Discrete Probability Distributions Questions and Answers

Answer: D

The expectation is 3/2.

\(P(X = x) = \frac{2}{3}\left(\frac{1}{3}\right)^{x-1}\) is the geometric distribution counting trials up to the first success, with success probability \(p = \frac{2}{3}\). Its mean is \(1/p\):

\[ \begin{aligned} E(X) &= \sum_{x \ge 1} x\cdot\tfrac{2}{3}\left(\tfrac{1}{3}\right)^{x-1} \\ &= \tfrac{2}{3}\cdot\frac{1}{(1 - 1/3)^{2}} = \tfrac{2}{3}\cdot\tfrac{9}{4} = \tfrac{3}{2} \end{aligned} \]

The option 2/3 is \(p\) itself, 2/9 is \(p(1-p)\), and 1 would mean the variable always equals 1. Since \(X \ge 1\) with positive probability of larger values, its mean must exceed 1, which already rules out three options.

Answer: B

The distribution with mean equal to variance is the Poisson.

A Poisson distribution with parameter \(\lambda\) has mean \(\lambda\) and variance \(\lambda\), which is used as a quick check on whether count data (accidents, colonies on a plate, defects) follow it.

The binomial has mean \(np\) and variance \(npq\), which is always smaller than the mean because \(q \lt 1\). In the normal distribution the mean \(\mu\) and variance \(\sigma^{2}\) are independent parameters. A chi-square distribution with \(k\) degrees of freedom has mean \(k\) and variance \(2k\).

Answer: D

The probability is \(e^{-3.75}\).

The mean number of errors per page is \(390/520 = 0.75\), so in 5 pages the mean is \(m = 5 \times 0.75 = 3.75\). For a Poisson variable, the chance of zero events is \(e^{-m}\):

\[ P(X = 0) = \frac{e^{-m} m^{0}}{0!} = e^{-3.75} \]

The options with positive exponents are greater than 1 and cannot be probabilities. \(e^{-4.75}\) adds 1 to the mean by mistake.

Answer: A

Mean equals variance for a Poisson distribution.

For a Poisson distribution with parameter \(\lambda\), both the mean and the variance equal \(\lambda\). This is often used as a quick test of whether count data are Poisson.

For the binomial distribution, mean \(np\) and variance \(np(1 - p)\) differ unless \(p = 0\). A normal distribution has independent mean and variance parameters. A uniform distribution on \([a, b]\) has mean \(\frac{a+b}{2}\) and variance \(\frac{(b-a)^{2}}{12}\), equal only by coincidence.

Answer: C

The mean is \(\sqrt{6}\), about 2.45.

For a Poisson distribution with mean \(\lambda\), \(P(x) = e^{-\lambda}\lambda^{x}/x!\). Setting the probabilities of 1 and 3 equal:

\[ \begin{aligned} \frac{e^{-\lambda}\lambda^{1}}{1!} &= \frac{e^{-\lambda}\lambda^{3}}{3!} \\ \lambda &= \frac{\lambda^{3}}{6} \\ \lambda^{2} &= 6,\quad \lambda = \sqrt{6} \end{aligned} \]

The mean must be positive, so the negative root is rejected. \(\sqrt{2}\) and \(\sqrt{3}\) come from using 2 or 3 instead of 3! = 6 in the denominator. 7 has no basis in the working. Since a Poisson mean equals its variance, the variance is also \(\sqrt{6}\).

Answer: D

The probability is 0.99863.

For a binomial distribution, mean \(np = 4\) and variance \(npq = 4/3\). Dividing:

\[ \begin{aligned} q &= \frac{4/3}{4} = \frac{1}{3} \\ p &= 1 - q = \frac{2}{3} \\ n &= \frac{4}{2/3} = 6 \end{aligned} \]

\[ \begin{aligned} P(x \ge 1) &= 1 - P(x = 0) \\ &= 1 - q^{6} \\ &= 1 - \frac{1}{729} = 0.99863 \end{aligned} \]

0.00137 is \(P(x = 0) = 1/729\) itself, the answer to the complementary question. 0.71256 and 0.81343 do not correspond to this distribution.

Answer: D

The first moment of a distribution about its mean is zero. The key takes the moments in the usual central form, in which the first central moment is the average deviation from the mean, and this is always zero whatever the mean and standard deviation.

The mean of 9 would be the first moment about the origin, not the central one, and 3 and 1 do not come from the values given (the standard deviation is 1 and the variance is also 1). The answer is option D, as the key gives.

Answer: A

The second moment about the mean of the Poisson distribution is 4.

For a Poisson distribution the mean and the variance are equal, and the variance is the second moment about the mean. With a mean of 4, the variance is also 4.

The values 2, 1 and 0 are not equal to the mean, so they do not follow.

If the raw second moment about the origin were meant, it would be the variance plus the square of the mean, that is \(4+16=20\), which is not among the options, so the question means the central second moment, the variance, which is 4.