Corrected exercises: binomial and normal distributions (201-SN1-RE)

Pitfall: forgetting $\binom{9}{6}=84$, which counts the ways of placing the 6 successes among the 9 trials.

“At most 1” groups 0 and 1:

$E(X)=np=48\times 0.25=12$ and $V(X)=np(1-p)=12\times 0.75=9$, so $\sigma=\sqrt{9}=3$.

Pitfall: giving $9$ as the standard deviation; that is the variance.

Direct reading of the table: $P(Z<0.83)=\Phi(0.83)=0.7967$.

By the symmetry of the curve, the area to the right of $-1.46$ equals the area to the left of $1.46$:

Check: the area to the right of a negative number is more than $0.5$.

An area between two bounds is a difference of areas to the left:

$P(Z>z)=0.10$ leaves $0.90$ to the left. In the body of the table, $\Phi(1.28)=0.8997$ is the closest to $0.90$: $z\approx 1.28$.

$z=\dfrac{75-68}{5}=1.40$, then $P(X>75)=1-\Phi(1.40)=1-0.9192=0.0808$.

$\Phi(z)=0.20$: $\Phi(-0.84)=0.2005$ is the closest, so $z\approx -0.84$.

Pitfall: taking $z=+0.84$; a value below the mean has a negative z-score.

Conditions: $np=24\ge 5$ and $n(1-p)=56\ge 5$. Approximating normal: $\mu=24$, $\sigma=\sqrt{16.8}\approx 4.099$.

“At most 20” contains the bar of 20, which ends at $20.5$:

$z=\dfrac{20.5-24}{4.099}\approx -0.85$, $P(X\le 20)\approx\Phi(-0.85)=0.1977$

The exact value, from the binomial, is $0.1978$: the half unit provides most of the accuracy.

The four conditions, checked against the situation:

1) $n=10$ seeds, a number fixed before sowing;

2) each seed germinates (success) or does not (failure);

3) each seed comes at random from the same lot: $p=0.85$ for each one;

4) each seed has its own pot: what happens to one does not change what happens to another, so the trials are independent.

The sampling is without replacement, but 10 seeds out of $40{,}000$ are $0.025\%$ of the lot. Removing one viable seed changes the proportion of viable seeds from $0.85$ to $\dfrac{33{,}999}{39{,}999}\approx 0.849996$: an invisible difference. We therefore accept $X\sim\mathcal{B}(10,\,0.85)$, as a very good approximation.

Pitfall: reciting the four conditions without checking them against the situation; the checking is what earns the marks.

“At least 9” groups two values:

More than half the time, at least 9 seeds out of 10 germinate.

Pitfall: forgetting $\binom{10}{8}=45$, which counts the ways of choosing the 8 seeds that germinate; without it, you get the probability of ONE single sequence, $0.0061$.

“Fewer than 8” stops at 7. By the complement, with the values from b):

The two numbers differ by exactly $P(X=8)\approx 0.2759$: a variable that COUNTS gives a nonzero probability to each integer, and the sign, $<$ or $\le$, decides whether the bar of 8 is included in the calculation (figure of the solution).

This will be the central difference with the normal distribution, which MEASURES: for it, a single value has probability zero and both signs give the same number (exercise 3).

Pitfall: computing $P(X\le 7)$ term by term, eight terms; the complement needs only two, already computed.

No. Without replacement in a packet of 20, the probability changes from one seed to the next: after one viable seed, 16 viable seeds remain out of 19, and the next probability is $\dfrac{16}{19}\approx 0.842$, no longer $\dfrac{17}{20}=0.85$. The draws are not independent.

Here 6 seeds out of 20 are $30\%$ of the packet, far beyond the $5\%$ rule of thumb that would allow the binomial.

Exact probability, by counting: the 6 seeds are chosen among the 17 viable ones.

Same number from the tree: $\dfrac{17}{20}\times\dfrac{16}{19}\times\dfrac{15}{18}\times\dfrac{14}{17}\times\dfrac{13}{16}\times\dfrac{12}{15}\approx 0.3193$.

The binomial would give $(0.85)^{6}\approx 0.3771$, an overestimate of $18\%$: without replacement, each viable seed drawn makes the next one slightly less likely.

