International University - VNUHCM
School of Industrial Engineering and Management (IEM) Midterm Examination
Date: Tuesday, Nov 09th 2021; Duration: 120 minutes
Open book; Online, Laptops/Cell-phone/… are not allowed.
SUBJECT: Engineering Probability & Statistics (ID: IS004IU)
Approval by the School/Department Lecturer: Signature Signature Full name: Dr. Tran Duc Vi, Dr. Phan Nguyen Ky Phuc
Full name: Hà Thị Xuân Chi Proctor 1 Proctor 2 Signature Signature Full name: Full name: STUDENT INFO
Student name: Huỳnh Thúy Vy
Student ID: IELSIU18178
INSTRUCTIONS: the total of point is 100 (equivalent to 30% of the course) 1. Purpose: •
Able to understand, calculate, and present basic statistics for a given dataset including numerical and (G1) •
Able to understand calculate probability and other attributes of discrete and continuous random variables (G2) •
Able to conduct the simple data exploration (G4) 2. Requirement: •
Discussion and material transfer are strictly prohibited. •
Submit your exam on Blackboard right after the exam time is over. HCMC National Unive rs ity
Student Name:………...………………………. Inte rnational Unive rs ity
Student ID:……………………………. - - - - - - - - - - - - QUESTIONS Question 1 (25 points)
Following are GPA of 20 students admitted to graduate program in School of Industrial Engineering and Management (IEM)
3.55, 3.65, 3.57, 3.83, 3.76, 3.78, 3.63, 3.88, 3.65, 3.59
3.53, 3.58, 3.65, 3.82, 3.81, 3.64, 3.57, 3.69, 3.80, 3.78
a. Represent the data in a stem and leaf plot (5 pts)Answer: Stem Leaf 3.5 5 7 9 3 8 7 3.6 5 3 5 5 4 9 3.8 8 3 2 1 0 3.7 6 8 8
b. Calculate the same mean . (5pts) = = 3.688
c. Calculate the sample standard deviation s. (5pts) = = 0.109
d. Determine the proportion of the data values that lies within ± 1.5 and compare with the lower
bound given by Chebyshev’s inequality. (5pts) = 0.109 ± 1.5 + 1.5s = - 1.5s = (3.5245,3,8515)
There are 19 out of 20 values lies within the range, that is of value lies between that points. By Chebyshev’sinequality: 1 - , k = 1.5 1 -
e. Draw the box plot from the data (5pts) Question 2 (20 points)
A man has 2 coins, one is fair coin and one is 2-head coin. He randomly picks a coin and toss it.
a) Find the probability that the face up value is Head (5 pts) 2/3 HCMC National Unive rs ity
Student Name:………...………………………. Inte rnational Unive rs ity
Student ID:……………………………. - - - - - - - - - - - -
Let C1 be the event that he chooses a regular coin and let C2 be the event that he chooses
the two-headed coin. Note that C1 and C2 form a partition of the sample space. We already
know that P(H|C1)=0.5, P(H|C2)=1.
Thus, we can use the law of total probability to write
P(H) = P(H|C1) x P(C1) + P(H|C2) x P(C2) = =
b) Given the face up value is Head what is the probability that it comes from the 2-head coin?
If the result is Head he will pick an item from factory 1. If the result is Tail, he will pick an
item from factory 2. The ratios of defective item in factory 1 and 2 are 50% and 60%. (5pts)
Applying the Bayes' rule for P(C2|H) P(C2|H) = =
c) Find the probability that the item he picks is defective. (5 pts)
d) Given the item he picks is defective, find the probability that this result is come from the toss of fair coin. (5pts) Question 3 (15 points)
We roll two fair 6-sided dice. Each one of the 36 possible outcomes is assumed to be equally likely.
(a) Given that the roll results in a sum of 4 or less, find the conditional probability that doubles arerolled. (5 pts)
Let A be the event that doubles were rolled and B be the event that the roll resulted in a sum of 4 or less. Compute: Pr[A|B] =
Each of the 6 doubles is equally likely, and only (1,1) and (2,2) sum to less than 4, so Pr[B|A] = .
There are 6 ways to roll and get sum ≤ 4, so Pr[B] = = . So Pr[A|B] = = .
(b) Find the probability that at least one die roll is a 5. (5 pts)
When only first dice has 5, the possible cases are (5,1) (5,2) (5,3) (5,4) (5,6)
When only second dice has 5, the possible cases are (1,5) (2,5) (3,5) (4,5) (6,5)
When both have 5,then the only case is (5,5), so number of favorable cases are 11
Total cases are 36, so probability that at least one die roll is a 5 =
(c) Given that the two dice land on different numbers, find the conditional probability that at least onedie roll is a 5. (5 pts)
Let A be the event that at least one die is a 5 and B be the event that two dice land on different numbers.
We are going to compute Pr[A|B] =
Simply enumerating out the possibilities, we see that Pr[A∩B] = = and Pr[B] = = . So, Pr[A|B] = = Question 4 (10 points)
At beginning at each month, a store always refills its stock so that the number of TVs in the store is
equal to 6. The target for the sale man is to sell 4 TVs / month. If he can sale more than 4, he will
receive a reward of 3 for each extra TV sale. Assume that the demand of televisions in a month follows
binomial distribution with n=6, p=0.6.
a) Find the probability that sale man can have the reward. (5 pts)
b) Find the expected reward of the sale man (5 pts) Question 5 (15 points)
Assume that the number of defect screws in a box follows Poisson distribution with = 2. The box will
pass the quality test if number of defect screw is smaller or equal to 3.
a) Find the probability passing the quality test of a box. (7pts)
b) In stead of checking each box, now two boxes are combined as a package. A package
willpass the quality test if number of defect screws is smaller or equal to 6. Find the
probability passing the quality test of a package. (8pts) Question 6 (15 pts)
Weekly demand for a product has a normal distribution with mean 1,500 and standard deviation of 100.
The current on hand inventory is 2,500 and no deliveries will be occurring in the next two weeks.
Assuming that the demands in different weeks are independent,
a) What is the probability that the demand in each of the next 2 weeks is less than 1,400? (7 pts)
Xi is a normal random variable (weekly demand), with parameters ,
The probability that the demand in each of the next 2 weeks is less than 1,400 P(X1X2): P(X1X2) = P = P = P = P = P ( = P ( = ( = 0.025
b) What is the probability that the total of the demands in the next 2 weeks exceeds 2,000? (8
pts)The probability that the total of the demands in the next 2 weeks exceeds 2,000 P(X1 + X2):
P(X1 + X2) = P (X1 + X2 > 2000) = P (X1 + X2 - > 2000 - ) = P ( > ) = P( > ) = P (z > -7.071) = 1 - P (z < -7.071) - END –