s t a n b u l K ü l t ü r U n i v e r s i t y Faculty of Engineering
IE 3101 - IE310 - MAT003 - MBT5001
FINAL EXAM
Fall 2012-2013
SOLUTIONS
Directions You have 90 minutes to complete the exam. Please do not leave the examination room in the rst 30 minutes of the exam. There are six questions, of varying credit (100 points total). Indicate clearly your nal answer to each question. You are allowed to use a calculator. During the exam, please turn o your cell phone(s).
You cannot use the book or your notes. You have one page for cheat-sheet notes at the end of the exam papers. The answer key to this exam will be posted on Department of Mathematics and Computer Science board after the exam.
Good luck! Emel Yavuz Duman, PhD.
Question 1. Question 4.
Question 2. Question 5.
Question 3. Question 6.
TOTAL
stanbul Kültür University
Que ion 1. 15points
When coin 1 is ipped, it lands heads with probability 0.4; when coin 2 is ipped, it lands heads with probability 0.7. One of these coins is randomly chosen and ipped 10 times. What is the probability that exactly 7 of the 10 ips land on heads?
Answer. The number of heads obtained when ipping coin 1 or coin 2 ten times is a Binomial random variable. So the probability that exactly 7 of the 10 ips land on heads is
1 2
(10 7
)
0.47(1− 0.4)10−7+1 2
(10 7
)
0.77(1− 0.7)10−7
= 1 2
10!
7!· 3!
(0.47· 0.63+ 0.77· 0.33)
= 0.15465.
Que ion 2. 20points
Ten percent of the population is left-handed. Use the normal approximation to the Binomial distribution to nd the probability that there are at least 60 left-handed students in a school of 400 students.
Answer. Let X be the number of left-handed students. Since
nθ = 400× 0.10 = 40 > 5 and n(1 − θ) = 400(1 − 0.10) = 400 × 0.9 = 360 > 5 then the binomial random variable X can be approximated by a normal distribution using a continuity correction. Therefore, for
µ = nθ = 400× 0.10 = 40 and σ =√
nθ(1− θ) =√
400× 0.10(1 − 0.10) = 6 we have
P (X ≥ 60) = P (X > 59.5) = P (
Z > 59.5− 40 6
)
= P (Z > 3.25)
= 0.5− 0.4994 = 6 × 10−4.
Que ion 3. 15points
Vehicles pass through a junction on a busy road at an average rate of 780 per hour.
Find the probability that only two passes in two minutes.
Answer. Let X be the number of vehicles. It is clear that X is a Poisson random variable. On the other hand, since the average number of vehicles per minute is
780 60 = 13, then
λ = 13× 2 = 26.
So, the probability that only two passes in two minutes is
P (X = 2) = p(2; 26) = 262× e−26
2! = 1.7269× 10−9.
IE 3101 - IE310 - MAT003 - MBT5001 2
stanbul Kültür University
Que ion 4. 20points
At the Baltimore aquarium, there are a total of 40 penguins 15 of which are King Penguins. Everyday at 2 pm, a group of 5 penguins is selected to perform at the Penguin Show. What is the probability that on a given day at least 2 King Penguins are selected for the Penguin Show?
Answer. Let X be the number of King Penguins that are selected for the show. Then X is a Hypergeometric random variable with parameters
N = 40, n = 5 and M = 15.
So, the probability that on a given day at least 2 King Penguins are selected for the Penguin Show is
P (X ≥ 2) = 1 − [P (X = 0) + P (X = 1)]
= 1− [h(0; 5, 40, 15) + h(1; 5, 4015)]
= 1− [(15
0
)(25
5
) (40
5
) + (15
1
)(25
4
) (40
5
) ]
= 1− 0.36911
= 0.63089.
Que ion 5. 15points
Susan is a sales representative who has a history of making a successful sale from about 80% of her sales contracts. If she makes 12 successful sales this week, she will get a bonus. Let X be a random variable representing the number of contracts needed for Susan to get the 12th sale. What are the expected value and standard deviation of X.
Answer. Since X has a negative Binomial distribution with parameters k = 12 and θ = 0.80
we have
µ = k
θ = 12
0.80 = 15, and
σ =
√ k θ
(1 θ − 1
)
=
√ 12 0.80
( 1 0.80 − 1
)
=
√15
2 = 1.9365.
Que ion 6. 15points
Suppose that X is a continuous random variable with probability density function f (x) =
{ 4
x5, x≥ 1, 0, x < 1.
Find P (0 < X < 1.2|0.5 < X < 1.5).
Answer.
P (0 < X < 1.2|0.5 < X < 1.5) = P (0 < X < 1.2 and 0.5 < X < 1.5) P (0.5 < X < 1.5)
= P (0.5 < X < 1.2)
P (0.5 < X < 1.5) = P (1 ≤ X < 1.2) P (1 ≤ X < 1.5)
=
∫1.2
1 4 x5dx
∫1.5 1
4
x5dx = −x−4|1.21
−x−4|1.51 = 1.2−4− 1−4 1.5−4− 1−4
= 671
1040 = 0.64519.
IE 3101 - IE310 - MAT003 - MBT5001 3