Pitfall: applying $\mathcal{B}(6,\,0.85)$ because “85% of the seeds are viable”; the proportion is right, the model is wrong.

For a seed taken alone, by the law of total probability:

For the 10 seeds, we condition only once on the state of the tray, which they all share:

The distribution $\mathcal{B}(10,\,0.85)$ would give $(0.85)^{10}\approx 0.1969$.

The condition that fails is INDEPENDENCE: all the seeds share the same soil. If one seed fails to germinate, the tray is more likely to be contaminated, which makes the failure of the others more likely: successes come together, and so do failures.

Yet each seed's probability really is $0.85$: checking that $p$ is the same is not enough, you also need a reason to believe in independence, such as one pot per seed.

Pitfall: concluding that the distribution is binomial as soon as each trial has the same probability of success.

The binomial theorem states that $(a+b)^{n}=\sum_{k=0}^{n}\binom{n}{k}a^{k}b^{n-k}$. With $a=p$ and $b=q$:

The $n+1$ probabilities of the binomial distribution are exactly the terms of the expansion of $(p+q)^{n}$, hence the name of the distribution.

For the 16 plants, the 17 bars of the figure therefore add up to 1.

Expand $G$ by the binomial theorem, with $a=pt$ and $b=q$:

Differentiate both expressions. On the left, by the chain rule, $G'(t)=np\,(q+pt)^{n-1}$. On the right, term by term, since the sum is finite: $G'(t)=\sum_{k=1}^{n}k\,P(X=k)\,t^{k-1}$.

At $t=1$, the right side gives $\sum_{k=0}^{n}k\,P(X=k)$, which is the definition of $E(X)$, and the left side gives $np\,(q+p)^{n-1}=np$:

The term $k=0$ disappears when differentiating, which changes nothing since it equals $0\times P(X=0)$.

Pitfall: “proving” it by writing $E(X)=n\times p$ “because we do $n$ trials with probability $p$”; that is the intuition, not a proof. The proof starts from the definition $E(X)=\sum k\,P(X=k)$.

Second derivative, on both sides:

At $t=1$: $E\big(X(X-1)\big)=n(n-1)p^{2}$. Now $E\big(X(X-1)\big)=E(X^{2})-E(X)$, so $E(X^{2})=n(n-1)p^{2}+np$.

With $V(X)=E(X^{2})-E(X)^{2}$:

Pitfall: taking $G''(1)$ for $E(X^{2})$; the second derivative brings down $k(k-1)$, not $k^{2}$.

$E(X)=16\times 0.75=12$ plants.

$V(X)=16\times 0.75\times 0.25=3$, so $\sigma=\sqrt{3}\approx 1.73$ plants.

Over a large number of trials of the treatment, 12 plants out of 16 recover on average, with a typical spread of less than 2 plants around 12.

Pitfall: giving $V(X)=3$ as the standard deviation; the variance is in “plants squared”, the standard deviation, its square root, is in the unit of $X$.

In the quotient, $p^{k}q^{n-k}$ cancels and the ratio of the coefficients remains:

The probability increases from $k$ to $k+1$ as long as this ratio exceeds 1:

Here $np-q=12-0.25=11.75$: the distribution rises for $k=0$ to $11$, hence up to $k=12$, then falls. The mode is $12$.

The figure confirms it: the bar of 12 is the tallest, a little above $0.22$, followed by that of 13, $P(X=13)\approx 0.2079$.

Here the mode coincides with the expected value because $np=12$ is an integer; in general, the mode is an integer between $np-q$ and $np+p$.

$\varphi'(z)=\dfrac{1}{\sqrt{2\pi}}\,(-z)\,e^{-z^{2}/2}$, which has the sign of $-z$, since the exponential is strictly positive.

$\varphi$ increases on $(-\infty, 0]$ and decreases on $[0, +\infty)$: its maximum is reached at $0$.

Second derivative, by the product rule:

It vanishes and changes sign at $z=-1$ and $z=1$: the curve is concave down on $[-1, 1]$ and concave up outside. The inflection points have $y$-coordinate $\varphi(\pm 1)=\dfrac{e^{-1/2}}{\sqrt{2\pi}}\approx 0.2420$.

