Kingdom of Saudi Arabia
Ministry of Higher Education
Majmaah University
College of Engineering
Department: Basic Engineering Sciences
Exam Information
Course code:
Instructor:
Section number:
Exam time:
Exam hall:
Number of pages:
First Exam
STAT 201
Course Title:
Dr. Mukhtar Salah
1
Exam date:
From:
4:00 pm
To:
3D38
8
Total score:
Statistics and Probability
Mon. 27/10/2014
6:00 pm
20 Points
Student’s Information
Name:
ID:
Exam Instructions:
 Read the questions carefully and several times. Writing clearly will make
it possible to assess your answers correctly.
 Any form of cheating, intention to cheating, or disruption of the exam will
subject you to a disciplinary action as specified by the university’s
regulations.
 Carrying or using the mobile phone during the exam is not allowed. Using
the phone during the exam will carry a disciplinary action.
 The calculator is only permitted by the exam supervisor.
 No student can leave the exam room during the first half hour of the
exam. Any student coming after the first half hour of the exam will not be
permitted to enter the exam.
 A student leaving the exam, even for legitimate reasons, may not be given
a makeup exam.
Points Earned:
Question
Score
1
2
3
4
5
6
Total score in the exam:
Dear student, we encourage you to answer the exam the best you can and we
wish you all success.
1
Question 1:( 3pts)
A tire manufacturer wants to determine the inner diameter of a certain
grade of tire. Ideally, the diameter would be 570 mm. The data are as
follows:
572, 572, 573, 568, 569, 575, 565, 570.
(a) Find the sample mean and median.
(b) Find the sample variance, standard deviation, and range.
(c) Using the calculated statistics in parts (a) and (b), can you comment
on the quality of the tires?
2
Question 2:( 4points)
Following are measurements of electrical conductivity (in microsiemens
per centimeter) for 20 water samples given by the table below.
Class
10-20 21-31 32-42 43-53 54-64
Frequency
3
6
7
3
1
1) Draw the histogram for the data.
2)Find the standard deviation for the given data.
3
Question 3( 4points):
The following table gives a two-way classification, based on gender and
employment status, for a sample of 800 adults
Employed Unemployed
Male
332
30
Female 393
45
If one person is selected at random from this sample, find the probability
that this person is
a. unemployed
b. a female
c. employed given the person is male
d. female given that the person is unemployed.
4
Question 4 (3points) :
Ali is going to graduate from an industrial engineering department in a
university by the end of the semester. After being interviewed at two
companies he likes, he assesses that his probability of getting an offer
from company A is 0.8, and his probability of getting an offer from
company B is 0.6. If he believes that the probability that he will get offers
from both companies is 0.5.
1) what is the probability that he will get at least one offer from these two
companies?.
2) what is the probability that he will get the offer from company A but
not from company B.
5
Question 5 (3points) :
A certain industrial process is brought down for recalibration whenever
the quality of the items produced falls below specifications. Let X
represent the number of times the process is recalibrated during a week,
and assume that X has the following probability mass function.
0
1
2
3
4
𝑥
0.35
0.25
0.2
0.15
0.05
𝑃(𝑥)
1) Find the mean of 𝑋.
2) Find the variance of 𝑋.
3)What is the probability that X is less than 3.
6
Question 6 (3points) :
Elongation (in percent) of steel plates treated with aluminum are random
with probability density function
 x

f ( x)   250
0
20  x  30
otherwise
a. Find the variance of the elongations.
b. A particular plate elongates 28%. What proportion of plates elongate
less than this?
=============================================================
End of the Exam
Good Luck
Dr. Mukhtar Salah
7
Reference formulas
Discrete Random Variable
𝜇
Continuous Random Variable
∞
∑ 𝑥𝑃(𝑋 = 𝑥)
∫ 𝑥𝑓(𝑥)𝑑𝑥
𝑥
𝜎
2
2
∑ 𝑥 𝑃(𝑋 = 𝑥) − 𝜇
∞
2
−∞
∫ 𝑥 2 𝑓(𝑥)𝑑𝑥 − 𝜇 2
𝑥
−∞
𝑛
1
𝑆 =
(∑ 𝑥𝑖2 𝑓𝑖 − 𝑛(𝑥̅ )2 ) ,
𝑛−1
2
𝑥̅ =
∑𝑛𝑖=1 𝑥𝑖 𝑓𝑖
𝑛
𝑥̅ =
∑𝑛𝑖=1 𝑥𝑖
𝑛
𝑖=1
𝑛
1
𝑆 =
(∑ 𝑥𝑖2 − 𝑛(𝑥̅ )2 ) ,
𝑛−1
2
𝑖=1
𝑃(𝐴|𝐵) =
𝑃(𝐴 ∩ 𝐵)
𝑃(𝐵)
𝑃(𝐴 ∩ 𝐵 𝐶 ) = 𝑃(𝐴) − 𝑃(𝐴 ∩ 𝐵)
𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)
𝑍=
𝑋−𝜇
𝜎
𝑓(𝑥) = 𝜆𝑒 −𝜆𝑥
𝑇=
𝑋̅ − 𝜇
𝑠/√𝑛
∑𝑛𝑖=1 𝑥𝑖 𝑦𝑖 −
𝑟=
√∑𝑛𝑖=1 𝑥𝑖2 −
𝛽̂1 =
2
(∑𝑛
𝑖=1 𝑥𝑖 )
𝑛
𝑛
∑𝑛
𝑖=1 𝑥𝑖 ∑𝑖=1 𝑦𝑖
𝑛
𝑛
√∑𝑛𝑖=1 𝑦𝑖2 − (∑𝑖=1 𝑦𝑖 )
𝑛
∑𝑛𝑖=1 𝑥𝑖 𝑦𝑖 −
𝑛
∑𝑛
𝑖=1 𝑥𝑖 ∑𝑖=1 𝑦𝑖
∑𝑛𝑖=1 𝑥𝑖2 −
𝑛
(∑𝑛
𝑖=1 𝑥𝑖 )
𝛽̂0 = 𝑦̅ − 𝛽̂1 𝑥̅
8
𝑛
2
2