“An Aggie does not lie, cheat, or steal or tolerate those who do.”
Answer Key for Homework 4 (Collected: February 28, 2005, Monday in class)
1.
The hydrogenation of benzene to cyclohexane is promoted with finely divided porous nickel catalyst.
The catalyst particles can be considered to be spheres of various sizes. All particles have masses
between 10 and 70 g. Let X be the mass of a randomly chosen particle. The probability density
function of X is given by
 x  10
,

f ( x)   1800
0,
10  x  70
otherwise
(a) What proportion of particles have masses less than 50g?
50

x  10
1  x2
  10 x  =800/1800=0.4444
P( X  50)  
dx
1800 1800  2
 10
10
50
(b) Find the mean mass of the particles.
70

x  10
1  x3
  5 x 2  =50
E( X )   x 
dx 
1800
1800  3
 10
(c) Find the standard deviation of the particle masses.
70
x  10
1  x4
x3 
  10  =2700
E( X )   x 
dx 
1800
1800  4
3  10
2
2
Var(X)=2700-502=200
Then the standard deviation is 14.1421
(d) Find the cumulative distribution function of the particle masses.
x  10
0,
x
x


 x  10
1  x2
1  x2
  10 x  
  10 x  50 
F ( x)   
dx 
10  x  70
1800  2
 10 1800  2

10 1800
1,
x  70

(e) Find the median of the particle masses.
F (median) 

1  median 2

 10median  50   0.5 then solve for the median. Notice that
1800 
2

the roots are -32.4264 and 52.4264. So the median is 52.4264.
(f) Find the 90th percentile of the particle masses.
Let’s call it x*.
F ( x*) 

1  ( x*) 2

 10 x * 50   0.9 then solve for the x*. Notice that the roots are
1800  2

-46.9210 and 66.9210. So the median is 66.9210.
2.
Exercise 4.11 (Exercise 4.11 and 4 13 as the parts (f)(g)(h) in the 5th edition)
“An Aggie does not lie, cheat, or steal or tolerate those who do.”
a)
b)
c)
d)
e)
12
 0.25
4
12 0.5 2
F(1)-F(0.5)=

 0.1875
4
4
0.5 2
1-F(0.5)= 1 
 0.9375
4
median 2
If
 0.5 then median= 2 =1.4142 because x values defined in between 0 and 2,
4
- 2 cannot be possibility.
d
( F ( x))  x/2, 0x<2 and 0 otherwise
f(x)=
dx
F(1)=
2
2
x3
(
0
.
5
)
f) E(X)=  x(0.5 x)dx 
=1.3333
3 0
0
2
2
x4
2
(
0
.
5
)
g) Var(X)=  x (0.5 x)dx -[E(X)] =
-[1.3333]2=0.2222 then the standard deviation is
4
0
0
2
0.4714
2
2
x4
h) E(X2)=  x (0.5 x)dx  (0.5)
=2
4
0
0
2
3.
Exercise 4.27
(a) (c)=0.9838. The area on the left of c is 0.9838 then c=2.14.
(b) P(0Zc)=0.291 then the area on the left of c is 0.5+0.291=0.791 which gives c as 0.81.
(c) P(Z  c)=0.121. The area on the right of c is 0.121 then the area on the left of c is 0.879 which
gives c as 1.17.
(d) P(-c  Z  c)=0.668. If the area between -c and c is 0.668, the area on the left of -c is (10.668)/2=0.166 and the area on the right of c is 0.166 by symmetry. The c value is 0.97.
(e) P(|Z|  c) =P(Z  -c or Z  c)=0.016. The sum of the areas on the left of -c and on the right of c is
0.016. The area on the left of -c is 0.016/2=0.008 and the area on the right of c is 0.008 by symmetry.
The c value is 2.41.
4.
Exercise 4.32
X: concentration ~N(=0.30,2=0.062)
0.25  0.30 

  P(Z >-0.83)=1-0.2033=0.7967
0.06 

0.10  0.30 

(b) P(X ≤ 0.10)= P Z 
  P(Z  -3.33)=0.0004
0.06 

(a) P(X > 0.25)= P Z 
“An Aggie does not lie, cheat, or steal or tolerate those who do.”


