Exponential Distribution MCQs

1. The length of time a person speaks over the phone follows an exponential distribution with a mean of 6 minutes. What is the probability that the person will talk for more than 8 minutes?

a) 0.28 b) 0.27 c) 0.39 d) None

Correct Answer: b) 0.27

Explanation:
Mean = 6 minutes, so λ = 1/6
P(X > 8) = e^(-λ*8) = e^(-(1/6)*8) = e^(-4/3) ≈ 0.27
Short Trick: For exponential distribution, P(X > x) = e^(-λx)

2. The exponential distribution is:

a) Symmetric b) Always decreasing c) Always increasing d) Bell-shaped

Correct Answer: b) Always decreasing

Explanation:
The exponential PDF is f(x) = λe^(-λx) for x ≥ 0
This function is highest at x = 0 and continuously decreases as x increases
Short Trick: The graph of exponential PDF always slopes downward from left to right

3. The exponential distribution is a special case of which distribution?

a) Binomial b) Normal c) Gamma d) Poisson

Correct Answer: c) Gamma

Explanation:
Exponential distribution is a gamma distribution with shape parameter α = 1
If X ~ Gamma(α, λ) and α = 1, then X follows Exp(λ)
Short Trick: Exponential is Gamma with shape parameter = 1

4. What is the MGF of an exponential distribution with rate parameter λ?

a) 1/(1 - λt) b) λ/(λ - t) c) e^(-λt) d) λe^t

Correct Answer: a) 1/(1 - λt)

Explanation:
MGF of exponential distribution is M(t) = ∫e^(tx)λe^(-λx)dx
= λ∫e^((t-λ)x)dx = λ/(λ-t) = 1/(1-t/λ) = 1/(1-λt) for t < λ
Short Trick: Memorize this standard MGF formula for exponential distribution

5. What is the mean of an exponential distribution with rate λ?

a) λ b) 1/λ c) λ^2 d) 1/λ^2

Correct Answer: b) 1/λ

Explanation:
Mean of exponential distribution = 1/λ
Can be calculated from the MGF: M'(0) = λ/(λ-t)² evaluated at t=0 = 1/λ
Short Trick: Mean = 1/rate parameter

6. What is the variance of an exponential distribution with rate λ?

a) λ b) 1/λ c) 1/λ² d) λ²

Correct Answer: c) 1/λ²

Explanation:
Variance of exponential distribution = 1/λ²
Can be calculated from the MGF: M''(0) - [M'(0)]² = 2/(λ²) - (1/λ)² = 1/λ²
Short Trick: Variance = (Mean)² = (1/λ)²

7. The exponential distribution is often used to model:

a) The number of successes in a fixed time b) Time between successive events c) Binomial failures d) Continuous symmetrical outcomes

Correct Answer: b) Time between successive events

Explanation:
Exponential distribution is commonly used for modeling waiting times
Specifically, the time between events in a Poisson process follows an exponential distribution
Short Trick: When events occur randomly over time (Poisson process), the time between events is exponentially distributed

8. Which of the following statements is true about the exponential distribution?

a) It has memory b) It has no memory (memoryless property) c) It is symmetric d) It is only used for normal processes

Correct Answer: b) It has no memory (memoryless property)

Explanation:
The exponential distribution has the memoryless property: P(X > s+t | X > t) = P(X > s)
This means the future waiting time doesn't depend on how long you've already waited
Short Trick: Exponential is the only continuous distribution with the memoryless property

9. A bank receives customers at an average rate of 3 customers per hour. The arrival of customers follows an exponential distribution. What is the probability that the next customer will arrive after 30 minutes?

a) 0.4451 b) 0.6691 c) 0.2231 d) 0.1141

Correct Answer: a) 0.4451

Explanation:
Rate = 3 customers/hour = 3/60 = 0.05 customers/minute
P(X > 30) = e^(-λ*t) = e^(-0.05*30) = e^(-1.5) ≈ 0.4451
Short Trick: Convert λ to the same units as time (minutes), then use e^(-λt)

10. Using the same data as above, what is the probability that the next customer arrives within 20 minutes?

a) 0.6321 b) 0.4351 c) 0.7321 d) 0.2361

Correct Answer: c) 0.7321

Explanation:
Rate = 3 customers/hour = 0.05 customers/minute
P(X ≤ 20) = 1 - P(X > 20) = 1 - e^(-0.05*20) = 1 - e^(-1) ≈ 1 - 0.3679 = 0.6321
Note: There seems to be an error in the options, correct answer should be close to 0.6321

11. The time (in minutes) between buses arriving at a station follows an exponential distribution with a mean of 12 minutes. What is the probability that a passenger will have to wait more than 15 minutes?

