Your Answer : (Not Answered)

Correct Answer : B

Both means are \(3/5\): \(\frac{3}{5}\) and \(\frac{3}{5}\).

Integrate over the unit square, splitting each integrand into products of one-variable integrals:

\[ \begin{aligned} E(X) &= \tfrac{6}{5}\int\!\!\int x(x + y^{2}) \\ &= \tfrac{6}{5}\left(\tfrac{1}{3} + \tfrac{1}{2}\cdot\tfrac{1}{3}\right) = \tfrac{6}{5}\cdot\tfrac{1}{2} = \tfrac{3}{5} \end{aligned} \]

\[ \begin{aligned} E(Y) &= \tfrac{6}{5}\int\!\!\int y(x + y^{2}) \\ &= \tfrac{6}{5}\left(\tfrac{1}{2}\cdot\tfrac{1}{2} + \tfrac{1}{4}\right) = \tfrac{6}{5}\cdot\tfrac{1}{2} = \tfrac{3}{5} \end{aligned} \]

The density integrates to \(\frac{6}{5}(\frac{1}{2} + \frac{1}{3}) = 1\), as it must. Both variables lie in [0, 1], so a mean of \(6/5\) is impossible, which rules out two options at once; \(2/5\) for \(E(X)\) is an arithmetic slip.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct Answer : A

The variance is 1.587.

For an F-distribution with \(\nu_{1}\) and \(\nu_{2}\) degrees of freedom (here 5 and 9; the variance exists only when \(\nu_{2}\) is greater than 4):

\[ \begin{aligned} V &= \frac{N}{D} \\ N &= 2\nu_{2}^{2}(\nu_{1} + \nu_{2} - 2) \\ D &= \nu_{1}(\nu_{2} - 2)^{2}(\nu_{2} - 4) \end{aligned} \]

\[ \begin{aligned} V &= \frac{2(81)(12)}{5(49)(5)} \\ &= \frac{1944}{1225} = 1.587 \end{aligned} \]

The mean of this distribution is \(\nu_{2}/(\nu_{2} - 2) = 9/7 = 1.286\), which none of the options gives; 1.378, 2.578 and 1.498 do not follow from the formula.

Your Answer : (Not Answered)

Correct Answer : C

The probability is \(\frac{7}{9}\).

With two dice there are 36 equally likely outcomes. A sum of 7 occurs in 6 ways and a sum of 11 in 2 ways (5-6 and 6-5), so 8 outcomes are excluded:

\[ \begin{aligned} P &= 1 - \frac{6 + 2}{36} \\ &= \frac{28}{36} = \frac{7}{9} \end{aligned} \]

The chance that the sum is 7 or 11 is \(\frac{8}{36} = \frac{2}{9}\); the question asks for the complement. \(\frac{1}{9}\), \(\frac{3}{9}\) and \(\frac{5}{9}\) do not match either count, so only \(\frac{7}{9}\) fits.

Your Answer : (Not Answered)

Correct Answer : D

The probability is \(\frac{8}{70}\).

There are \(3 + 4 + 2 + 1 = 10\) people. One from each of the four categories can be chosen in \(3 \times 4 \times 2 \times 1\) ways.

\[ \begin{aligned} \text{total} &= \binom{10}{4} = 210 \\ \text{favourable} &= 3 \times 4 \times 2 \times 1 = 24 \\ P &= \frac{24}{210} = \frac{8}{70} \end{aligned} \]

\(\frac{8}{70}\) is \(\frac{4}{35}\), left unreduced to match the options. The other options correspond to 6, 12 and 18 favourable committees out of 210, all undercounts of the 24 ways to pick one person from each category.

Your Answer : (Not Answered)

Correct Answer : D

The quartiles are 30 and 70.

Bowley's coefficient of skewness, based on quartiles, is:

\[ S_{k} = \frac{Q_{3} + Q_{1} - 2M}{Q_{3} - Q_{1}} \]

\[ \begin{aligned} 0.6 &= \frac{100 - 2(38)}{Q_{3} - Q_{1}} \\ &= \frac{24}{Q_{3} - Q_{1}} \\ Q_{3} - Q_{1} &= 40 \end{aligned} \]

With \(Q_{3} + Q_{1} = 100\), adding and subtracting gives \(Q_{3} = 70\) and \(Q_{1} = 30\). The pairs 40 and 60 and 20 and 80 add to 100 but give differences of 20 and 60, so skewness of 1.2 and 0.4. The pair 30 and 60 does not even add to 100.

Your Answer : (Not Answered)

Correct Answer : C

The coefficient of variation is 8.33 (per cent).

Karl Pearson's coefficient of skewness is \((\text{mean} - \text{mode}) / \sigma\). Solve it for the standard deviation, then divide by the mean:

\[ \begin{aligned} -0.3 &= \frac{120 - 123}{\sigma} \Rightarrow \sigma = 10 \\ CV &= \frac{10}{120} \times 100 = 8.33 \end{aligned} \]

The neighbouring values 6.33, 7.33 and 9.33 come from arithmetic slips; only \(\sigma = 10\) is consistent with the given skewness.

