CHAPTER 6
DISCRETE PROBABILITY DISTRIBUTIONS
1.
Mean = 1.3
Variance = 0.81 found by:
   XP( X )  0(.20)  1(.40)  2(.30)  3(.10)  13
.
 2  ( X   ) 2 P ( X )
 (0  1.3)2 (0.2)  (1  1.3)2 (0.4)  (2  1.3) 2 (0.3)  (3  1.3) 2 (0.1)
 0.81
3.
a.
b.
c.
The second or middle, one
0.20, 0.40, 0.90
 = 14.5, 5(.1)  10(.3)  15(.2)  20(.4)  14.5
Variance = 27.25
  (5  14.5) (.1)  (10  14.5) (.3)  (15  14.5) (.2)  (20  14.5) 2 (.4)  27.25
2
2
2
 = 5.22 found by
2
27.25
5.
4 g pill with the standard deviation of 0.2. The other option gives a standard deviation of 0.89 for
the 4 pills.
7.
a.
b.
c.
Probability
d.
0.6612, found by (0.4351+0.2261)
 = 3.0402 , found by (2) 0.4351+(3) 0.2261+(4) 0.2307+(5) 0. 0838+(6)) 0.0202+(7)
0.0041
2 = 1.25 , found by (2-3.0402)2 0.4351+(3-3.0402)2 0.2261+(4-3.0402)2 0.2307+(53.0402)2 0.0838+(6-3.0402)2 0.0202 +(7-3.0402)2 0.0041),  =1.12, found by 1.25
0.6
0.4
0.2
0.02
1
2
3
4
5
Persons per Family
6-1
6 8
9.
11.
P(2) 
4!
(.25) 2 (.75) 4  2  0.2109
2!(4  2)!
a.
P(2) = 0.2109, found by
b.
P(3) = 0.0469, found by P(3) 
a.
x
P(X)
0
0.064
1
0.288
2
0.432
3
0.216
Mean = 1.8, found by   0(0.064) ...3(0.216)  18
.
b.
4!
(.25) 3 (.75) 4  3  0.0469
3!(4  3)!
Variance =0.72, found by  2  (0  18
. ) 2 0.064...(3  18
. ) 2 0.216  0.72
Standard deviation = 0.8485, found by
13.
a.
b.
c.
d.
19.
21.
12!
(0.1)0 (0.9)12
0!(12  0)!
12!
0.3765, found by P(1) 
(0.1)1 (0.9)11
1!(12  1)!
12!
0.2301, found by P(2) 
(0.1) 2 (0.9)10
2!(12  2)!
 = 1.2, found by 12(0.1),  2  1.08,
  1.04
0.2824, found by P(0) 
n!
25!
p x (1  p) ( nx ) 
.0.210 (1  0.20) ( 2510)  0.0118
x!(n  x)!
10!(25  10)!
15
17.
  0.72  0.8485
b.
P(X≤ 5) = P(X=0)+P(X=1)+ . . .+P(X=5)
= 0.122+0.270+0.285+0.190+0.090+0.032) = 0.989
P(X≥ 2) = 1- [P(X=0)+P(X=1)] = 1- [0.122+0.270] =0.608
a.
0.0008, found by
b.
0.315, found by p(X=5)+P(X=6)+P(X=7) = 0.026+0.088+0.201 =0.315 (Binomial
Distribution table)
a.
n1
10!
p x (1  p) ( n  x ) 
.0.83 (1  0.80) (103)
x!(n  x)!
3!(10  3)!
a.
 =10.5, found by 15(0.7),  = 1.7748, found by (15)( 0.7)( 0.3)
b.
0.2061, found by
c.
0.4247, found by 0.2061+0.2186
15!
(0.7)10 (0.3) 5
10!5!
6-2
d.
0.5154, found by 0.2186+0.1700+0.0916+0.0305+0.0047
 x e 
23.
25.

0.4 0 e 0.4
 0.6703
0!
a.
0.6703, found by P(X=0) =
b.
0.3297, found by 1-P(X=0) = 1-0.6703
a.
x!
P(3) = 0.0521, found by P(3) =
 x e 
x 
b.
P(0) = 0.0009, found by
 e
x!
x!

