1. Management of an airline knows that 0.5% of the airline's passengers lose their luggage
on domestic flights. Management also know that the average value claimed for a lost
piece of luggage on domestic flights is $600.00. The company is considering increasing
fares by an appropriate amount to cover expected compensation to passengers who lose
their luggage. By how much should the airline increase fares? Why? Explain, using the
ideas of a random variable and its expectation.
Solution
Given that 0.5% of the airline's passengers lose their luggage on domestic flights.
That is five among the 100 pasengers lose their luggages. We also have the additional
information that the average value claimed for a lost piece of luggage on domestic flights
is $600.00. average is nothing but the total of the amount claimed for five lost luggages
divided by five.
Let x1,x2,x3, …………x100 be the random variables denoting the hundred passengers.
And let us assume that let x1,x2,x3,x4,x5 be the five random variables denoting the
passengers who lose their luggages
According to the mathematical expectation E(x)=xi*pi ,1=1,2,3,4,5
Pi is nothing but the probability corresponding to the five passengers.
Which is nothing but 1/5 i.e p1=p2=p3=p4=p5=1/5
Now
E(x)=x1*1/5+x2*1/5+x3*1/5+x4*1/5+x5*1/5
={x1+x2+x3+x4+x5}/5
From the problem we know that E(x)=$600
{x1+x2+x3+x4+x5}/5= 600
{x1+x2+x3+x4+x5}=600*5=3000
The lost luggage compensation is $3000 which has to be shared among the 100
passengers as fare .
3000/100=30
Hence the $30 has to be increased in the total fare to compensate the fare for lost
luggages.
2. A large shipment of computer chips is known to contain 10% defective chips. If 100 chips are
randomly selected, what is the expected number of defective ones? What is the standard
deviation of the number of defective chips? Use Chebyshev's theorem to give bounds such that
there is at least a 0.75 chance that the number of defective chips will be within the two bounds.
Solution
Given n=100
P=1/10
Q=1-(1/10)=1/9
Mean μ=np=100*1/10=10
s.d  =npq =100*1/10*9/10=9
by Chebyshev's theorem
p[ |X- μ |<k ]  1-(1/k2)
p[ |X-10 |<9k ]  1-(1/k2)
p{10-9k< X<10+9k }  1-(1/k2)
since in the problem it is given that there is at least a 0.75 chance that the number of
defective chips
we have p{10-9k< X<10+9k }  0.75
i.e 1-(1/k2) =0.75
1/k2=1-.75
1/k2=0.25
Therefore
k2=4
K=2
Now we have p{10-9*2< X<10+9*2 }  0.75
p{-8< X<18 }  0.75
hence the solution to the problem.
3. A mainframe computer in a university crashes on the average 0.71 times in a semester.
a. What is the probability that it will crash at least two times in a given semester?
b. What is the probability that it will not crash at all in a given semester?
c. The MIS administrator wants to increase the probability of no crash at all in a semester to at
least 90%. What is the largest u (mu) that will achieve this goal?
Solution:
Given
=np =0.71
Let the random variable x denotes the no. of crashes in a given semester
Then
a). probability that it will crash at least two times in a given semester
is given by the poisson distribution
p(X=x)=e-  x / x
p(X<=2)=p(X=0)+P(X=1)+P(X=2)
e-0.71
= 0.491644
p(X<=2)= 0.491644+0.34906+0.12391 =0.9646
b).the probability that it will not crash at all in a given semester IS GIVEN BY
P(x=0)= e-0.71
= 0.491644
c). the largest u (mu) that will achieve this goal will be
=np =0.11
since bypoisson distribution
which is a limiting case of binomial dist under the following conditions
1.n the no of trials is infinitely large.
2 p,the constant probability of success for each trial is indefinitely small i.e p tends to 0
3 =np is finite.
So achieve this goal p has to be minimized for the given value of n,hence by taking the
mean value to be 0.11 the probability of no crashes at all in a given semester is 90%
Hence the required value of mean is 0.11
4. A recent survey published in Business Week concludes that Gatorade commands an
80% share of the sports drink market versus 11% for Coco-Cola's Power Ade and 3% for Pepsi's
All Sport. A market research firm wants to conduct a new taste test for which it needs Gatorade
drinkers. Potential participants for the test are selected by random screening of drink users to find
Gatorade drinkers. What is the probability that
a. The first randomly selected drinker qualifies
b. Three soft drink users will have to be interviewed to find the first Gatorade drinker.
Solution:
Using Bayes theorem of probability this problem can be efficiently solved.
Let c1 c2 c3 denote command of Gatorade, Coco-Cola's Power& Pepsi's All Sport.
Let A denotes the drink users.
Then p(c1)=p(c2)=p(c3)=1/3
P(A/c1)=0.80
P(A/c2)=0.11
P(A/c3)=0.03
a).p(c1/A)={p(c1)*p(A/c1)}/ { P(ci)*P(A/ci)} where i=1,2,3
therefore
p(c1/A)={0.33*0.80}/{0.33*0.80+0.33*0.11+0.33*0.03}
=0.2640/0.3102 = 0.85107
b).let c denote the total probability to find Three soft drink users will have to be
interviewed to find the first Gatorade drinker.
P(C/A)= P(c1/A)+P(c2/A)+p(c3/A)
p(c1/A)={0.33*0.80}/{0.33*0.80+0.33*0.11+0.33*0.03}=0.85107
p(c2/A)={0.33*0.11}/{0.33*0.80+0.33*0.11+0.33*0.03}=0.0363/0.3102=0.1170
p(c3/A)={0.33*0.03}/{0.33*0.80+0.33*0.11+0.33*0.03}=0.0093/0.3102=0.0028
P(C/A)= P(c1/A)+P(c2/A)+p(c3/A)=0.85107+0.1170+0.0028=0.97095
Hence the total probability is 0.97095