Category: Probability & Statistics Objective Questions

250+ TOP MCQs on Mathematical Expectation and Answers

Probability and Statistics Multiple Choice Questions & Answers (MCQs) on “Mathematical Expectation”.

1. The expectation of a random variable X(continuous or discrete) is given by _________
a) ∑xf(x), ∫xf(x)
b) ∑x2 f(x), ∫x2 f(x)
c) ∑f(x), ∫f(x)
d) ∑xf(x2), ∫xf(x2)
Answer: a
Clarification: The expectation of a random variable X is given by the summation (integral) of x times the function in its interval. If it is a continuous random variable, then summation is used and if it is discrete random variable, then integral is used.

2. Mean of a random variable X is given by _________
a) E(X)
b) E(X2)
c) E(X2) – (E(X))2
d) (E(X))2
Answer: a
Clarification: Mean is defined as the sum of the function in its domain multiplied with the random variable’s value. Hence mean is given by E(X) where X is a random variable.

3. Variance of a random variable X is given by _________
a) E(X)
b) E(X2)
c) E(X2) – (E(X))2
d) (E(X))2
Answer: c
Clarification: Variance of a random variable is nothing but the expectation of the square of the random variable subtracted by the expectation of X (mean of X) to the power 2. Therefore the variance is given by E(X2) – (E(X))2.

4. Mean of a constant ‘a’ is ___________
a) 0
b) a
c) a/2
d) 1
Answer: b
Clarification: Let f(x) be the pdf of the random variable X.
Now, E(a) = ∫af(x)
= a∫f(x)
= a(1) = a.

5. Variance of a constant ‘a’ is _________
a) 0
b) a
c) a/2
d) 1
Answer: a
Clarification: V(a) = E(a2) – (E(X))2
= a2 – a2
= 0.

6. Find the mean and variance of X?

x 0 1 2 3 4
f(x) 1/9 2/9 3/9 2/9 1/9

a) 2, 4/3
b) 3, 4/3
c) 2, 2/3
d) 3, 2/3
Answer: a
Clarification: Mean = (E(X) = ∑f(x) = 0(frac{1}{9}) + 1(frac{2}{9}) + 2(frac{3}{9}) + 3(frac{2}{9}) + 4(1/9) )
= 2
Variance ( = E(X^2)-(E(X))^2 = (0 + frac{2}{9} + frac{12}{9} + frac{28}{9} + frac{26}{9}) – 4 )
( = frac{4}{3} ).

7. Find the expectation of a random variable X?

x 0 1 2 3
f(x) 1/6 2/6 2/6 1/6

a) 0.5
b) 1.5
c) 2.5
d) 3.5
Answer: b
Clarification: (E(X) = 0(frac{1}{6}) + 1(frac{2}{6}) + 2(frac{2}{6}) + 3(frac{1}{6}) = 1.5. )

8. Find the expectation of a random variable X if f(x) = ke-x for x>0 and 0 otherwise.
a) 0
b) 1
c) 2
d) 3
Answer: b
Clarification: (int_0^∞ ke^{-x} dx = 1 )
kГ(1) = 1
k = 1
Now, (E(X) = int_0^∞ xe^{-x} dx = Г(2) = 1.)

9. Find the mean of a random variable X if f(x) = x – 5⁄2 for 0a) 3.5
b) 3.75
c) 2.5
d) 2.75
Answer: b
Clarification: (E(X) = int_0^1 (x-5/2)dx+∫_1^2(2x)dx+0 )
(= (frac{x^3}{3} – frac{5x^2}{4}) ) {from 0 to 1} ( + (frac{2x^3}{3}) ) {from 1 to 2}
(= frac{1}{3} – frac{5}{4} + frac{16}{3} – frac{2}{3} )
= 3.75.

10. Find the mean of a continuous random variable X if f(x) = 2e-x for x>0 and -ex for x<0.
a) 0
b) 1
c) 2
d) 3
Answer: d
Clarification: (E(X) = int_0^∞ 2xe^{-x} dx + int_{-∞}^0 xe^x dx )
= 2 Г(2) + Г(2) = 3.

11. What is moment generating function?
a) Mx(t) = E(etx)
b) Mx(t) = E(e-tx)
c) Mx(t) = E(e2tx)
d) Mx(t) = E(et)
Answer: a
Clarification: Moment generating function is nothing but the expectation of etX. So, the function is multiplied with etX before performing the integration or summation.

