The Normal Distribution: Practice

Turn the tail area into an area to the left, then find its z-score in the table.

$1 - 0.0228 = 0.9772 = \Phi(2.0) \;\Longrightarrow\; z = 2.0$

So $0.33$ s sits exactly two standard deviations above the mean. Solve the z-score equation for $\sigma$.

$2.0 = \frac{0.33 - 0.25}{\sigma} = \frac{0.08}{\sigma} \;\Longrightarrow\; \sigma = \frac{0.08}{2.0} = 0.04 \text{ s}$

The value $0.08$ s is the distance from the mean to the cutoff, which is two standard deviations, not one.

The area to the left of $47$ is $0.9332$, and that entry appears in the table at $z = 1.5$.

$z = 1.5 \;\Longrightarrow\; 47 = \mu + 1.5(6) = \mu + 9$

Solve for the mean.

$\mu = 47 - 9 = 38$

Since most of the distribution lies below $47$, the mean must sit well below $47$, which rules out $56$ immediately.

An area of exactly $0.5$ to the left means the value sits at the centre, since the curve is symmetric and the mean splits it in half.

$\Phi(0) = 0.5 \;\Longrightarrow\; \mu = 50$

The second condition gives $\Phi(z) = 0.9332$, which the table places at $z = 1.5$. So $80$ is $1.5$ standard deviations above $50$.

$80 = 50 + 1.5\sigma \;\Longrightarrow\; 1.5\sigma = 30 \;\Longrightarrow\; \sigma = 20$

Taking $\sigma = 30$ would put $80$ only one standard deviation up, which corresponds to $\Phi(1) = 0.8413$, not $0.9332$.

Convert the negative endpoint with symmetry, then subtract the two areas to the left.

$\Phi(-1.5) = 1 - \Phi(1.5) = 1 - 0.9332 = 0.0668$

$P(-1.5 < Z < 0.5) = \Phi(0.5) - \Phi(-1.5) = 0.6915 - 0.0668 = 0.6247$

A common slip is to add the two table entries instead of subtracting, which double-counts the left half of the curve and gives an answer larger than it should be.

The middle band is symmetric, so the leftover area splits evenly between the two tails.

$\frac{1 - 0.8664}{2} = \frac{0.1336}{2} = 0.0668 \text{ in each tail}$

The area to the left of the upper cutoff is therefore $1 - 0.0668 = 0.9332$, which the table places at $z = 1.5$. The band runs from $z = -1.5$ to $z = +1.5$.

$x = 500 \pm 1.5(100) \;\Longrightarrow\; 350 \text{ to } 650$

The range $400$ to $600$ is only one standard deviation each way, holding about $68\%$.

The curve itself has a perfectly good formula, $\varphi(z) = \frac{1}{\sqrt{2\pi}} e^{-z^2/2}$, so the difficulty is not in describing it. The difficulty is in finding the area underneath it.

$\text{no elementary function has } e^{-z^2/2} \text{ as its rate of change}$

That is a theorem, not a gap in anyone's cleverness: no combination of powers, roots, exponentials, logarithms and trigonometric functions produces the area. So the areas are computed numerically, once, by adding up enormous numbers of very thin strips, and the results are recorded in a table.

Standardizing has already removed $\mu$ and $\sigma$ from the picture, so the standard curve's areas are the same for every data set.

Both endpoints are negative, so convert both with the symmetry rule before subtracting.

$\Phi(-0.5) = 1 - 0.6915 = 0.3085, \qquad \Phi(-2.5) = 1 - 0.9938 = 0.0062$

Now take the area left of the upper endpoint and remove the area left of the lower one.

$P(-2.5 < Z < -0.5) = 0.3085 - 0.0062 = 0.3023$

The value $0.3085$ on its own is the whole tail below $-0.5$, which stretches out forever and is not cut off at $-2.5$.

Standardize both endpoints.

$z_1 = \frac{118 - 120}{2} = -1.0, \qquad z_2 = \frac{123 - 120}{2} = 1.5$

Find the area of the band, using symmetry on the negative endpoint.

$P(-1.0 < Z < 1.5) = \Phi(1.5) - \Phi(-1.0) = 0.9332 - 0.1587 = 0.7745$

Now apply that fraction to the batch.

$4000 \times 0.7745 = 3098$

So about $3{,}098$ rods fall in the range. The figure $3{,}732$ comes from using $\Phi(1.5)$ alone and forgetting to cut off the tail below $118$ mm, and $902$ counts the rods outside the band.

For a continuous quantity, probability is defined as area under the density curve, and the region standing above one single point is infinitely thin.

$\text{width } 0 \;\Longrightarrow\; \text{area } 0 \;\Longrightarrow\; P(X = 3) = 0$

Probability zero is not the same as impossible: the curve is perfectly positive at $x = 3$, so values land near $3$ all the time, and a measurement must land somewhere. What is impossible is landing on a value with infinitely many specified decimal places, which is what "exactly $3$" demands. The height of the curve at $3$ is not zero, and $3$ may sit anywhere relative to the mean.

A percentile is the area to the left, so search the $\Phi$ column for $0.9772$. It appears exactly at $z = 2.0$.

$z = 2.0$

Now start at the mean and walk two standard deviations up.

$x = \mu + z\sigma = 60 + 2.0(12) = 60 + 24 = 84 \text{ thousand}$

A salary of $72$ thousand is only one standard deviation up, which is the $84.13$th percentile, and $96$ thousand is three standard deviations up, the $99.87$th.

The event covers two tails, one beyond $+1.6$ and one below $-1.6$, and by symmetry they have equal area.

$P(Z > 1.6) = 1 - \Phi(1.6) = 1 - 0.9452 = 0.0548$

Double it to collect both tails.

$P(|Z| > 1.6) = 2(0.0548) = 0.1096$

The value $0.0548$ counts only the upper tail, and $0.8904$ is the complement, the central band $|Z| < 1.6$.

The mathematics of the normal curve is not in question. What is in question is whether the curve fits the data, and here it does not.

$\text{beyond } 3\sigma \text{ under a normal curve: about } 0.3\% \text{ of the time}$

Real market returns produce far more of these extreme days than a bell curve allows, so the distribution has heavier tails than the normal. A model that assumes normality will therefore price a crash as a once-in-millennia event and be wrong.

The Central Limit Theorem is untouched by this: it says that sums of many independent small effects tend to normality, and market moves on a crash day are anything but independent.