ISE 130 - Engineering Probability and Statistics| Questions And Answers | With Complete Solution

ISE 130 (Fall 2025)


ISE 130 - Engineering Probability and Statistics
Homework 2
Due: Sep 20, 2025 @ 11:59 PM

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Problem 1 (10 points):
The sample space of a random experiment is {a, b, c, d, e, f}, and each outcome is equally
likely. A random variable is defined as follows:
Outcome a b c d e f
X 0 0 1.5 1.5 2 3
Determine the probability mass function of random variable X. Use the probability mass
function to determine the following probabilities.
(a) P(X = 1.5), (b) P(0.5< X < 2.7), (c) P(X > 3), (d) P(0 ≤ X < 2), and (e) P(X = 0 or X = 2)

X 𝒇(𝒙𝒊 ) = 𝑷(𝑿 = 𝒙𝒊 )
Solutions:
0 1/3
𝑓𝑥 (0) = 𝑃(𝑋 = 0) = 1/6 + 1/6 = 1/3
1.5 1/3
𝑓𝑥 (1.5) = 𝑃(𝑋 = 1.5) = 1/6 + 1/6 = 1/3
𝑓𝑥 (2) = 𝑃(𝑋 = 2) = 1/6 2 1/6
𝑓𝑥 (3) = 𝑃(𝑋 = 3) = 1/6 3 1/6


(a) P(X = 1.5) = 1/3
(b) P(0.5< X < 2.7) = P(X = 1.5) +P(X = 2) = 1/3 + 1/6 = 1/2
(c) P(X > 3) = 0
(d) P(0 ≤ X < 2) = P(X = 0) +P(X = 1.5) = 1/3 + 1/3 = 2/3
(e) P(X = 0 or X = 2) = 1/3 + 1/6 = 1/2


Problem 2 (5 points):
Determine the cumulative distribution function of a random variable in Problem 1.


Solutions:
0 𝑥<0
1/3 0 ≤ 𝑥 < 1.5
𝐹(𝑥) = 𝑃(𝑋 ≤ 𝑥𝑖 ) = 2/3 1.5 ≤ 𝑥 < 2
5/6 2≤𝑥<3
{1 𝑥≥3


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, ISE 130 (Fall 2025)


Problem 3 (5 points):
Let f(x)= 2x/20,for x=0,1,2,3,4 be the probability mass function of a random variable X.
(a) P (X = 2), (b) P(X ≤ 2), (c) P(1 ≤ X < 3), and (d) P(X > -8)


Solutions:
(a) P(X = 2) = 2*2/20 = 1/5
(b) P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0 + 2/20 + 4/20 = 3/10
(c) P(1 ≤ X < 3)= P(X = 1) + P(X = 2) = 3/10
(d) P(X > -8) = 1



Problem 4 (10 points):
Errors in an experimental transmission channel are found when the transmission is checked by
a certifier that detects missing impulses. The number of errors found in an eight bit-byte is a
random variable with the following distribution:
0.0 𝑥<1
0.6 1≤𝑥<3
𝐹(𝑥) = {
0.9 3 ≤𝑥<7
1.0 7≤𝑥
(a) P(X ≤ 4), (b) P(X > 7), (c) P(X ≤ 3), (d) P(X > 3), (e) P(2 < X ≤ 4)

X 𝒇(𝒙𝒊 ) = 𝑷(𝑿 = 𝒙𝒊 )
Solutions:
1 0.6
𝑓𝑥 (1) = 𝑃(𝑋 = 1) = 0.6
3 0.3
𝑓𝑥 (3) = 𝑃(𝑋 = 3) = 0.9 − 0.6 = 0.3
7 0.1
𝑓𝑥 (7) = 𝑃(𝑋 = 7) = 1 − 0.9 = 0.1



(a) P(X ≤ 4) = P(X = 1) + P(X = 3) = 0.6 + 0.3 = 0.9
(b) P(X > 7) = 0
(c) P(X ≤ 3) = P(X = 1) + P(X = 3) = 0.6 + 0.3 = 0.9
(d) P(X > 3) = P(X = 7) = 0.1
(e) P(2 < X ≤ 4) = P(X = 3) = 0.3




This study source was downloaded by 100000820853758 from CourseHero.com on 12-03-2025 06:03:46 GMT -06:00


https://www.coursehero.com/file/251905289/HW2-Solutionspdf/