12. Find the Moment Generating Function of f(x) = x for 0a) ((frac{e^t-1}{t})^2 )
b) ((frac{e^{-t}-1}{t})^2 )
c) ((frac{e^{2t}-1}{t})^2 )
d) ((frac{e^{2t}-1}{t^2}) )
Answer: a
Clarification: Mx(t) = E(etx) = (int_0^1 xe^{tx} dx+int_1^2 (2-x) e^{tx} dx + 0 = (frac{e^t-1}{t})^2. )

13. E(X) = npq is for which distribution?
a) Bernoulli’s
b) Binomial
c) Poisson’s
d) Normal
Answer: b
Clarification: In binomial distribution, probability of success is given by p and that of failure is given by q and the event is done n times. The mean of this distribution is given by npq.

14. E(X) = λ is for which distribution?
a) Bernoulli’s
b) Binomial
c) Poisson’s
d) Normal
Answer: c
Clarification: In Poisson’s distribution, there is a positive constant λ which is the mean of the distribution and variance of the distribution.

15. E(X) = μ and V(X) = σ2 is for which distribution?
a) Bernoulli’s
b) Binomial
c) Poisson’s
d) Normal
Answer: d
Clarification: In Normal distribution, the mean and variance is given by μ and σ2 respectively. In case of standard normal distribution the mean is 0 and the variance is 1.

250+ TOP MCQs on Binomial Distribution and Answers

Probability and Statistics Multiple Choice Questions & Answers (MCQs) on “Binomial Distribution”.

1. In a Binomial Distribution, if ‘n’ is the number of trials and ‘p’ is the probability of success, then the mean value is given by ___________
a) np
b) n
c) p
d) np(1-p)
Answer: a
Clarification: For a discrete probability function, the mean value or the expected value is given by
Mean (μ)=(sum_{x=0}^{n}xp(x))
For Binomial Distribution P(x)=nCx px q(n-x), substitute in above equation and solve to get
µ = np.

2. In a Binomial Distribution, if p, q and n are probability of success, failure and number of trials respectively then variance is given by ___________
a) np
b) npq
c) np2q
d) npq2
Answer: b
Clarification: For a discrete probability function, the variance is given by
Variance (V)=(sum_{x=0}^n x^2p(x)-mu^2)
Where µ is the mean, substitute P(x)=nCx px q(n-x) in the above equation and put µ = np to obtain
V = npq.

3. If ‘X’ is a random variable, taking values ‘x’, probability of success and failure being ‘p’ and ‘q’ respectively and ‘n’ trials being conducted, then what is the probability that ‘X’ takes values ‘x’? Use Binomial Distribution
a) P(X = x) = nCx px qx
b) P(X = x) = nCx px q(n-x)
c) P(X = x) = xCn qx p(n-x)
d) P(x = x) = xCn pn qx
Answer: b
Clarification: It is the formula for Binomial Distribution that is asked here which is given by P(X = x) = nCx px q(n-x).

4. If ‘p’, ‘q’ and ‘n’ are probability pf success, failure and number of trials respectively in a Binomial Distribution, what is its Standard Deviation?
a) (sqrt{np})
b) (sqrt{pq})
c) (np)2
d) (sqrt{npq})
Answer: d
Clarification: The variance (V) for a Binomial Distribution is given by V = npq
Standard Deviation = (sqrt{variance} = sqrt{npq}).

5. In a Binomial Distribution, the mean and variance are equal.
a) True
b) False
Answer: b
Clarification: Mean = np
Variance = npq
∴ Mean and Variance are not equal.

6. It is suitable to use Binomial Distribution only for ___________
a) Large values of ‘n’
b) Fractional values of ‘n’
c) Small values of ‘n’
d) Any value of ‘n’
Answer: c
Clarification: As the value of ‘n’ increases, it becomes difficult and tedious to calculate the value of nCx.

7. For larger values of ‘n’, Binomial Distribution ___________
a) loses its discreteness
b) tends to Poisson Distribution
c) stays as it is
d) gives oscillatory values
Answer: b
Clarification: P(x)=(lim_{nrightarrowinfty}c_x^n p^x q^{n-x}=frac{e^{-m}m^x}{x!})
Where m = np is the mean of Poisson Distribution.

8. In a Binomial Distribution, if p = q, then P(X = x) is given by?
a) nCx (0.5)n
b) nCn (0.5)n
c) nCx p(n-x)
d) nCn p(n-x)
Answer: a
Clarification: If p = q, then p = 0.5
Substituting in P(x)=nCx px q(n-x) we get nCn (0.5)n.

