1. Hypergeometric Distribution is Continuous Probability Distribution.
a) True
b) False
Explanation: Hypergeometric Distribution is a Discrete Probability Distribution. It defines the probability of k successes in n trials from N samples
2. 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
Explanation: 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.
3. 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
Explanation: 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.
4. 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
Explanation: 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.
5. If ‘m’ is the mean of a Poisson Distribution, then variance is given by ___________
a) m2
b) m1⁄2
c) m
d) m⁄2
Explanation: 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.
6. 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) \(\frac{e^m m^x}{x!}\)
Explanation: 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.
7. 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
Explanation: The variance of a Poisson distribution with mean ‘m’ is given by V = m, hence
Standard Deviation = \(\sqrt{variance} = \sqrt{m}\)
8. In a Poisson Distribution, the mean and variance are equal.
a) True
b) False
Explanation: Mean = m
Variance = m
∴ Mean = Variance.
9. 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) \(\frac{x!}{e^{(x-m)}}\)
Explanation: P(x)=\(\frac{e^{-m}m^x}{x!}\)
Put m = e, and get correct solution.
10. Poisson distribution is applied for ___________
a) Continuous Random Variable
b) Discrete Random Variable
c) Irregular Random Variable
d) Uncertain Random Variable
Explanation: 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.