chapter

Continuous and Standard Probability Distributions Practice Questions for GATE CS

Solve 48+ Continuous and Standard Probability Distributions practice questions for GATE CS with answers and detailed solutions. Free sample questions below.

Try a question

Answer it here to see how it works. Nothing is recorded until you sign in.

Question 1

Level 1: Warm-up

For a normal random variable X \sim N (0, 9), the standard deviation of X is ______.

Question 2

Level 1: Warm-up

The mean lifetime of a component is 4 hours. If the lifetime follows an exponential distribution, the variance of the lifetime (in hours^{2}) is ______.

Question 3

Level 1: Warm-up

The probability density function of a random variable X is given by f (x) = \frac{1}{σ 2 π} e^{- \frac{( x - μ ) ^{2}}{2 σ ^{2}}} . This represents which distribution?

Question 4

Level 1: Warm-up

Let X be an exponential random variable with parameter λ = 4 . The value of the probability density function f (x) at x = 0 is ______.

Question 5

Level 1: Warm-up

Given the probability density function f (x) = C \cdot exp (- \frac{( x - 5 ) ^{2}}{50}) for x \in (- \infty, \infty), the mean of the random variable X is ______.

Question 6

Level 1: Warm-up

For a normal random variable X \sim N (5, 16), the variance of X is ______.

Question 7

Level 1: Warm-up

Let X be an exponential random variable with parameter λ = 5 . The value of the probability density function f (x) at x = - 1 is ______.

Question 8

Level 1: Warm-up

A continuous random variable X has PDF f (x) = C x for 0 \leq x \leq 2 and 0 otherwise. The value of the constant C is ______.

Question 9

Level 1: Warm-up

The maximum value of the probability density function f (x) = 2 e^{- 2 x} for x \geq 0 is ______.

Question 10

Level 1: Warm-up

Let f (x) = 2 x for 0 \leq x \leq 1 and 0 otherwise. The probability P (0.5 \leq X \leq 1) is ______.

Free preview ends here

Login to view the complete practice questions and solutions

Creating an account is free. You get the rest of this chapter, step-by-step solutions, and a study plan built around the topics you are actually weak at.

Why MastersUp

Personalised first. High quality throughout.

Most platforms hand everyone the same content. Here the content moves with your performance, topic by topic.

Built around you, not around a syllabus PDF

Every answer you give moves your topic-level intelligence rate. The next question, the next revision card and tomorrow's plan all change with it.

Revision that hits your weak spots

We only revise topics you have actually attempted and are still below the safe bar on — never the same chapter on repeat.

Questions calibrated to the real exam

Each question carries a measured toughness. You are served a rung above your current level, so practice keeps stretching you.

Notes written for recall, not for volume

Full lesson cards for first study, curated short-note cards for the last mile — with derivations, traps and exam patterns marked.

One place for everything

Notes, chapter practice, previous-year questions, test series and full-length papers — all feeding one picture of your preparation.

Honest progress

No vanity streaks. Progress here means chapters mastered and accuracy that held up on harder questions.

Unlock the whole course

Full notes and short notes, the complete question bank with worked solutions, mock tests, full-length papers, and an adaptive plan that rebuilds itself as you improve.

View all plans Start free

Continuous and Standard Probability Distributions Practice Questions for GATE CS

Solve 48+ Continuous and Standard Probability Distributions practice questions for GATE CS with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Continuous and Standard Probability Distributions

Chapter Journey

Continuous & Standard Probability Distributions

What you will master: Understanding the shape, parameters, and applications of key continuous distributions used in engineering and computer science.

1. Exponential Lifetime Distribution Modeling time between events, memoryless property. Key for reliability contexts.
2. Normal Distribution Identification Recognizing the Gaussian PDF, mean, and variance. High weightage.
3. Density Normalization & Interval Probabilities Finding constants for valid PDFs and integrating over intervals. Fundamental skill.

Topic Hero: Exponential Lifetime Distribution

Topic 1 of 3

Exponential Lifetime Distribution

The exponential distribution models the time elapsed until a specific event occurs. It is widely used in reliability engineering and queuing theory to represent "waiting times" or component lifespans.