9. Binomial Distribution is a ___________
a) Continuous distribution
b) Discrete distribution
c) Irregular distribution
d) Not a Probability distribution
Answer: b
Clarification: It is applied to a discrete Random variable, hence it is a discrete distribution.

250+ TOP MCQs on Hypergeometric Distributions and Answers

Probability and Statistics Multiple Choice Questions & Answers (MCQs) on “Hypergeometric Distributions”.

1. The mean of hypergeometric distribution is _____________
a) n*k / N-1
b) n*k-1 / N
c) n-1*k / N
d) n*k / N
Answer: d
Clarification: The mean of hypergeometric distribution is given as
E(X) = n*k /N where,
n is the number of trials, k is the number of success and N is the sample size.

2. The probability of success and failures in hypergeometric distribution is not fixed.
a) True
b) False
Answer: a
Clarification: In Binomial Distribution the probability of success and failures has to be fixed.
On the other hand hypergeometric distribution probability of success and failures is not fixed.

3. Consider selecting 6 cards from a pack of cards without replacement. What is the probability that 3 of the cards will be black?
a) 0.3320
b) 0.3240
c) 0.4320
d) 0.5430
Answer: a
Clarification: The given Experiment follows Hypergeometric distribution with
N = 52 since there are 52 cards in a deck.
k = 26 since there are 26 black cards in a deck.
n = 6 since we randomly select 6 cards from the deck.
x = 3 since 3 of the cards we select are black.
h(x; N, n, k) = [kCx] [N-kCn-x] / [NCn]
h(3; 52, 6, 26) = [26C3] [26C3] / [52C6]
h(3; 52, 6, 26) = 0.3320
Thus, the probability of randomly selecting 6 black cards is 0.3320.

4. Suppose we draw eight cards from a pack of 52 cards. What is the probability of getting less than three spades?
a) 0.985
b) 0.785
c) 0.685
d) 0.585
Answer: c
Clarification: The Random Experiment follows hypergeometric distribution with,
N = 52 since there are 52 cards in a deck.
k = 13 since there are 13 spades in a deck.
n = 8 since we randomly select 8 cards from the deck.
x = 0 to 2 since we want less than 3 spades.
h(x<3; N, n, k) = [kCx] [N-kCn-x] / [NCn]
where x ranges from 0 to 2.
h(x<3; 52, 8, 13) = [13C0] [39C8] / [52C8] + [13C1] [39C7] / [52C8] + [13C2] [39C6] / [52C2]
h(x<3; 52, 8, 13) = 0.685.

5. The Variance of hypergeometric distribution is given as __________
a) n * k * (N – k) * (N – 1) / [N2 * (N – 1)]
b) n * k * (N – k) * (N – n) / [N2 * (N – k)]
c) n * k * (N – 1) * (N – n) / [N2 * (N – 1)]
d) n * k * (N – k) * (N – n) / [N2 * (N – 1)]
Answer: d
Clarification: The variance of hypergeometric distribution is given as n * k * (N – k) * (N – n) / [N2 * (N – 1)] where,
n is the number of trials, k is the number of success and N is the sample size.

6. Hypergeometric probability of hypergeometric distribution function is given by the formula _________
a) h(x; N, n, k) = [kCx] [NCn-x] / [NCn]
b) h(x; N, n, k) = [kCx] [N-kCn-x] / [NCn]
c) h(x; N, n, k) = [kCx] [N-kCn] / [NCn]
d) h(x; N, n, k) = [kCx] [N-kCn-x] / [N-kCn]
Answer: b
Clarification: Hypergeometric probability of hypergeometric distribution function is given by the formula:
h(x; N, n, k) = [kCx] [N-kCn-x] / [NCn]
Where n is the number of trials, k is the number of success and N is the sample size.

7. Find the Variance of a Hypergeometric Distribution such that the probability that a 3-trial hypergeometric experiment results in exactly 2 successes, when the population consists of 7 items.
a) 0.6212
b) 0.6612
c) 0.6112
d) 0.6122
Answer: d
Clarification: The Variance of hypergeometric distribution is given as,
n * k * (N – k) * (N – 1) / [N2 * (N – 1)] where,
n is the number of trials, k is the number of success and N is the sample size.
Hence n = 3, k = 2, N = 7.
Var(X) = 0.6122.