Your Answer : (Not Answered)

Correct Answer : C

The Leslie matrix is \(\begin{bmatrix}0&0&f_{3}\\0.25&0&0\\0&0.5&0\end{bmatrix}\).

A Leslie matrix has the age-specific fecundities in its first row and the survival probabilities on the sub-diagonal, from age class 1 to 2 and from 2 to 3.

The fish breed only in the third year, so the fecundities of the first two classes are 0 and only \(f_{3}\) stays. Of the newborns 25% survive to class 2, and 50% of those reach class 3, giving 0.25 and 0.5 below the diagonal.

The first two options give fecundity to classes 1 and 2, which do not breed. The last puts the survival rates on the main diagonal, which would mean fish staying in the same age class instead of ageing.

Your Answer : (Not Answered)

Correct Answer : C

The skewness is the same, but the coefficients of variation are not.

Both distributions have the probabilities 0.1, 0.2, 0.4, 0.2, 0.1 placed symmetrically about their middle value: X about 3, and Y about 35 (35 is 15 above 20 and 15 below 50, 25 above 10 and 25 below 60). A symmetric distribution has zero skewness, so the skewness is equal.

\[ \begin{aligned} \sigma_X^2 &= 2(0.1 \cdot 4) + 2(0.2 \cdot 1) = 1.2 \\ CV_X &= \sqrt{1.2}/3 \approx 0.365 \\ \sigma_Y^2 &= 2(0.1 \cdot 625) + 2(0.2 \cdot 225) = 215 \\ CV_Y &= \sqrt{215}/35 \approx 0.419 \end{aligned} \]

The coefficients of variation differ because Y is not a rescaling of X (its values are not evenly spaced). That rules out every option saying the coefficients are the same, and every option saying the skewness differs.

Your Answer : (Not Answered)

Correct Answer : A

The mutual information is 0.

The joint distribution puts probability 1/6 on each of the 2 × 3 = 6 pairs. The marginals are therefore uniform:

\[ \begin{aligned} P(X = i) &= 3 \times \tfrac{1}{6} = \tfrac{1}{2} \\ P(Y = j) &= 2 \times \tfrac{1}{6} = \tfrac{1}{3} \end{aligned} \]

Since \(\tfrac{1}{2} \times \tfrac{1}{3} = \tfrac{1}{6}\) for every pair, \(P(X, Y) = P(X)P(Y)\): X and Y are independent, and independent variables share no information.

\[ \begin{aligned} I(X, Y) &= H(X) + H(Y) - H(X, Y) \\ &= \log 2 + \log 3 - \log 6 = 0 \end{aligned} \]

The distractors are the separate entropies: log(2) is H(X), log(3) is H(Y), and log(6) is the joint entropy H(X, Y).

Your Answer : (Not Answered)

Correct Answer : D

The process standard deviation is 0.576.

The process standard deviation is estimated from the average range as \(\hat{\sigma} = \bar{R}/d_{2}\):

\[ \begin{aligned} \bar{R} &= \frac{1.2 + 1.5 + 1.1 + 1.4 + 1.5}{5} = 1.34 \\ \hat{\sigma} &= \frac{1.34}{2.326} \approx 0.576 \end{aligned} \]

\(A_{2}\) is used for the control limits of the \(\bar{X}\) chart, not for \(\sigma\); for example \(A_{2}\bar{R} = 0.577 \times 1.34 \approx 0.773\). None of 0.511, 2.463 or 2.322 equals \(\bar{R}/d_{2}\).

Your Answer : (Not Answered)

Correct Answer : C

With n attributes there are \(2^{n}\) ultimate class frequencies.

An ultimate class is specified by stating, for every one of the n attributes, whether it is present or absent, as in ABC, ABγ, AβC and so on. Each attribute gives two choices, so the number of ultimate classes is

\[ 2 \times 2 \times \cdots \times 2 = 2^{n} \]

For two attributes there are four: (AB), (Aβ), (αB) and (αβ). The total number of all class frequencies, of every order, is \(3^{n}\). \(n\) and \(2n\) are far too few, and \(4^{n}\) too many.

Your Answer : (Not Answered)

Correct Answer : C

With α = 5%, 5% of the time we will say that there is a real difference when there is no difference.

α is the probability of a Type I error: rejecting H₀ when it is true. Here H₀ says the mean is 10, so rejecting it amounts to declaring a real difference from 10. If H₀ is in fact true and the test is repeated many times, it will wrongly declare a difference in about 5% of cases.

Saying there is no real difference when there is none is the correct decision, which happens 95% of the time, not 5%. An incorrect inference 95% of the time reverses the figures, and α says nothing about how often the null hypothesis is true, which rules out the last option.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct Answer : C

The numbers are 4, 6, 9.