This is the geometric meaning of the standard deviation: on every normal curve, the inflection points are one standard deviation on either side of the mean (first figure of the solution).

Pitfall: looking for the maximum by setting $\varphi''$ equal to zero; the second derivative gives the changes of concavity, not the extrema.

$P(Z\le -a)=\int_{-\infty}^{-a}\varphi(t)\,dt$. With $u=-t$, $dt=-du$, and the bounds $-\infty$ and $-a$ become $+\infty$ and $a$:

since $\varphi$ is even: $\varphi(-u)=\varphi(u)$. Two symmetric tails have the same area.

Pitfall: writing $P(Z>-0.62)=1-0.7324=0.2676$. The check that prevents it: the area to the right of a NEGATIVE number contains the whole right half, so it is more than $0.5$.

An area between two bounds is a difference of two areas to the left:

If the table only gives positive $z$: $\Phi(-1.25)=1-\Phi(1.25)=1-0.8944=0.1056$.

Check on the figure: the shaded region contains the peak and about one standard deviation on each side; about two thirds of the total area is consistent with $0.68$.

Pitfall: adding the two areas to the left, or subtracting in the wrong order and getting a negative number.

We look for the area in the BODY of the table, then read $z$ in the margin.

$P(Z<z)=0.90$: $\Phi(1.28)=0.8997$ and $\Phi(1.29)=0.9015$; the closer one gives $z\approx 1.28$.

$P(-z<Z<z)=0.95$ leaves $0.05$ in the two tails, so $0.025$ on each side (second figure of the solution): $\Phi(z)=0.975$, and the table gives exactly $z=1.96$.

$P(Z>z)=0.975$: the area to the right is more than $0.5$, so $z<0$. The area to the left is $0.025$ and, by symmetry, $z=-1.96$.

Pitfall: looking up $0.95$ in the table for the centred interval; you get $1.645$, the bound of a single $5\%$ tail.

$P(Z=0.80)=\int_{0.80}^{0.80}\varphi(t)\,dt=0$: an integral over an interval of length zero.

So $P(Z\le 0.80)=P(Z<0.80)+P(Z=0.80)=P(Z<0.80)=0.7881$: for a continuous variable, $<$ and $\le$ give the same number.

In exercise 1 c), $P(X\le 8)$ and $P(X<8)$ differed by $P(X=8)\approx 0.2759$. A binomial COUNTS, each integer carries a mass; a normal MEASURES, a single value carries no area.

A probability of zero does not mean “impossible”: here we ask for the value $0.80$ exactly, with infinitely many decimal places. A real measurement is always an interval (exercise 7).

In $\int_{a}^{b}\dfrac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^{2}}dx$, let $z=\dfrac{x-\mu}{\sigma}$, so $x=\mu+\sigma z$ and $dx=\sigma\,dz$.

The bounds $x=a$ and $x=b$ become $z_{a}=\dfrac{a-\mu}{\sigma}$ and $z_{b}=\dfrac{b-\mu}{\sigma}$, in the same order since $\sigma>0$.

The $\sigma$ from $dx$ cancels the one in the denominator: what remains is the density of $Z$. So $P(a\le X\le b)=P(z_{a}\le Z\le z_{b})$, and a single table works for every normal distribution.

Pitfall: dividing by the variance $\sigma^{2}=9$ instead of $\sigma=3$; the proof shows that it is $\sigma$ that comes out of $dx$.

$z_{1}=\dfrac{244-250}{3}=-2.00$ and $z_{2}=\dfrac{256-250}{3}=2.00$.

About $95.4\%$ of the tablets are accepted, so $4.56\%$ are rejected, close to 46 per 1000.

This is the two-standard-deviation rule, recovered from the table; “$95\%$” is its rounded form.

$z=\dfrac{251.8-250}{3}=0.60$, then $P(X>251.8)=1-\Phi(0.60)=1-0.7257=0.2743$.

A little more than one tablet in four is heavier than $251.8$ mg.

Pitfall: reading $0.7257$ and stopping there; the table gives the area to the left, the question asks for “heavier than”.

We look for $x$ such that $P(X>x)=0.01$, so $\Phi(z)=0.99$. The table gives $\Phi(2.32)=0.9898$ and $\Phi(2.33)=0.9901$: $z\approx 2.33$.