Answer Key for Homework 4 (Due date: February 21, 2003)

Notice that you are working alone for this homework. If I were you, I would start doing it ASAP
even though you have lots of time. It will save you from last minute frustrations.
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penalty of 5 pt. if you do not staple them together.
Each question worth different with the total 100 pt. for the homework.



1.
Suppose the reaction temperature X (in C) in a certain chemical process has a uniform distribution
with a=-3 and b=7.
(a) write down the pdf of X,
1
 1
 , -3  x  7

f ( x)   7  (3) 10
0,
otherwise

(b) write down the cdf of X,
x  3
0,
x  3

F ( x)  
, -3  x  7
10

x7
1,
(c) Compute the median for X,
median

3
1
median  3
which is 0.5 then solve for median. Median=2
dx 
10
10
(d) Compute the 10th percentile for X.
r ( 0.10)

3
1
r (0.10)  3
which is 0.1 then solve for r(0.10). r(0.10)=-2 which is the 10th
dx 
10
10
percentile for X.
(e) Compute P(-1.5<X<1.8).
1.8
1
 10 dx 
1.5
1.8  1.5
 0.33
10
(f) For k satisfying –3<k<k+1<7, compute P(k<X<k+1)
k 1
1
 10 dx 
k
(k  1)  k
 0.10
10
(g) Compute E(X).
7
1
x2
x
dx

E(X)= 
10
20
3
7
49  9
2
20

3
(h) Compute Var(X).
7
1
x3
E( X )   x
dx 
10
30
3
2
7

2
3
7 3  (3) 3
 12.3333
30
Var(X)= E ( X )  ( E ( X ))  12.3333  2
(i) Compute E(5X3). (Hint: h(X)= 5X3 )
2
2
2
 8.3333
7
 x4 
 7 4  (3) 4 
1
  290
E (5 X )  5  x
dx  5   5
10
40
 40  3


3
7
3
2.
3
Exercise 4.12
(a) P(X<0)=F(0)=
1 3
03 
  4(0)    0.5
2 32 
3
1
(b) P(-1 < X < 1)=F(1)-F(-1)= 
2
1
(c) P(X > 0.5)=1-F(0.5)=1- 
2
(d) F’(x)=


3 
13   1 3 
(1) 3 
 4(1)       4(1) 
 =0.6875
32 
3   2 32 
3 
3 
0 .5 3  
 4(0.5) 
 =0.3164
32 
3 
3
3x 2 
 4 
  0.09375(4  x 2 )
32 
3 
(e) Since P(X<0)=P(X0)=0.5 then median=0 using part (a). Alternative method is F(median)=0.5
then solve for median.
3.
In each case, determine the value of the constant c that makes the probability statement correct.
(a) P(0  Z  c)=0.4649 then c=1.81
(b) P(Z  c)=0.0022 then c=2.85
(c) P(|Z|  c)=0.0044 then c=2.85
4.
Find the following percentiles for the standard normal distribution.
(a) 85th
P(Zz)=0.85 then the 85th percentile of Z is z=1.04
(b) 15th
P(Zz)=0.15 then the 15th percentile of Z is z=-1.04
5.
Let Z be the standard normal random variable and calculate the following probabilities.
(a) P(0  Z 1.9)=0.4713
(b) P(Z 1.9)=0.0287
(c) P(|Z| 1.9)=0.9426
6.
Exercise 4.35
(a) X~ N(25, 52)
P(X  x*)= 0.91 = P(Z  (x*-25)/5) then (x*-25)/5=1.34 and x*=31.7
(b) P(X  x*)= 0.06 = P(Z  (x*-25)/5) then (x*-25)/5=-1.555 and x*=17.225
(c) X~ N(3, 0.152)
P(X > x*)= 0.10 is the same as P(X  x*)=0.90= P(Z  (x*-3)/0.15) then (x*-3)/0.15=1.28 and
x*=3.192
7.
Exercise 4.39
10.256  
10.256   
 1.28
  0.1 . Using the table



9.671  
9.671   

 1.645
P(X<9.671)=0.05 then P Z 
  0.05 . Using the table






P(X>10.256)=0.10 then P Z 
We have two equations and two unknowns  and . If we solve for the unknowns, =10 and =0.2.
8.
Exercise 4.40
(a) P(|X- |1.5)=P(-1.5Z1.5)=0.8664
(b) P(|X- |2.5)=1-P(-2.5Z2.5)=0.0124
9.
Exercise 4.49 (notice that this is normal approximation to the binomial)
p: P(all drivers in a certain state regularly wear a seatbelt) =0.40
n=500 drivers selected
X: number of drivers regularly wear seatbelt ~ Binomial(n=500, p=0.40)
E(X)=np=500(0.40)=200
Var(X)=np(1-p)=500(0.40)(0.60)=120 then
np(1  p) =10.9545
(a) P( 180  X  230) = P(X  230) - P(X <180) = P(X  230)-P(X  179)
230  0.5  200 
179  0.5  200 


 P z 
  P z 
 =P(Z2.78)- P(Z-1.87)
10.9545
10.9545




= 0.9973 - 0.0307 =0.9666
174  0.5  200 

 = P(Z  -2.33) = 0.0099
10.9545


149

0
.
5

200


P(X<150) = P(X  149)= P z 
 = P(Z  -4.61) = 0
10.9545


(b) P(X<175) = P(X  174)= P z 
10. Exercise 4.53 (a)(b)
(a) (6)  (6  1)! = 5! = 120
(b) (5 / 2)  (3 / 2)(3 / 2)  (3 / 2)(1 / 2)(1 / 2) 
3 
= 1.3293
4
11. Exercise 4.55
X:time taken to mow the lawn ~Gamma (=2,=1/2) then X/ ~Gamma (=2,=1)
(a) P(X  1)=P(X/  2) = 0.594
(b) P(X  2)=P(X/  4) =1-P(X/ < 4) = 1 - 0.908 = 0.092
(c) P(0.5  X  1.5 ) = P(X/  3) - P(X/ < 1) =0.801 - 0.264 = 0.537
12. Exercise 4.92
X: the breakdown voltage ~N(=40, 2=1.52)
(a) P(39X42)= P(-0.67Z1.33)=0.9082-0.2514=0.6568
(b) P(X>x*)=0.15 then x*=?. First transform it to z and then look at the table.
x*=40+1.04(1.5)=41.56
(c) P(at least one exceeds 42)=1-P(all four does not exceed 42)=1-(P(X42))4=1-(P(Z1.33))4
=1-(0.9082)4=0.3197
13. Exercise 4.98. Additional to the question, (d) what is the value of the median path length?
X ~ Exponential (=1/=1/0.93) then f(x)= 0.93  e
P(Xx)= 1  e
P(X>x)= e
0.93 x
0.93 x
P(a < X < b) = e
0.93a
-e
0.93b
0.93 x
, x>0.
(a) If =0.93 then E(X)=1/0.93=1.0753 and also the standard deviation is 1.0753
P(X>3)= e
0.93( 3 )
(b) P(X>x)= e
=0.0614 and P(1 < X < 3) = e
0.93 x
0.93(1)
-e
0.93( 3 )
=0.10 then x=2.4759
(c) P(X>median)= e
0.93( median)
=0.5 then median=0.7453
=0.3331