1 / 6.
e. E(2Y  1)  1 / 3, E(1  2Y )  1 / 3, E(Y 2 )  1 / 6.
7.42 Let Y denote the sample mean strength of 100 random selected pieces of glass.
Thus, the quantity ( Y – 14.5)/.2 has an approximate standard normal distribution.
a. P( Y > 14) ≈ P(Z > 2.5) = .0062.
b. We have that P(–1.96 < Z < 1.96) = .95. So, denoting the required interval as
(a, b) such that P(a < Y < b) = .95, we have that –1.96 = (a – 14)/.2 and 1.96
= (b – 14)/.2. Thus, a = 13.608, b = 14.392.
7.48 a. Although the population is not normally distributed, with n = 35 the sampling
distribution of Y will be approximately normal. The probability of interest is
P(| Y   |  1)  P( 1  Y    1) .
In order to evaluate this probability, the population standard deviation σ is needed.
Since it is unknown, we will estimate its value by using the sample standard
deviation s = 12 so that the estimated standard deviation of Y is 12/ 35 = 2.028.
Thus,
1
1
P(| Y   |  1)  P( 1  Y    1)  P(  2.028
 Z  2.028
)  P( .49  Z  .49)
= .3758.
c. No, the measurements are still only estimates.
7.52 Let Y denote the average resistance for the 25 resistors. With μ = 200 and σ = 10
ohms,
a. P(199 ≤ Y ≤ 202) ≈ P(–.5 ≤ Z ≤ 1) = .5328.
b. Let X = total resistance of the 25 resistors. Then,
P(X ≤ 5100) = P( Y ≤ 204) ≈ P(Z ≤ 2) = .9772.