Random Variables Questions and Answers

Answer: D

The true statement is \(\bar{z} \leq \bar{x}\).

Because \(0 \leq y \leq 1\) and \(x \geq 0\), every outcome has \(z = xy \leq x\). Taking means preserves the inequality, so \(\bar{z} \leq \bar{x}\) always.

The other relations depend on the dependence between x and y, since \(\bar{z} - \bar{x}\bar{y}\) is the covariance. If \(y = x\), \(\bar{z}\) is the mean of \(x^{2}\), which is at least \(\bar{x}^{2}\), so '\(\leq\)' fails. If \(y = 1 - x\), the covariance is negative, so '\(\geq\)' fails, and equality holds only for uncorrelated variables.

Answer: B

The true statement is \(CV_{A} \lt CV_{B}\) and \(s_{A} \gt s_{B}\).

Brand A has mean 30 and deviations \(-3, -2, 0, 2, 3\); Brand B has mean 20 and deviations \(-3, -1, 0, 1, 3\). With divisor \(n - 1 = 4\):

\[ \begin{aligned} s_{A}^{2} &= \frac{9 + 4 + 0 + 4 + 9}{4} = 6.5, \ s_{A} \approx 2.55 \\ s_{B}^{2} &= \frac{9 + 1 + 0 + 1 + 9}{4} = 5, \ s_{B} \approx 2.24 \end{aligned} \]

So \(s_{A} \gt s_{B}\). The coefficient of variation divides by the mean:

\[ \begin{aligned} CV_{A} &\approx \frac{2.55}{30} = 0.085 \\ CV_{B} &\approx \frac{2.24}{20} = 0.112 \end{aligned} \]

Brand A varies more in absolute terms but less relative to its much larger mean, so \(CV_{A} \lt CV_{B}\). The conclusion is the same with divisor \(n\), since it scales both standard deviations equally.

Answer: C

The arithmetic mean is 50.

The first moment about a point A is the mean of the deviations from A:

\[ \begin{aligned} \mu_{1}' &= \frac{\sum (x - A)}{n} = \bar{x} - A \\ 40 &= \bar{x} - 10 \\ \bar{x} &= 50 \end{aligned} \]

This holds for any distribution, normal or not. 40 forgets to add back the point about which the moment was taken. 30 subtracts 10 instead of adding it, and 20 has no basis in the data. As a check, the first moment about the mean itself is always zero.

Answer: B

The root mean square deviation is the Standard deviation.

Standard deviation is found by squaring each deviation from the mean, taking the mean of those squares, and then the square root: literally the root of the mean square deviation.

\[ \sigma = \sqrt{\frac{\sum (x - \bar{x})^2}{n}} \]

Variance is the mean square deviation before the root is taken. Mean deviation averages the absolute deviations without squaring, and quartile deviation is half the interquartile range.

Answer: C

The entropies are the same, but the coefficients of variation are not.

Entropy depends only on the probabilities, \(H = -\sum p_i \log p_i\), not on the values the variable takes. X and Y have the same probabilities (0.1, 0.2, 0.4, 0.2, 0.1), so their entropies are equal.

The coefficient of variation is the standard deviation divided by the mean, and it depends on the values:

\[ \begin{aligned} \mu_X &= 3,\ \sigma_X^2 = 1.2,\ CV_X \approx 0.365 \\ \mu_Y &= 35,\ \sigma_Y^2 = 215,\ CV_Y \approx 0.419 \end{aligned} \]

Y's values (10, 20, 35, 50, 60) are not a fixed multiple of X's, since 10X would give 10 to 50, so the CVs differ. Options saying the entropies differ are wrong, and so is the option saying the CVs match.

Answer: C

The mean is \(\frac{(n+1)}{2}\).

Each value from 1 to n occurs once, so the mean is the sum of the first n natural numbers divided by n:

\[ \begin{aligned} \bar{x} &= \frac{1 + 2 + \dots + n}{n} \\ &= \frac{n(n+1)/2}{n} = \frac{n+1}{2} \end{aligned} \]

For n = 5 that gives (1 + 2 + 3 + 4 + 5)/5 = 3 = 6/2.

\(\frac{n(n+1)}{2}\) is the sum, not the mean: it forgets to divide by n. \(\frac{n}{2}\) and \(\frac{(n-1)}{2}\) are each half a unit off, as the n = 5 check shows (2.5 and 2).

Answer: B

The value of k is 0.125.

The probabilities of all values of a random variable must add up to 1:

\[ \begin{aligned} 0 + k + 2k + 2k + 3k &= 1 \\ 8k &= 1 \\ k &= 0.125 \end{aligned} \]

0.166 is 1/6, which comes from missing one of the 2k terms and adding the rest to 6k. 0.100 and 0.150 would make the total 0.8 and 1.2, which cannot be the total of a probability distribution.