ISE 130 Practice Exam #1 Fall 2025
Instructions: Closed Notes and Book
Total Points = 80
Name: ______________________
Points: ______________________
1. True (T) or false (F). (10)
(F) If P(A∩B) = P(A)P(B). then events A and B are mutually exclusive.
(T) For a discrete random variable, the probability mass function (PMF) assigns a
probability to each possible value.
(F) The geometric distribution has a fixed number of trials.
(T) In a normal distribution, about 68% of the data falls within one standard deviation
of the mean.
(T) Normal distribution is good for approximating binomial distribution when np > 5
and n(1-p) > 5.
2. Disks of polycarbonate plastic from a supplier are analyzed for scratch and shock
resistance. The results from 100 disks are summarized as follows:
Let A denote the event that a disk has high shock resistance, and let B denote the event that
a disk has high scratch resistance. If a disk is selected at random, determine the following
probabilities: (a) P(B); (b) P(A'); (c) P(AB); (d) P(AB); (e) P(A’B) (10)
Solutions:
(a) P(B) = 79/100 = 0.79
(b) P(A') = 14/100 = 0.14
(c) P(AB) = 70/100 = 0.70
(d) P(AB) = (70+9+16)/100 = 0.95
(e) P(A’B) = (70+9+5)/100 = 0.84
This study source was downloaded by 100000820853758 from CourseHero.com on 12-03-2025 06:00:53 GMT -06:00
https://www.coursehero.com/file/251905288/Practice-Exam1-Fall25-Solutionpdf/
Instructions: Closed Notes and Book
Total Points = 80
Name: ______________________
Points: ______________________
1. True (T) or false (F). (10)
(F) If P(A∩B) = P(A)P(B). then events A and B are mutually exclusive.
(T) For a discrete random variable, the probability mass function (PMF) assigns a
probability to each possible value.
(F) The geometric distribution has a fixed number of trials.
(T) In a normal distribution, about 68% of the data falls within one standard deviation
of the mean.
(T) Normal distribution is good for approximating binomial distribution when np > 5
and n(1-p) > 5.
2. Disks of polycarbonate plastic from a supplier are analyzed for scratch and shock
resistance. The results from 100 disks are summarized as follows:
Let A denote the event that a disk has high shock resistance, and let B denote the event that
a disk has high scratch resistance. If a disk is selected at random, determine the following
probabilities: (a) P(B); (b) P(A'); (c) P(AB); (d) P(AB); (e) P(A’B) (10)
Solutions:
(a) P(B) = 79/100 = 0.79
(b) P(A') = 14/100 = 0.14
(c) P(AB) = 70/100 = 0.70
(d) P(AB) = (70+9+16)/100 = 0.95
(e) P(A’B) = (70+9+5)/100 = 0.84
This study source was downloaded by 100000820853758 from CourseHero.com on 12-03-2025 06:00:53 GMT -06:00
https://www.coursehero.com/file/251905288/Practice-Exam1-Fall25-Solutionpdf/