8. Suppose we draw 4 cards from a pack of 52 cards. What is the probability of getting exactly 3 aces?
a) 0.9999
b) 0.9997
c) 0.0009
d) 0.0007
Answer: d
Clarification: The Random Experiment follows hypergeometric distribution with,
N = 52 since there are 52 cards in a deck.
k = 4 since there are 4 aces in a deck.
n = 4 since we randomly select 4 cards from the deck.
x = 3 since we want 3 aces.
h(x; N, n, k) = [kCx] [N-kCn-x] / [NCn]
h(3; 52, 4, 4) = [4C3] [48C1] / [52C4]
h(3; 52, 4, 4) = 0.0007.

9. The trials conducted in Hypergeometric distribution are done without replacement of the drawn samples.
a) True
b) False
Answer: a
Clarification: The trials conducted in Hypergeometric distribution are done without replacement of the drawn samples hence the probability of success and failure is not fixed.

10. Consider Nick draws 3 cards from a pack of 52 cards. What is the probability of getting no kings?
a) 0.8762
b) 0.7826
c) 0.8726
d) 0.7862
Answer: b
Clarification: The Random Experiment follows hypergeometric distribution with,
N = 52 since there are 52 cards in a deck.
k = 4 since there are 4 kings in a deck.
n = 3 since we randomly select 3 cards from the deck.
x = 0 since we want no kings.
h(x; N, n, k) = [kCx] [N-kCn-x] / [NCn]
h(0; 52, 4, 3) = [4C0] [48C3] / [52C3]
h(0; 52, 4, 3) = 0.7826.

11. Find the Variance of a Hypergeometric Distribution such that the probability that a 6-trial hypergeometric experiment results in exactly 4 successes, when the population consists of 10 items.
a) 14.4
b) 144
c) 1.44
d) 0.144
Answer: c
Clarification: The Variance of hypergeometric distribution is given as,
n * k * ( N – k ) * ( N – 1 ) / [N2 * (N – 1)] where,
n is the number of trials, k is the number of success and N is the sample size.
Hence n = 6, k = 4, N = 10.
Var(X) = 1.44.

12. Hypergeometric Distribution is Continuous Probability Distribution.
a) True
b) False
Answer: b
Clarification: Hypergeometric Distribution is a Discrete Probability Distribution. It defines the probability of k successes in n trials from N samples.

13. Emma likes to play cards. She draws 5 cards from a pack of 52 cards. What is the probability of that from the 5 cards drawn Emma draws only 2 face cards?
a) 0.0533
b) 0.0753
c) 0.0633
d) 0.6573
Answer: a
Clarification: The Random Experiment follows hypergeometric distribution with,
N = 52 since there are 52 cards in a deck.
k = 36 since there are 36 face cards in a deck.
n = 5 since we randomly select 5 cards from the deck.
x = 2 since we want 2 face cards.
h(x; N, n, k) = [kCx] [N-kCn-x] / [NCn]
h(2; 52, 5, 36) = [36C2] [12C3] / [52C5]
h(2; 52, 5, 36) = 0.0533.

14. Find the Expectation of a Hypergeometric Distribution such that the probability that a 4-trial hypergeometric experiment results in exactly 2 successes, when the population consists of 16 items.
a) 1/2
b) 1/4
c) 1/8
d) 1/3
Answer: a
Clarification: In Hypergeometric Distribution the Mean or Expectation E(X) is given as
E(X) = n*k /N
Here n = 4, k = 2, N = 16.
Hence E (X) = 1/2.

250+ TOP MCQs on Poisson Distribution and Answers

Probability and Statistics Multiple Choice Questions & Answers (MCQs) on “Poisson Distribution”.

1. In a Poisson Distribution, if ‘n’ is the number of trials and ‘p’ is the probability of success, then the mean value is given by?
a) m = np
b) m = (np)2
c) m = np(1-p)
d) m = p
Answer: a
Clarification: For a discrete probability function, the mean value or the expected value is given by
Mean(μ)=(sum_{x=0}^n xp(x))
For Poisson Distribution P(x)=(frac{e^{-m}m^x}{x!}) substitute in above equation and solve to get µ = m = np.

2. If ‘m’ is the mean of a Poisson Distribution, then variance is given by ___________
a) m2
b) m1⁄2
c) m
d) m⁄2
Answer: c
Clarification: For a discrete probability function, the variance is given by
Variance (v) = (sum_{x=0}^n x^2p(x)-mu^2)

Where µ is the mean, substitute P(x)=(frac{e^{-m}m^x}{x!}), in the above equation and put µ = m to obtain
V = m.

