Corrected practice final exam: probability and statistics (201-SN1-RE)

Population: the black spruces of the stand, the trees the inventory is about.

Sample: the $50$ spruces of the plot.

Variable: the DBH, quantitative CONTINUOUS: it is a length, which can take any value in an interval; the tape that rounds to the millimetre does not change its nature.

Caveat: a plot is a cluster sample; neighbouring trees resemble each other more than trees drawn one by one across the whole stand, and the estimate is less precise than $n=50$ would suggest.

Class midpoints: $10$, $14$, $18$, $22$, $26$, $30$.

$\overline{x}=\dfrac{900}{50}=18.0$ cm

$s^{2}=\dfrac{1344}{49}=27.43$, so $s=5.24$ cm.

Pitfall: dividing by $n=50$ gives $5.18$ cm, the standard deviation of a population; these $50$ trees are a sample.

The modal class is $[16, 20)$, the tallest bar of the histogram ($15$ trees).

Cumulative frequencies: $6$, $19$, $34$. The observation of rank $\frac{n}{2}=25$ falls in $[16, 20)$, with $F=19$ and $n_{c}=15$.

$\text{median}=16+\dfrac{25-19}{15}\times 4=17.6$ cm

$\overline{x}=18.0$ cm is slightly above the median, $17.6$ cm.

The histogram stretches further to the right: from the midpoint of the modal class ($18$ cm), there are $14$ cm to the boundary $32$ cm, against $10$ cm to the boundary $8$ cm, and the two large trees of $[28, 32)$ pull the mean up.

The distribution is slightly skewed to the right. The gap between mean and median, $0.4$ cm, is small compared with $s=5.24$ cm: the skewness is weak.

$\frac{90}{100}\times 50=45$; the cumulative frequencies are $34$ before $[20, 24)$ and $43$ before $[24, 28)$: the 45th tree is in $[24, 28)$, with $F=43$ and $n_{c}=5$.

$P_{90}=24+\dfrac{45-43}{5}\times 4=25.6$ cm

About $90\%$ of the spruces in the sample have a DBH below $25.6$ cm.

$|z|>2$: the tree is unusual according to this benchmark, one of the largest in the plot.

Caveat: the $|z|>2$ benchmark comes from bell-shaped distributions; here, by interpolation in $[28, 32)$, about $48.5$ trees out of $50$, that is $97\%$, have a DBH below $29.0$ cm, which confirms an extreme position without making it a measurement error.

We choose a SET of $5$ sites out of $12$: order does not matter (the cameras are identical) and there is no repetition (at most one camera per site). This is a combination.

Model 1 has $12$ possible sites, model 2 has $11$, and so on: order now matters, this is a permutation (arrangement).

Each set of $5$ sites gives $5!=120$ ways to distribute the models.

We use the complement: “no wetland site” means choosing the $5$ sites among the $8$ others.

Pitfall: counting “one wetland site, then $4$ arbitrary sites”, $4\times\binom{11}{4}=1320$, counts several times the same choice when it has two or more wetland sites.

Two sites among the $4$ wetland sites AND three among the $8$ others: multiplication principle.

Every choice of $5$ sites contains exactly $k$ wetland sites, for a single $k$ between $0$ and $4$: the five cases form a partition of the $\binom{12}{5}$ choices.

For a fixed $k$, we choose $k$ wetland sites and $5-k$ others: $\binom{4}{k}\binom{8}{5-k}$ choices (multiplication principle).

Addition principle: $56+280+336+112+8=792=\dbinom{12}{5}$.

The $792$ choices are equally likely, and $336$ of them are favourable.

A sequence is a permutation of $6$ objects some of which are identical: $3$ deer, $2$ foxes, $1$ lynx.

Pitfall: $6!=720$ would count as different two sequences that differ only by swapping two deer, which are indistinguishable on the list of species.

Definition: $P(C\mid L)=\dfrac{P(C\cap L)}{P(L)}$, and likewise $P(C\cap L)=P(C)P(L\mid C)$ (multiplication rule).

$R$, $C$, $D$ form a partition of the sample space: they are pairwise mutually exclusive and their union is certain. Hence $L=(L\cap R)\cup(L\cap C)\cup(L\cap D)$, a union of mutually exclusive events.

Law of total probability: $P(L)=P(R)P(L\mid R)+P(C)P(L\mid C)+P(D)P(L\mid D)$.

Substituting the numerator and the denominator gives the stated formula, which assumes $P(L)>0$.

$P(C\mid L)=\dfrac{0.270}{0.317}\approx 0.8517$ and $P(R\mid L)=\dfrac{0.044}{0.317}\approx 0.1388$

Nearly $14\%$ of the photos labelled “Canada lynx” actually show a bobcat: the software rarely mislabels a bobcat ($8\%$), but bobcats are almost twice as numerous.

Pitfall: confusing $P(L\mid C)=0.90$ with $P(C\mid L)=0.85$.

Mutually exclusive: yes, an animal is not of both species at once, $P(C\cap R)=0$.

Independent: no, because $P(C)P(R)=0.30\times 0.55=0.165\neq 0=P(C\cap R)$.

Two mutually exclusive events with nonzero probabilities are always dependent: knowing that it is a bobcat RULES OUT the Canada lynx.

For each species, the probability of two labels $L$ is the square of $P(L\mid\cdot)$ (independence given the species).

The second label raises the probability from $0.85$ to $0.99$: Bayes applied twice, the posterior distribution after the first photo serving as the prior for the second.

Caveat: two successive photos of the same animal share the angle, the light and the distance; their errors are probably linked, and $0.9855$ is an optimistic bound, valid within the assumed model.

