1. p 125 # 88
1.
Let X = the number of diodes on a board that fail.
a.
E(X) = np = (200)(.01) = 2, V(X) = npq = (200)(.01)(.99) = 1.98, X = 1.407
b.
X has approximately a Poisson distribution with  = np = 2,
so P(X  4) = 1 – P(X  3) = 1 – F(3;2) = 1 - .857 = .143, pretty close
c.
P(board works properly) = P(all diodes work) = P(X = 0) = F(0;2) = .135
NOW ANOTHER BINOMIAL using above as success prob and n= 5 fixed trials.
Let Y = the number among the five boards that work, a binomial r.v. with n = 5 and p = .135. Then
P(Y  4)
= P(Y = 4 ) + P(Y = 5) =
 5
 5
 (.135) 4 (.865)   (.135) 5 (.865) 0
 4
 5
= .00144 + .00004 = .00148
2. p 126 # 92
2.
a.
P(X = 10 and no violations) = P(no violations | X = 10)  P(X = 10)
= (.5)10  [F(10;10) – F(9;10)]
= (.000977)(.125) = .000122
b.
P(y arrive and exactly 10 have no violations)
= P(exactly 10 have no violations | y arrive)  P(y arrive)
(10) y
y!
y
10
y
(10)
e (5)

y!
10!( y  10)!
= P(10 successes in y trials when p = .5) 
=
 y  10
 (.5) (.5) y 10 e 10
10 
e 10
e 10 (5) y
c. P(exactly 10 without a violation) = 
y 1010!( y  10)!

e 10  510
=
10!
10
10
(5) y 10
e10  510   (5)u  5 5 e  5
 e5
=

e e =

10!  u 0 (u)! 
10!
y 10 ( y  10)!

since sum of Poisson over all values equals 1
=
e 5  510
= p(10;5). = .986 - .968 = 0.018 from using table in back of book or directly as
10!
.018133
In fact, generalizing this argument shows that the number of “no-violation” arrivals within the hour has
a Poisson distribution with parameter 5; the 5 results from p = 10(.5). said differently , we could have
considered an new effective Poisson process with rate p
3. p. 135 # 2
1.
f(x) =
a.
1
10
for –5  x  5, and = 0 otherwise, can integrate or use area of a rectangle = base*height
P(X < 0) =

0
1
5 10
dx  .5
b.
P(–2.5 < X < 2.5) =
c.
P(–2  X  3) =
d.
P(k < X < k + 4) =


2.5
1
 2.5 10
3
dx  .5
1
 2 10

dx  .5
k 4
k
1
10
dx  10x k
k 4
 101 [(k  4)  k ]  .4
4. p. 135 # 6
2.
a.
0.8
0.7
0.6
f(x)
0.5
0.4
0.3
0.2
0.1
0.0
2.0
2.5
3.0
3.5
4.0
4.5
x

4
1=
c.
P(X > 3) =
d.
P114  X  134   
e.
P( |X–3| > .5) = 1 – P( |X–3|  .5) = 1 – P( 2.5  X  3.5)
2
1

4
3
3 4
[1  ( x  3) 2 ]dx  .5 by symmetry of the p.d.f
13 / 4
3
11 / 4 4
[1  ( x  3) 2 ]dx 
=1–
5. p 135 # 8
a
4
3
k 
3
4
1
k[1  ( x  3) 2 ]dx   k[1  u 2 ]du 
b.

.5
1/ 4
3
4 1 / 4
3
.5 4

[1  (u ) 2 ]du 
[1  (u ) 2 ]du 
47
 .367
128
5
 .313
16
0.20
f(y)
0.15
0.10
0.05
0.00
0
2
4
6
8
10
y
5
10
y2   2
1 2 
y 
b.  f ( y)dy   ydy   (  y)dy =
  y

5
50  0  5
50  5
1 
1  1 1
=  ( 4  2)  (2  )    1
2 
2  2 2

5
10
1
0 25
2
5
1
25
3
c. P(Y  3) =

3
1
0 25
d. P(Y  8) = 
y2 
9
 .18
ydy 
 
50  0 50

5
1
0 25
8
ydy   ( 52  251 y )dy 
5
e. P( 3  Y  8) = P(Y  8) – P(Y < 3) =
P(Y < 2 or Y > 6) = 
f.
2
1
0 25

23
 .92
25
46 9 37


 .74
50 50 50
10
ydy   ( 52  251 y )dy 
6
6. p. 143 # 12
a.
P(X < 0) = F(0) = .5
b.
P(–1  X  1) = F(1) – F(–1) =
c.
P(X > .5) = 1 – P(X  .5) = 1 – F(.5) = 1 – .6836 = .3164
11
16
 .6875
2
 .4
5
3
3x 2 
d 1 3 
x3 
   4x   = 0   4 
  .09375 4  x 2

dx  2 32 
3  
32 
3 

d.
F(x) = F(x) =
e.
F ~   .5 by definition. F(0) = .5 from a above, which is as desired.