3. The p.d.f of Poisson Distribution is given by ___________
a) (frac{e^{-m}m^x}{x!})
b) (frac{e^{-m}x!}{m^x})
c) (frac{x!}{m^xe^{-m}})
d) (frac{e^m m^x}{x!})
Answer: a
Clarification: This is a standard formula for Poisson Distribution, it needs no explanation.
Even though if you are interested to know the derivation in detail, you can refer to any of the books or source on internet that speaks of this matter.

4. If ‘m’ is the mean of a Poisson Distribution, the standard deviation is given by ___________
a) (sqrt{m})
b) m2
c) m
d) m⁄2
Answer: a
Clarification: The variance of a Poisson distribution with mean ‘m’ is given by V = m, hence
Standard Deviation = (sqrt{variance} = sqrt{m})

5. In a Poisson Distribution, the mean and variance are equal.
a) True
b) False
Answer: a
Clarification: Mean = m
Variance = m
∴ Mean = Variance.

6. In a Poisson Distribution, if mean (m) = e, then P(x) is given by ___________
a) (frac{e^{(x-m)}}{x!})
b) (frac{e^{(m-x)}}{x!})
c) (frac{x!}{e^{(m-x)}})
d) (frac{x!}{e^{(x-m)}})
Answer: b
Clarification: P(x)=(frac{e^{-m}m^x}{x!})
Put m = e, and get correct solution.

7. Poisson distribution is applied for ___________
a) Continuous Random Variable
b) Discrete Random Variable
c) Irregular Random Variable
d) Uncertain Random Variable
Answer: b
Clarification: Poisson Distribution along with Binomial Distribution is applied for Discrete Random variable. Speaking more precisely, Poisson Distribution is an extension of Binomial Distribution for larger values ‘n’. Since Binomial Distribution is of discrete nature, so is its extension Poisson Distribution.

8. If ‘m’ is the mean of Poisson Distribution, the P(0) is given by ___________
a) e-m
b) em
c) e
d) m-e
Answer: a
Clarification: P(x)=(frac{e^{-m}m^x}{x!})
Put x = 0, to obtain e-m.

9. In a Poisson distribution, the mean and standard deviation are equal.
a) True
b) False
Answer: b
Clarification: In a Poisson Distribution,
Mean = m
Standard Deivation = m1⁄2
∴ Mean and Standard deviation are not equal.

10. For a Poisson Distribution, if mean(m) = 1, then P(1) is?
a) 1/e
b) e
c) e/2
d) Indeterminate
Answer: a
Clarification: P(x)=(frac{e^{-m}m^x}{x!})
Put m = x = 1, (given) to obtain 1/e.

11. The recurrence relation between P(x) and P(x +1) in a Poisson distribution is given by ___________
a) P(x+1) – m P(x) = 0
b) m P(x+1) – P(x) = 0
c) (x+1) P(x+1) – m P(x) = 0
d) (x+1) P(x) – x P(x+1) = 0
Answer: c
Clarification: P(x)=(frac{e^{-m}m^x}{x!})
p(x+1) = e-1 mx+1 /(x + 1)!
Divide P(x+1) by P(x) and rearrange to obtain (x+1) P(x+1) – m P(x) = 0.

250+ TOP MCQs on Normal Distribution and Answers

Probability and Statistics Multiple Choice Questions & Answers (MCQs) on “Normal Distribution”.

1. Normal Distribution is applied for ___________
a) Continuous Random Distribution
b) Discrete Random Variable
c) Irregular Random Variable
d) Uncertain Random Variable
Answer: a
Clarification: This is the rule on which Normal distribution is defined, no details on this as of why For more knowledge on this aspect, you can refer to any book or website which speaks on the same.

2. The shape of the Normal Curve is ___________
a) Bell Shaped
b) Flat
c) Circular
d) Spiked
Answer: a
Clarification: Due to the nature of the Probability Mass function, a bell shaped curve is obtained.

3. Normal Distribution is symmetric is about ___________
a) Variance
b) Mean
c) Standard deviation
d) Covariance
Answer: b
Clarification: Due to the very nature of p.m.f of Normal Distribution, the graph appears such that it is symmetric about its mean.

4. For a standard normal variate, the value of mean is?
a) ∞
b) 1
c) 0
d) not defined
Answer: c
Clarification: For a normal variate, if its mean = 0 and standard deviation = 1, then its called as Standard Normal Variate. Here, the converse is asked.