(c) P(X ≤ x*)= P Z 
x * 0.30 
  0.95 where largest 5% is the same as the smallest 95% then
0.06 
x * 0.30
 1.645 . The largest 95% is x*=0.3987.
0.06
The largest 5% of the concentration values are above 0.3987.
5.
Exercise 4.34 (Hint: compute the probability between 2.9 and 3.1 for each machine. The machine
with higher probability is more likely to produce an acceptable cork)
3.1  3 
 2.9  3
Z
  P 1  Z  1
0.1 
 0.1
Machine 1: P2.9  X  3.1  P
=0.8413-0.1587=0.6826
3.1  3.04 
 2.9  3.04
Z
  P 7  Z  3
0.02 
 0.02
Machine 2: P2.9  X  3.1  P
=0.9987-0=0.9987
Machine 2 is more likely to produce an acceptable cork.
6.
7.
Exercise 4.37
X: Opening altitude ~N(=200, 2=302)
Damage to the equipment of parachute opens when X<100
P(X<100)=P(z<-3.33)=0.0004
P(damage if at least 1 of the 5 independently dropped parachutes)=1-P(none of 5)=1-[p(x100)]5= 1[0.9996)]5=0.002
Exercise 4.38 (Hint:
P  0.1  X    0.1  0.95 is the given information. Transform
it to z, use the symmetry and solve for )
0.1 
  0.1
P  0.1  X    0.1  P
Z
  0.95 then
 
 
 0. 1 
0.1 


by symmetry P Z 
  0.025 and P Z 
 =0.025 . When you look at the standard
 
 


normal table, (-0.1/)=-1.96 which will give =0.051
8.
Exercise 4.42
X: Rockwell harness of a particular alloy ~ N(=70, 2=32)
75  70 
 67  70
z
  P(1  z  1.67) =0.9525-0.1587=0.7938
3 
 3
(a) P(67<X<75)= P
(b) P(70-c<X<70+c)=0.95 then c=?
70  c  70 
c
c
 70  c  70
c

P
z
 z    1  2 P z 
  P

3
3
3
3 


 3

c

Then P z 
  (1  0.95) / 2 =0.025
3 

Using the table –c/3=-1.96 then c=3(1.96)=5.88
(c) E(X)=np=10 P(67<X<75)=10(0.7938)=7.938


(d) p=P(X<73.84)= P z 
73.84  70 
  P( z  1.28)  0.8997
3

“An Aggie does not lie, cheat, or steal or tolerate those who do.”
10 
(0.8997) 9 (0.1003)  (0.8997)10 =0.265
9 
P(at most 8)=1-P(exactly 9)-P(exactly 10)= 1  
Exercise 4.60 (Exercise 4.59 in the 5th edition)
9.
X ~ Exponential (=1/=1/0.01386) then f(x)= 0.01386  e
P(Xx)= 1  e
P(X>x)= e
0.01386 x
, x>0.
0.01386 x
0.01386 x
P(a < X < b) = e
0.01386a
-e
0.01386b
(a) P(X  100) = 1  e
0.01386(100)
P(X  200) = 1  e
0.01386( 200)
= 0.7499
= 0.9375
P(100  X  200) = P(X  200) - P(X < 100) = e
0.01386(100)
- e
0.01386( 200)
(b) P(X >  + 2 )=P(X > (1/0.01386)+2(1/0.01386))=P(X > 216.4502)= e
where the mean and the standard deviation of X are 1/.
(c) P(X > median) = 0.50 then e
10.
0.01386median
= 0.1876
0.01386( 216.4502)
= 0.0498
=0.5 and median =-ln(0.5)/0.01386=50
Exercise 4.97 (Exercise 4.93 in the 5th edition)
X:actual contents of a jar ~N(137.2 , 1.62 )
Actual stated content=135


(a) P(X>135)= P Z 
135  137.2 
 =P(Z>-1.38)=0.9162
1.6

(b) Y: number among 10 contains more than the stated amount
Y~Binomial(n=10, p=P(X>135)=0.9162)
P(Y8)=P(Y=8)+P(Y=9)+P(Y=10)
10 
10 
10 
(0.9162) 8 (0.0838) 2   (0.9162) 9 (0.0838)1   (0.9162)10 (0.0838) 0
8 
9 
10 
= 
=0.1569+0.3812+0.4168=0.9549


(c) =137.2 and P ( X  135)  P Z 
table
135  137.2

135  137.2 
  0.95 then by using the standard normal


 1.645 which will give us =1.3374