7. p. 143 # 18
3.
f ( x) 
1
1
 for –1 ≤ x ≤ 1
1  (1) 2
P(Y = .5) = P(X ≥ .5) =
a.
1

.5
1
dx = .25
2
P(Y = –.5) = .25 as well, due to symmetry. For –.5 < y < .5, F(y) = .25 +
b.
.5y. Since Y ≤ .5, F(.5) = 1 and F(y) = 1 for y > .5 as well. That is,
y  .5
 0

F ( y)  .5  .5 y  .5  y  .5
 1
.5  y

1.0
0.8
F(y)
0.6
0.4
0.2
0.0
-1.0
-0.5
0.0
y
0.5
1.0
8. p143 # 20
a.
For 0  y  5, F(y) =
For 5  y  10, F(y) =

y
0


y
0
1
y2
udu 
25
50
5
y
f (u)du   f (u)du   f (u)du
0
5
y 2
1
u 
2
y2
    du  y 
1
2 0  5 25 
5
50

y
1
dx = .25 + .5(y + .5) = .5 +
.5 2
1.0
0.8
F(y)
0.6
0.4
0.2
0.0
0
2
4
6
8
10
y
For 0 < p  .5, p = F(yp) =
b.
For .5 < p  1, p =
c.
y 2p
50
 y p  50 p 
1/ 2
2
p
y
2
yp 
 1  y p  10  5 2(1  p)
5
50
E(Y) = 5 by straightforward integration (or by symmetry of f(y)),
50
 4.1667 . For the waiting time X for a single bus,
12
25
E(X) = 2.5 and V(X) =
since Y = X1 + X2
12
and similarly V(Y)=
9. p144 # 26
a.
1=


0
2.5k
(1  x / 2.5) 6
6
k (1  x / 2.5) 7 dx =

=
0
k
 k = 2.4
2 .4
b.
2.5
2.0
f(x)
1.5
1.0
0.5
0.0
0
c.
let u = (1  x / 2.5)
1
2
3
4
x
5
6
7
8
E(X) =




0
x  2.4(1  x / 2.5) 7 dx =
x 2  2.4(1  x / 2.5) 7 dx =
0
and σX =


1


1
2.5(u  1)  2.4u 7  2.5du = 0.5, or $500. Similarly, E(X2) =
(2.5(u  1)) 2  2.4u 7  2.5du = 0.625, so V(X) = 0.625 – (0.5)2 = 0.375,
0.375 = 0.612, or $612.
d.
The maximum out-of-pocket expense, $2500, occurs when $500 + 20%(X – $500) equals $2500;
this accounts for the $500 deductible and the 20% of costs above $500 not paid by the insurance plan.
Solve: $2,500 = $500 + 20%(X – $500)  X = $10,500. At that point, the insurance plan has already paid
$8,000, and the plan will pay all expenses thereafter.
Recall that the units on X are thousands of dollars. If Y denotes the expenses paid by the company
(also in $1000s), Y = 0 for X ≤ 0.5; Y = .8(X – 0.5) for 0.5 ≤ X ≤ 10.5; and Y = (X – 10.5) + 8 for X > 10.5.
From this,
E(Y) =


10.5


0
y  2.4(1  x / 2.5) 7 dx =

0. 5
0
0 dx +
10.5

0.5
.8( x  0.5)  2.4(1  x / 2.5) 7 dx +
( x  10 .5)  2.4(1  x / 2.5) 7 dx = 0 + 0.16024 + .00013, or $160.37.
10. p154 # 30
a.
(c) = .9100  c  1.34 (.9099 is the entry in the 1.3 row, .04 column)
b.
9th percentile = –91st percentile = –1.34
c.
(c) = .7500  c  .675 since .7486 and .7517 are in the .67 and .68 entries, respectively.
d.
25th = –75th = –.675
e.
(c) = .06  c  .–1.555 (.0594 and .0606 appear as the –1.56 and –1.55 entries, respectively).