Download Exam 1 Solution 1. (10 pts) The following circuit operates if and only

Exam 1 Solution
1. (10 pts) The following circuit operates if and only if there is a path of functional devices
from left to right. The probability that each device functions is as shown. Assume that the
probability that a device is functional does not depend on whether or not other devices
are functional. What is the probability that the circuit operates?
Solution:
1-[1-(.9)(.8)(.7)][1-(.95)(.95)(.95)]=.9294
2.
x
a.
b.
c.
d.
e.
( 20 pts) (Controlling water hyacinth) Entomological engineers are continually searching
for new biological agents to control one of the world’s worst aquatic weeds, the water
hyacinth. An insect that naturally feeds on water hyacinth is the delphacid. The number
of delphacid eggs of water hyacinth blades is a random variable, X, with the following
probability distribution:
f(x)
F(x)
xf(x)
x2 f(x)
1
.40
.40
.40
2
.50
.90
1.00
3
?.06
.96
.18
4
.04
1.00
.16
Total
1.74
3.58
(4pts) Find the probability of having three eggs, f(3)? =0.06
(4ts) Compute the cumulative distribution function F(2)=.90
(4pts) Compute the probability P(1<X 3)=.96-.40=.56 or .50+.06=.56
(4pts) Calculate the expectation of X=1.74
( 4 pts) Calculate the variance of X= 3.58-(1.74)2=.5524
3.
(14 pts) In an automated filling operation, when the process is run at slow speed, the
probability of an incorrect fill is .001. When the process is run at high speed, the
probability of an incorrect fill is 0.01. The process is run at high speed 30% of the time,
and at low speed 70% of the time. If an incorrectly filled container is found, what is the
probability that it was filled during high-speed operation?
Let I={process runs incorrect fil}, L={process runs at low speed}, H={process runs at high
speed}. Then P(I|L)=0.001, P(I|H)=0.01. P(H)=0.30 and P(L)=.70. We are looking for P(H|I).
(a) First we compute the P(I) using the total probability rule:
P(I)=P(I|H)P(H)+P(I|L)P(L)=(.01)(.30)+(.001)(.70)=.003+.0007=.0037.
P(H|I)=P(I|H)P(H)]/P(I)=(.01)(.30)/.0037=3/3.7=.8108
4. (10 pts) An oil drilling company ventures into various locations, and its success or
failure is independent from one location to another. Suppose the probability of a success
at any specific location is 0.25.
(a) ( 5 pts) What is the probability that a driller drills 10 locations and finds 1 success?
Solution: Let X be the number of drills needed to find the first location with oil. Then X is
geometric distributed with success probability 0.25.
P(X=10)=(.75)9(.25)=0.0188
(b) ( 5 pts) What is the expected number of locations to drill to find oil?
E(X)=1/p=1/.25=4 locations
5. (10 pts) It is conjectured that an impurity exists in 30% of all drinking wells in a certain
rural community. In order to gain some insight on this problem, it is determined that
some tests should be made. It is too expensive to test all of the many wells in the area, so
10 were randomly selected for testing. Assume the wells are independent.
(a) What is the probability that exactly three wells selected have the impurity
assuming that the conjecture is correct?
Let X be the number of wells selected having impurity. Then X is binomial distributed
with n=10 and p=.30.
P(X=3)=F(3)-F(2)=.6496-.3828=.2668
(b) What is the probability that more than three wells are impure?
P(X>3)=1-P(X<=3)=1-.6496=.3504.
6. (20 pts) The number of customers arriving per hour at a certain automobile service
facility is assumed to follow a Poisson distribution with mean =7.
(a) What is the mean number of arrivals during a 2-hour period?
Let X be the number of customers arriving in a two hour period, the mean number of
arrivals is 2 times 7=14 customers and X~Poisson (14).
(b) Compute the probability at least one arrivals in a 2-hour period.
P(X>=1)=1-P(X=0)=1-e-14 (140)/0!=.9999
7. (12 points) The tensile strength of paper is modeled by a normal distribution with a mean
of 35 pounds per square inch and a standard deviation of 2 pounds per square inch. Use Ztable to answer the following questions
(a) What is the probability that the tensile strength of a randomly selected piece of paper
is less than 40 lbs/in2?
P(X<40)=P(Z<(40-35)/2=P(Z<2.50)=0.993790
(b) What is the 90th percentile of the tensile strength of paper?
Step 1: The 90th percentile of Z is z=1.28
Step 2: The 90th percentile of X is x=35+2(1.28)=37.56 lb/in2.