$P(L_{1}\cap L_{2})=0.24658$ (part e) and $P(L_{1})P(L_{2})=0.317^{2}\approx 0.1005$.

The two numbers differ: $L_{1}$ and $L_{2}$ are NOT independent.

A first label $L$ makes a Canada lynx more likely, hence a second label $L$ as well. Independence given the species does not imply independence outright.

$X$ takes only two values: $1$ (negative mixture) or $6$ (positive mixture).

The mixture is negative if all five swabs are: by independence, $P(X=1)=0.95^{5}\approx 0.7738$.

$P(X=6)=1-0.95^{5}\approx 0.2262$; the sum is indeed $1$.

$E(X)=1\times 0.7738+6\times 0.2262=1+5\times 0.2262\approx 2.1311$ tests per pool.

Per swab: $\dfrac{2.1311}{5}\approx 0.4262$ test, instead of $1$.

Pooling saves on average about $57\%$ of the tests, in this model.

$\sigma(X)=\sqrt{4.3761}\approx 2.092$ tests.

Check: a variable with two values $5$ apart has variance $5^{2}\times 0.7738\times 0.2262\approx 4.3761$.

Let $x_{1},\dots,x_{n}$ be the values of $X$ and $p_{1},\dots,p_{n}$ their probabilities, with $\sum p_{i}=1$. $aX+b$ takes the values $ax_{i}+b$ with the same probabilities.

The constant $b$ shifts all the values without spreading them: it disappears from the variance.

$C=30X+12$, in dollars.

$E(C)=30\times 2.1311+12\approx 75.93$ dollars

$\sigma(C)=|30|\,\sigma(X)\approx 30\times 2.0919\approx 62.76$ dollars

Pitfall: adding $12$ to the standard deviation; a fixed cost does not change the spread.

For a pool of $k$ swabs: $X=1$ with probability $0.95^{k}$, $X=k+1$ otherwise.

$E(X)=0.95^{k}+(k+1)(1-0.95^{k})=1+k(1-0.95^{k})$, and per swab $g(k)=\dfrac{E(X)}{k}=\dfrac{1}{k}+1-0.95^{k}$.

Values: $g(4)\approx 0.4355$; $g(5)\approx 0.4262$; $g(6)\approx 0.4316$; $g$ decreases from $k=2$ to $5$, then increases up to $10$.

The minimum is reached at $k=5$: small pools require many tests of mixtures, large pools are too often positive.

Caveat: the model assumes the swabs are independent and that mixing loses no sensitivity; birds from the same pond infect one another, and diluting a weakly positive swab can mask it.

A density is nonnegative with total integral $1$: $f\ge 0$ requires $k\ge 0$, then

$\dfrac{k}{3}=1$ gives $k=3$.

$F(x)=\displaystyle\int_{0}^{x}\frac{3}{(1+t)^{4}}\,dt=\left[-\frac{1}{(1+t)^{3}}\right]_{0}^{x}=1-\frac{1}{(1+x)^{3}}$ for $x\ge 0$ ($F(x)=0$ for $x<0$).

For a continuous variable, $P(X=0.5)=0$: including the endpoints or not changes nothing.

The median $m$ satisfies $F(m)=\frac{1}{2}$: $(1+m)^{3}=2$, so $m=\sqrt[3]{2}-1\approx 0.260$ hectare.

Half of the fires burn less than $0.26$ ha, that is about $2{,}600$ m².

$E(X)=\displaystyle\int_{0}^{+\infty}x\,\frac{3}{(1+x)^{4}}\,dx$. By parts, $u=x$ and $v'=\dfrac{3}{(1+x)^{4}}$, so $u'=1$ and $v=-\dfrac{1}{(1+x)^{3}}$:

The bracket tends to $0$ because $\dfrac{b}{(1+b)^{3}}\to 0$. $E(X)=0.5$ hectare.

By parts again, $u=x^{2}$ and $v=-\dfrac{1}{(1+x)^{3}}$; the bracket tends to $0$:

$\displaystyle E(X^{2})=\int_{0}^{+\infty}\frac{2x}{(1+x)^{3}}\,dx$, and with $w=1+x$: $\displaystyle 2\int_{1}^{+\infty}\frac{w-1}{w^{3}}\,dw=2\left[-\frac{1}{w}+\frac{1}{2w^{2}}\right]_{1}^{+\infty}=2\times\frac{1}{2}=1$

$V(X)=E(X^{2})-E(X)^{2}=1-0.25=0.75$ and $\sigma(X)=\sqrt{0.75}\approx 0.866$ hectare.

$E(X)=0.5$ ha is almost twice the median $0.26$ ha: the distribution is strongly skewed to the right. A few large fires weigh heavily in the mean, while the typical fire stays small.

Caveat: the density gives a nonzero probability to huge fires; in reality, the area is bounded by the zone, and the model only holds for the order of magnitude of the fires observed.

The event $X>2$ is contained in $X>1$, hence the intersection equals $X>2$.

Without the condition, $P(X>2)\approx 0.0370$: a fire that has already passed $1$ ha is eight times more likely to pass $2$ ha than a fire that is just starting.

Each fire exceeds $1$ ha with probability $q=0.125$ (part b), independently of the others: the number $Y$ of such fires follows the binomial distribution $\mathcal{B}(12,\,0.125)$.

$P(Y=0)=0.875^{12}\approx 0.2014$; $P(Y=1)=12\times 0.125\times 0.875^{11}\approx 0.3453$; $P(Y=2)=\dbinom{12}{2}\times 0.125^{2}\times 0.875^{10}\approx 0.2713$

Caveat: fires from the same storm, in dry and windy weather, have linked areas; independence is an assumption of the model.

About $92.7\%$ of smolts are at least $13$ cm long.