5. The area under a standard normal curve is?
a) 0
b) 1
c) ∞
d) not defined
Answer: b
Clarification: For any probability distribution, the sum of all probabilities is 1. Area under normal curve refers to sum of all probabilities.

6. The standard normal curve is symmetric about the value ___________
a) 0.5
b) 1
c) ∞
d) 0
Answer: d
Clarification: Normal curve is always symmetric about mean, for standard normal curve or variate mean = 0.

7. For a standard normal variate, the value of Standard Deviation is ___________
a) 0
b) 1
c) ∞
d) not defined
Answer: b
Clarification: If the mean and standard deviation of a normal variate are 0 and 1 respectively, it is called as standard normal variate. Here the converse is asked.

8. Normal Distribution is also known as ___________
a) Cauchy’s Distribution
b) Laplacian Distribution
c) Gaussian Distribution
d) Lagrangian Distribution
Answer: c
Clarification: Named after the one who proposed it. For further details, refer to books or internet.

9. Skewness of Normal distribution is ___________
a) Negative
b) Positive
c) 0
d) Undefined
Answer: c
Clarification: Since the normal curve is symmetric about its mean, its skewness is zero. This is a theoretical explanation for mathematical proofs, you can refer to books or websites that speak on the same in detail.

10. For a normal distribution its mean, median, mode are equal.
a) True
b) False
Answer: a
Clarification: It has a theoretical evidence that requires some serious background on several topics For more details you can refer to any book or website that speaks on the same.

11. In Normal distribution, the highest value of ordinate occurs at ___________
a) Mean
b) Variance
c) Extremes
d) Same value occurs at all points
Answer: a
Clarification: This is due the behaviour of the pdf of Normal distribution.

12. The shape of the normal curve depends on its ___________
a) Mean deviation
b) Standard deviation
c) Quartile deviation
d) Correlation
Answer: b
Clarification: This can be seen in the pdf of normal distribution where standard deviation is a variable.

13. The value of constant ‘e’ appearing in normal distribution is ___________
a) 2.5185
b) 2.7836
c) 2.1783
d) 2.7183
Answer: d
Clarification: This is a standard constant.

14. In Standard normal distribution, the value of mode is ___________
a) 2
b) 1
c) 0
d) Not fixed
Answer: c
Clarification: In a standard normal distribution, the value of mean is 0 and in normal distribution mean and mode coincide.

15. In Standard normal distribution, the value of median is ___________
a) 1
b) 0
c) 2
d) Not fixed
Answer: b
Clarification: In a standard normal distribution the value of mean is o and in normal distribution mean, median and mode coincide.

250+ TOP MCQs on Exponential Distribution and Answers

Probability and Statistics Multiple Choice Questions & Answers (MCQs) on “Exponential Distribution”.

1. The mean of exponential distribution is given as __________
a) 1/λ
b) λ
c) λ2
d) 1/λ2
Answer: a
Clarification: The mean of Exponential distribution is given as 1/λ and variance as 1/λ2.
E(X)=1/λ.

2. A mobile conversation follows a exponential distribution f (x) = (1/3)e-x/3. What is the probability that the conversation takes more than 5 minutes?
a) e-5/3
b) e-15
c) 5e-15
d) e-5/3
Answer: a
Clarification: f(x) = (1/3)e-x/3. The call should last more than 5 minutes so integrating from 5 till infinity we get
(frac{1}{3} ∫ (e^{-x/3}dx) = frac{1}{3}(frac{- e^{-5/3}}{-1/3}) )
= e-5/3.

3. Exponential distribution is bi-variate.
a) True
b) False
Answer: b
Clarification: Exponential distribution is uni-variate.
It is only defined for non-negative variables.

4. A random variable X has an exponential distribution with probability distribution function is given by
f(x)= 3e-3x for x>0 = 0 otherwise
Find probability that X is not less than 2.
a) e-3
b) e-6 – 3
c) e-6
d) e-6 – 1
Answer: c
Clarification: Probability of the function taking values from 2 to infinity.
P(X > 2) = 1 – P(X < 2) = Integrating the function from 0 to 2 we get
P(X = 1) = 1 – [3(e0 – e-6)/(-3)]
= e-6.

5. Consider a random variable with exponential distribution with λ=1. Compute the probability for P (X>3).
a) e-3
b) e-1
c) e-2
d) e-4
Answer: a
Clarification: The function takes values from 3 to infinity. This can be written alternatively as integrating from 0 to 3 and subtracting the whole from
P(X > 3) = 1 – P(X < 3)
= 1 – (1 – e-3)
= e-3.