Key Learning Points:

PDF and CDF formulas
Mean, Variance, and Memoryless Property
Solving classic exam problems

Continuous and Standard Probability Distributions: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Engineering Mathematics MCQ

For a normal random variable $X \sim N (0, 9)$ , the standard deviation of $X$ is ______.

A.
9
B.
3
C.
81
D.
\sqrt{3}

Correct Answer:

Step-by-Step Solution

Key idea: In the notation $N (μ, σ^{2})$ , the second parameter is the variance, not the standard deviation.

Step 1: Recall the notation. $X \sim N (μ, σ^{2})$ means:

Mean = $μ$
Variance = $σ^{2}$
Standard deviation = $σ = σ^{2}$

Step 2: Given $X \sim N (0, 9)$ :

Variance = $σ^{2} = 9$
Standard deviation = $σ = 9 = 3$

Answer: 3

Question 2 · Engineering Mathematics MCQ

The mean lifetime of a component is 4 hours. If the lifetime follows an exponential distribution, the variance of the lifetime (in hours $^{2}$ ) is ______.

A.
4
B.
8
C.
16
D.
2

Correct Answer:

Step-by-Step Solution

Key idea: For an exponential distribution, the variance equals the square of the mean.

Step 1: Recall the relationship. For $X \sim Exp (λ)$ :

$E [X] = \frac{1}{λ}, Var (X) = \frac{1}{λ ^{2}} = (E [X])^{2}$

Step 2: Given $E [X] = 4$ hours, compute the variance.

$Var (X) = 4^{2} = 16 hours^{2}$

Answer: 16

Question 3 · Engineering Mathematics MCQ

The probability density function of a random variable $X$ is given by $f (x) = \frac{1}{σ 2 π} e^{- \frac{( x - μ ) ^{2}}{2 σ ^{2}}}$ . This represents which distribution?

A.
Exponential
B.
Normal
C.
Poisson
D.
Uniform

Correct Answer:

Step-by-Step Solution

Key idea: The given formula is the standard definition of the Normal (Gaussian) probability density function.

Step 1: Identify the structure. The presence of $e^{- quadratic term}$ and the specific coefficient $\frac{1}{σ 2 π}$ are the unique signatures of the Normal distribution.

Step 2: Compare with other distributions. Exponential has $e^{- λ x}$ , Poisson is discrete, Uniform is constant. None match this form.

Answer: Normal

Question 4 · Engineering Mathematics MCQ

Let $X$ be an exponential random variable with parameter $λ = 4$ . The value of the probability density function $f (x)$ at $x = 0$ is ______.

A.
4
B.
0
C.
1
D.
0.25

Correct Answer:

Step-by-Step Solution

Key idea: The exponential PDF is $f (x) = λ e^{- λ x}$ for $x \geq 0$ .

Step 1: Identify the formula. For exponential distribution, $f (x) = λ e^{- λ x}$ .

Step 2: Substitute $x = 0$ and $λ = 4$ .

$f (0) = 4 \cdot e^{- 4 \cdot 0} = 4 \cdot e^{0} = 4 \cdot 1 = 4$

Answer: 4

Question 5 · Engineering Mathematics MCQ

Given the probability density function $f (x) = C \cdot exp (- \frac{( x - 5 ) ^{2}}{50})$ for $x \in (- \infty, \infty)$ , the mean of the random variable $X$ is ______.

A.
50
B.
25
C.
5
D.
10

Correct Answer:

Step-by-Step Solution

Key idea: In the normal PDF form $f (x) = C \cdot exp (- k (x - μ)^{2})$ , the mean $μ$ is the value that makes the squared term zero.

Step 1: Compare with the standard normal PDF form:

$f (x) = \frac{1}{σ 2 π} exp (- \frac{( x - μ ) ^{2}}{2 σ ^{2}})$

Step 2: Match the exponent. Given:

$- \frac{( x - 5 ) ^{2}}{50}$

This matches $- \frac{( x - μ ) ^{2}}{2 σ ^{2}}$ with $μ = 5$ .

Step 3: The mean is $μ = 5$ .

