Download SMAM 351 REVIEW FOR EXAM 2 1. In testing a certain kind of truck

SMAM 351
REVIEW FOR EXAM 2
1. In testing a certain kind of truck tire over a rugged terrain it is found that 20%
of the trucks fail to complete the test run without a blowout. Of the next 20 trucks
tested find
A. the probability that exactly 4 trucks have blowouts.
This is a binomial distribution with n = 20 and p = .20. Using the table
P(X=4) = P(X≤4) – P(X≤3) =.630 – .411 = .219
B. the probability that from 3 to 6 trucks inclusive have blowouts.
P(3≤X≤6) = P(X≤6) – P(X≤2) = .913 – .206 = .707
C. the probability that fewer than 14 trucks do not have blowouts.
Use n = 20 p =.80
P(X<14)=P(X≤13) = .087
D. the probability that more than 5 trucks have blowouts.
P(X>5) = 1– P(X≤5) = 1–.804 = .196
E. the mean and the variance of the number of trucks that do not have blowouts.
µ = 20(.8) =16 σ2 = 20(.8)(.2) = 3.2
2. The probability that a person living in a certain city has a cat is 0.4. What is
the probability that
A. the fifth person interviewed is the first to own a cat?
P[X=5] =(.6)4(.4) = .05184
B. the tenth person interviewed is the fifth to own a cat?
 9
P[X = 10] =   (.6)5 (.4)5 = .1003
 4
C. at most five people are interviewed before someone says he has a cat?
.4 +(.6)(.4) + (.6)2(.4) + (.6)3(.4) +(.6)4(.4) = .92224
or
.4(1− (.6)5 )
= 1− (.6)5 = .92224
1 – .6
3. A certain area of the United States is hit on the average by 5 hurricanes each
year. Find the probability that this area will be hit by
A. fewer than 4 hurricanes in a given year
P(X≤3) = .265
B. anywhere from 6 to 10 hurricanes inclusive. during a two year period (λ=10)
P(6≤X≤10) = P(X≤10) – P(X≤5) = .583 – .067 = .516
4. The probability that a person dies from a certain disease is .002. What is the
probability that at least 4 of the next 1000 infected will die?
λ = 1000(.002) =2
P(X≥4) = 1–P(X≤3) = 1– .857 = .143
5. A record collection consists of 5 jazz records, 2 classical records and 3 hard
rock records. For a sample of 4 records:
A. What is the p.m.f. of the number of jazz records?
 5  5 
 

 x  4 − x
f(x) =
, x = 0,1, 2, 3, 4
10
 
 4
B. What is the probability of at most one jazz record?
 5  5  5  5
     
 0  4  1  3
5
50
55
11
P(X ≤ 1) = f(0) + f(1) =
+
=
+
=
=
= .2619
210 210 210 42
10
10
 
 
 4
 4
C. What is the mean and the variance of the number of jazz records?
Using the formula on p 124
5
=2
10
10 − 4
5 5
V(X) =
⋅4⋅ ⋅
= .6667
9
10 10
E(X) = 4 ⋅
6. Given the probability mass function
x
0
1
2
3
4
p(x) 0.41 0.37 0.16 0.05 0.01
Find
A. P(X>2)
P(X>2)=p(3) + p(4) = .05+.01 =.06
B. The CDF
0
.41

.78
F(x) = 
.94
.99

 1
x<0
0≤ x <1
1≤ x < 2
2≤x<3
3≤x<4
x≥4
C. The mean and the variance of X
EX = 0(.41) + 1(.37) +2(.16) + 3(.05) + 4(.01) = .88
EX2 = 02(.41) + 12(.37) + 22(.16) + 33(.05) + 42(.01) = 1.62
σ2 = 1.62 – .882 =.8456
D. The mean and the variance of 3X+4.
E(3X+4) = 3EX+4 = 3(.88)+4 = 6.64
Var(3X+4) = 9 Var(x) = 9(.8456)= 7.6104
7. It is known from experience that 1.4 percent of the calls received by a
switchboard are wrong numbers. Use the Poisson approximation to the
binomial distribution to find the probability that among 150 calls received by the
switchboard two are wrong numbers.
λ = 150 (.014) = 2.1
e −2.1(2.1)2
P(X = 2) =
= .27
2!
8. Show that
A. If X is a geometric random variable P(X = x + n | x > n) = P(X = x)
P(X = x + n | X > n) =
P(X = x + n)
(1 − p)x + n −1p
= ∞
P(X > n)
∑ (1− p)x −1p
x = n +1
=
.
x + n −1
(1 − p)
p
= (1 − p)x −1p = P(X = x)
n
(1 − p) p / p
B. Show that p(x + 1,λ) =
parameter λ.
p(x + 1, λ ) =
λ
p(x,λ) where p(x,λ) is a Poisson distribution with
x +1
e − λ λx +1
λ e − λ λx
λ
=
⋅
=
p(x, λ )
(x + 1)! x + 1 x!
x +1
9. A tournament has five rounds. The rounds are played until one of the two
teams wins three games. What is the expected length of the tournament
assuming that the teams are evenly matched?
Let X be the number of games that are played
3
3
1
 1  1
P(X = 3) =   +   =
 2  2
4
3
3
 3  1  1  3  1  1 3
P(X = 4) =       +       =
 2  2   2   2  2   2  8
3
 4  1  1  4  1  1
P(X = 5) =       +       =
 2  2   2   2  2   2 
8
3
EX = 3 ⋅
2
2
3
1
3
3
+ 4 ⋅ + 5 ⋅ = 4.125games
4
8
8
10. The continuous random variable X has cumulative distribution function
0
x2
F(x) =
4
1
x<0
0 ≤x≤2
x>2
Find
A. The probability density function.
 x
f(x) =  2
 0
0≤x≤2
otherwise
B. P(0 .5 ≤ X ≤ 1 .5)=F(1.5) –F(.5)=.5625 –.0625=.5
C. P(X > 1) =1–F(1) = 1–.25 = .75
D. The mean and the variance of 2X+3.
EX =
∫
EX =
2
x2
x3
dx =
2
6
2
0
∫
2
0
σ2 = 2 −
2
=
0
x3
x4
dx =
2
8
8 4
=
6 3
2
=2
0
16 2
=
9
9
4
17
E(2X + 3) = 2EX + 3 = 2( ) + 3 =
3
3
2 8
Var(2X + 3) = 4 ⋅ =
9 9