a) 0.4856 b) 0.2865 c) 0.8845 d) 0.6867

Correct Answer: b) 0.2865

Explanation:
Mean = 12 minutes, so λ = 1/12
P(X > 15) = e^(-λ*15) = e^(-(1/12)*15) = e^(-1.25) ≈ 0.2865
Short Trick: Time/Mean = 15/12 = 1.25, so probability is e^(-1.25)

12. Using the same scenario above, what is the probability that the next bus arrives within 10 minutes?

a) 0.1654 b) 0.2654 c) 0.4654 d) 0.5654

Correct Answer: d) 0.5654

Explanation:
Mean = 12 minutes, so λ = 1/12
P(X ≤ 10) = 1 - P(X > 10) = 1 - e^(-(1/12)*10) = 1 - e^(-5/6) ≈ 1 - 0.4346 = 0.5654
Short Trick: Time/Mean = 10/12 = 5/6, so probability is 1 - e^(-5/6)

13. If a device has an exponentially distributed lifetime with a mean of 5 hours, what is the rate λ?

a) 0.2 b) 5 c) 1 d) 0.5

Correct Answer: a) 0.2

Explanation:
For exponential distribution, mean = 1/λ
If mean = 5 hours, then λ = 1/5 = 0.2 per hour
Short Trick: λ = 1/mean

14. If the time until failure for a machine is exponential with rate λ=0.1, what is the probability that the machine lasts more than 10 hours?

a) 0.3679 b) 0.9048 c) 0.1353 d) 0.5488

Correct Answer: a) 0.3679

Explanation:
P(X > 10) = e^(-λ*10) = e^(-0.1*10) = e^(-1) ≈ 0.3679
Short Trick: e^(-1) ≈ 0.368 is a common value to memorize

15. Which of the following is the CDF of an exponential distribution with rate λ?

a) 1 - e^(-λx) b) e^(λx) c) 1/(1 + λx) d) λe^(-x)

Correct Answer: a) 1 - e^(-λx)

Explanation:
CDF of exponential distribution is F(x) = ∫(from 0 to x) λe^(-λt)dt = 1 - e^(-λx) for x ≥ 0
Short Trick: CDF = 1 - [survival function], where survival function is e^(-λx)

16. What is the domain of the moment generating function for an exponential distribution with rate λ?

a) t < 0 b) t < λ c) t < ∞ d) t > λ

Correct Answer: b) t < λ

Explanation:
The MGF of exponential distribution is M(t) = λ/(λ-t)
This is defined only when λ-t > 0, which means t < λ
Short Trick: For exponential distribution, MGF exists only when t < λ

17. The nth moment about the origin of the exponential distribution can be found by:

a) Differentiating the CDF b) Differentiating the MGF n times and evaluating at t=0 c) Integrating the PDF d) Squaring the variance

Correct Answer: b) Differentiating the MGF n times and evaluating at t=0

Explanation:
The nth moment E[X^n] can be found by taking the nth derivative of the MGF and evaluating at t=0
M^(n)(0) = E[X^n]
Short Trick: This is a general property of moment generating functions for any distribution

18. If X ~ Exp(λ=2), what is the value of E[X²]?

a) 0.5 b) 1 c) 2 d) 0.25

Correct Answer: a) 0.5

Explanation:
For exponential distribution, E[X²] = 2/λ²
With λ = 2, E[X²] = 2/(2²) = 2/4 = 0.5
Short Trick: E[X²] = 2×(Mean)² = 2×(1/λ)²

19. What is the MGF of X ~ Exp(λ)?

a) λ/(λ - t) for t < λ b) λ/(λ + t) c) e^(λt) d) λ^t

Correct Answer: a) λ/(λ - t) for t < λ

Explanation:
The MGF of exponential distribution is M(t) = E[e^(tX)] = ∫e^(tx)λe^(-λx)dx
= λ∫e^((t-λ)x)dx = λ/(λ-t) for t < λ
Short Trick: This is the standard form of the MGF for the exponential distribution

20. If the MGF of a distribution is 2/(2 - t), what is the expected value of the distribution?

a) 0.5 b) 1 c) 2 d) 4

Correct Answer: a) 0.5

Explanation:
The MGF 2/(2 - t) corresponds to an exponential distribution with λ = 2
The expected value is E[X] = 1/λ = 1/2 = 0.5
Short Trick: Compare with standard form λ/(λ-t) to identify λ

📘 Conceptual and Basic Properties

📊 Real-Life Scenarios and Applications

🧠 Theoretical and MGF-Based Questions

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