Answer: 5

Question 6 · Engineering Mathematics MCQ

For a normal random variable $X \sim N (5, 16)$ , the variance of $X$ is ______.

A.
5
B.
4
C.
16
D.
256

Correct Answer:

Step-by-Step Solution

Key idea: In the notation $N (μ, σ^{2})$ , the second parameter represents the variance directly.

Step 1: Recall the notation convention. $X \sim N (μ, σ^{2})$ means Mean = $μ$ and Variance = $σ^{2}$ .

Step 2: Extract the values. Given $N (5, 16)$ :

$μ = 5$
$σ^{2} = 16$

Step 3: The question asks for variance, which is 16.

Answer: 16

Question 7 · Engineering Mathematics MCQ

Let $X$ be an exponential random variable with parameter $λ = 5$ . The value of the probability density function $f (x)$ at $x = - 1$ is ______.

A.
5e^5
B.
0
C.
5e^{-5}
D.
1

Correct Answer:

Step-by-Step Solution

Key idea: The exponential PDF is zero for $x < 0$ .

Step 1: Recall the piecewise definition:

$f (x) = {λ e^{- λ x} 0 x \geq 0 x < 0$

Step 2: Since $x = - 1 < 0$ , we use the second case.

$f (- 1) = 0$

Answer: 0

Question 8 · Engineering Mathematics MCQ

A continuous random variable $X$ has PDF $f (x) = C x$ for $0 \leq x \leq 2$ and $0$ otherwise. The value of the constant $C$ is ______.

A.
1/2
B.
1
C.
2
D.
1/4

Correct Answer:

Step-by-Step Solution

Key idea: Use the normalization condition $\int f (x) d x = 1$ to solve for the unknown constant.

Step 1: Set up the integral over the non-zero domain $[0, 2]$ .

$\int_{0}^{2} C x d x = 1$

Step 2: Evaluate the integral.

$C [\frac{x ^{2}}{2}]_{0}^{2} = 1$

$C (\frac{4}{2} - 0) = 1$

$2 C = 1$

Step 3: Solve for C.

$C = \frac{1}{2}$

Answer: 1/2

Question 9 · Engineering Mathematics MCQ

The maximum value of the probability density function $f (x) = 2 e^{- 2 x}$ for $x \geq 0$ is ______.

A.
2
B.
1
C.
0
D.
e^{-2}

Correct Answer:

Step-by-Step Solution

Key idea: The exponential PDF is a decreasing function, so its maximum occurs at the left boundary $x = 0$ .

Step 1: Recognize that $f (x) = 2 e^{- 2 x}$ is a decreasing exponential function for $x \geq 0$ .

Step 2: The maximum value occurs at the smallest $x$ , which is $x = 0$ .

$f (0) = 2 e^{- 2 \cdot 0} = 2 e^{0} = 2 \cdot 1 = 2$

Answer: 2

Question 10 · Engineering Mathematics MCQ

Let $f (x) = 2 x$ for $0 \leq x \leq 1$ and $0$ otherwise. The probability $P (0.5 \leq X \leq 1)$ is ______.

A.
0.25
B.
0.5
C.
0.75
D.
1

Correct Answer:

Step-by-Step Solution

Key idea: Calculate the area under the PDF curve between the specified limits.

Step 1: Set up the definite integral for the interval $[0.5, 1]$ .

$P (0.5 \leq X \leq 1) = \int_{0.5}^{1} 2 x d x$

Step 2: Evaluate the integral.

$[x^{2}]_{0.5}^{1}$

Step 3: Substitute limits.

$1^{2} - (0.5)^{2} = 1 - 0.25 = 0.75$

Answer: 0.75

Try a question

Login to view the complete practice questions and solutions

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

Continuous and Standard Probability Distributions Practice Questions for GATE CS

Chapter Roadmap: Continuous and Standard Probability Distributions

Chapter Journey

Topic Hero: Exponential Lifetime Distribution

Exponential Lifetime Distribution

Continuous and Standard Probability Distributions: Solved Questions with Step-by-Step Explanations (10 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

More practice questions in this unit