Test them against both means:

\[ \begin{aligned} GM &= \sqrt[3]{4 \times 6 \times 9} = \sqrt[3]{216} = 6.0 \\ AM &= \frac{4 + 6 + 9}{3} = \frac{19}{3} \approx 6.3 \end{aligned} \]

For 6, 6, 6 both means are exactly 6, since the AM equals the GM only when all values are equal. For 5, 6, 8 the AM is 6.33 but the GM is \(\sqrt[3]{240} \approx 6.21\), not 6.0. As 4, 6, 9 fits both, 'None of the above' is wrong.

Your Answer : (Not Answered)

Correct Answer : C

The maximum likelihood estimator is \(1/(\overline{X} - 1)\).

The density is an exponential shifted to start at 1. For a sample with every \(x_{i} \ge 1\) the log-likelihood is

\[ \begin{aligned} \ell(\theta) &= n\ln\theta - \theta\sum_{i=1}^{n}(x_{i} - 1) \\ \ell'(\theta) &= \frac{n}{\theta} - \sum_{i=1}^{n}(x_{i} - 1) = 0 \\ \hat{\theta} &= \frac{n}{\sum x_{i} - n} = \frac{1}{\overline{X} - 1} \end{aligned} \]

\(\ell''(\theta) = -n/\theta^{2} \lt 0\), so this is a maximum. \(1/\overline{X}\) is the estimator for the unshifted exponential, which ignores the shift by 1, and \((1/\overline{X}) - 1\) applies the shift in the wrong place. \(\overline{X}\) estimates a mean, not a rate.

Your Answer : (Not Answered)

Correct Answer : D

\(E\left(\frac{Y}{X}\right)\) is equal to \(\frac{5}{7}\).

For a chi-squared variable with \(k\) degrees of freedom, \(E(\chi_{k}^{2}) = k\) and, for \(k \gt 2\), \(E\left(\frac{1}{\chi_{k}^{2}}\right) = \frac{1}{k - 2}\). By independence:

\[ \begin{aligned} E\left(\tfrac{X}{Y}\right) &= \tfrac{m}{n - 2} = 3, \quad m + n = 14 \\ 14 - n &= 3n - 6 \Rightarrow n = 5,\ m = 9 \\ E\left(\tfrac{Y}{X}\right) &= \tfrac{n}{m - 2} = \tfrac{5}{7} \end{aligned} \]

Taking \(E(Y/X)\) as the reciprocal \(\frac{1}{3}\) is the common mistake: the expectation of a reciprocal is not the reciprocal of the expectation, and \(\frac{1}{3}\) is not offered.

Your Answer : (Not Answered)

Correct 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.

Your Answer : (Not Answered)

Correct Answer : A

The density of \(Y\) is \(\frac{1}{5}e^{-\frac{1}{5}(y-2)}\) for \(y \gt 2\), and 0 otherwise.

\(X\) is exponential with rate \(3/5\). The map \(y = 3x + 2\) is increasing, with inverse \(x = (y - 2)/3\) and \(dx/dy = 1/3\). By the change of variables formula:

\[ \begin{aligned} f_{Y}(y) &= f_{X}\left(\frac{y-2}{3}\right)\cdot\frac{1}{3} \\ &= \frac{3}{5}e^{-\frac{3}{5}\cdot\frac{y-2}{3}}\cdot\frac{1}{3} \\ &= \frac{1}{5}e^{-\frac{1}{5}(y-2)},\quad y \gt 2. \end{aligned} \]

Since \(x \gt 0\) corresponds to \(y \gt 2\), the density is 0 for \(y \le 2\). As a check, scaling an exponential by 3 divides its rate by 3, from \(3/5\) to \(1/5\). The option keeping rate \(3/5\) forgets the Jacobian, and the rates \(2/5\) and \(4/5\) come from no consistent working.

Your Answer : (Not Answered)

Correct Answer : C

The hypothesis can be rejected at 10% level of significance.

The interval is centred at \(\bar{x} = 22\) with half-width 1.51. With \(n = 10\) and \(\sigma\) unknown, the 98% interval uses \(t_{0.01, 9} = 2.821\), so the standard error is:

\[ \begin{aligned} s/\sqrt{n} &= 1.51/2.821 = 0.535 \\ t &= (22 - 20.5)/0.535 = 2.80 \end{aligned} \]

Since 20.5 lies just inside the 98% interval, \(H_{0}\) cannot be rejected at 2%, so the first option is wrong. But \(2.80\) exceeds \(t_{0.025, 9} = 2.262\) and \(t_{0.05, 9} = 1.833\), so \(H_{0}\) is rejected at 5% and at 10%. That makes 'cannot be rejected at 5%' and 'cannot be rejected at any level' wrong. The same conclusion follows if \(\sigma\) is treated as known (z = 2.31 against 1.96 and 1.645).