7 3 e 7

 0.0521
3!
7 0 e 7
0!
P(x4)=0.3528, found by  =3, (0.005(600)
27.
1-p(x3) = 1-(0.0498+0.1494+0.2240+0.2240)
29.
A random variable is a quantitative or qualitative outcome that results from a chance experiment. A
probability distribution includes the likelihood of each possible outcome.
31.
The characteristics of a binomial experiment are:
1.
The experiment consists of n Bernoulli trials
2.
The two possible outcomes of each trial are generally denoted as success (S) or
failure (F).
3.
The outcome of any trial is independent of the outcome of any other
trial.
4.
The probability of success (p) remains the same from trial to trial.
33.
 = 2, found by 0(0.1)+1(0.2)+2(.3)+3(0.4)
 = 1, found by
35.
(0 - 2) 2 (0.1)  . . .  (3 - 2) 2 (0.40 )
 = 1.3 found by 0(0.4)+1(0.2)+2(0.2)+3(0.1)+4(0.1),  =1.345, found by
(0 - 1.30) 2 (0.4)  . . .  (4 - 1.30) 2 (0.10 ) .
37.
39.
41.
a
b.
profit=27600, found by E(X) =500(0.3)+1000(0.5)+1500(0.2) =950
950(50)+(1000-950)(2)-20000 =27600
a.
b.
43.
0.001, found by Binomial Distribution table
0.002, found by 1-[P(X=0)+ P(X=1)+. . .+P(X=4)] =1-[0.349+0.387+0.194+0.057+0.011]
0.8486, found by
30!
30!
 30!

1- 
0.15 0 0.85 30 
0.15 10.85 29 
0.15 2 0.85 2 
1! 29!
2!28!
 0! 30!

30!
3
27
0.15 0.85
0.1703, found by
3!27!
c.
4.5 found by 30 (0.15)
a.
0.3679, found by
9!
0.15 10.85 8
1!8!
6-3
45.
47.
49.
51.
53.
55.
57.
9!
0.15 0 0.85 9
0!9!
b.
0.2316, found by
c.
1.35, found by 9(0.15)
a.
0.1311, found by
b.
c.
2.4, found by (0.15)(16)
0.2100, found by 1-[0.0743+0.2097+0.2775+0.2285]
a.
b.
c.
d.
0
0.0025
1
0.0207
2
0.0763
3
0.1665
4
0.2384
5
0.2340
6
0.1596
7
0.0746
8
0.0229
9
0.0042
10
0.0003
 = 4.5, found by 10(0.45) and  =1.5732, found by   10(0.45)(0.55)  1.5732
0.2384
0.5044, found by 0.0025 + 0.0207 + 0.0763 + 0.1665 + 0.2384
a.
b.
1.4, found by 700*0.002
0.2466, found by (1.40e-1.4)/0!
c.
P(X>2) = 0.938 found by 1  
a.
b.
P(X= 0) = 0.0025 (From Poisson table,  =6 and X = 0))
0.9875 = (1-0.0025)5
16!
(0.15) 4 (0.85)12
4!12!
 6 0 e 6

0!

61 e 6 6 2 e 6

1!
2!




 = 4, the following answers are based on Poisson Distribution Table
a.
b.
c.
d.
0.0183
0.1954
0.6289
0.5665
a.
0.1086, found by
b.
0061, found by
c.
0.9939, found by 1-(0.0061)
5.17 e 5.1
7!
5.10 e 5.1
0!
 = 8, x = 3
a.
p(X=3) = 8 3 e-8/3! =0.0286
6-4
b.
59.
61.
c.
P(X < 3) = P(X=0)+ P(X=1)+ P(X=2)
= 0.0003+0.0027+0.0107
(Poisson Distribution Table)
=0.0137
P(X > 4) = 1-[P(X= 0)+P(X=1)+P(X=2)+P(X =3)+P(X=4)]
=1-[0.0003+0.0027+0.0107+0.0286+0.0573]
=1- 0.0996
=0.9004
a.
 = 0.7, found by (196/280)
b.
P(X=3) = 0.0284, found by
c.
P(X1) = 0.5034, found by 1  
0.7 3 e 0.7
3!
 0.7 0 e 0.7 


0
!


Answer will depend on latest data on the web site.
6-5