The Normal Distribution: Additional Practice (Set 1)

Convert the negative endpoint with symmetry.

$\Phi(-0.6) = 1 - \Phi(0.6) = 1 - 0.7257 = 0.2743$

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

$P(-0.6 < Z < 1.2) = \Phi(1.2) - \Phi(-0.6) = 0.8849 - 0.2743 = 0.6106$

The value $0.1592$ is the trap worth naming. It is $\Phi(1.2) - \Phi(0.6) = 0.8849 - 0.7257$, which looks up $+0.6$ and forgets to convert it to $\Phi(-0.6)$ first, so it measures the band from $+0.6$ to $1.2$ instead of from $-0.6$ to $1.2$. The value $0.3894$ is the complement, the area outside the correct band.

Turn the right-tail area into an area to the left, because that is what the table stores.

$\Phi(c) = 1 - 0.0062 = 0.9938$

Search the $\Phi$ column for $0.9938$. It appears exactly at $z = 2.5$.

$c = 2.5$

The cutoff must be positive: only a tiny sliver of area lies above it, so it sits far out on the right. A negative $c$ would leave more than half the curve above it.

Standardize the cutoff.

$z = \frac{86 - 68}{12} = \frac{18}{12} = 1.5$

Find the area to the right.

$P(Z > 1.5) = 1 - \Phi(1.5) = 1 - 0.9332 = 0.0668$

Apply it to the cohort.

$3000 \times 0.0668 = 200.4$

So about $200$ students score above $86$. The figure $2{,}800$ counts the students below the cutoff, which is the complement of what was asked.

Read each area as a z-score. The entry $0.8413$ sits at $z = 1$, and $0.1587 = 1 - 0.8413$ sits at $z = -1$ by symmetry. So the two values are one standard deviation on either side of the mean.

$46 = \mu + \sigma, \qquad 34 = \mu - \sigma$

Add the two equations to eliminate $\sigma$.

$80 = 2\mu \;\Longrightarrow\; \mu = 40, \qquad \sigma = 46 - 40 = 6$

The mean is exactly halfway between the two symmetric cutoffs, which is a useful check: $34$ and $46$ straddle $40$ evenly.

Cut the region under the curve into thin vertical strips, each one close to a rectangle of width $w$ and height $h$. The horizontal move multiplies every width by $\sigma$, and the vertical move divides every height by $\sigma$.

$(\sigma w)\left(\frac{h}{\sigma}\right) = wh$

Every strip keeps exactly the area it had, so the whole region does too. This is the reason the total area stays at $1$ no matter what $\sigma$ is, and the reason corresponding pieces of area match up between any normal curve and the standard one. It is what makes a single table enough.

Mind the direction. This question runs from the standard curve out to $f$. Standardizing is the reverse trip, $X \to Z$, which divides horizontal distances by $\sigma$ and multiplies heights by $\sigma$. The area is unchanged either way, for exactly the same reason: the two scale factors always multiply to $1$.

A percentile is the area to the left, so look for that area in the $\Phi$ column.

$\Phi(z) = 0.9938$

The table places that entry exactly at $z = 2.5$.

$z = 2.5$

The number $0.9938$ is the area itself, not a z-score, and confusing the two is the trap: an area is always between $0$ and $1$, while a z-score is a count of standard deviations and can be any size.

Standardize the cutoff once and reuse it.

$z = \frac{53 - 50}{2} = 1.5 \;\Longrightarrow\; \Phi(1.5) = 0.9332$

Now evaluate each quantity. $P(X < 53) = 0.9332$. Because the weight is continuous, the single value $53$ carries zero probability, so $P(X \le 53) = 0.9332$ as well: including the endpoint adds nothing. And $1 - P(X > 53) = 1 - 0.0668 = 0.9332$, since the area to the right and the area to the left must make up the whole.

$P(X > 53) = 1 - 0.9332 = 0.0668$

So $P(X > 53)$ is the odd quantity out: it is the area to the right, while the other three are all names for the area to the left. The point of the item is that $<$ and $\le$ do not differ for a continuous distribution, so hunting for the difference there is a dead end.

A ball reaching a given slot must have taken some number of right-bounces out of the $12$ available, and the number of different routes that do this is a binomial coefficient.

$\text{routes to the slot with } k \text{ right-bounces} = \binom{12}{k}$

Those counts are $1, 12, 66, 220, 495, 792, 924, 792, \ldots$, huge in the middle and tiny at the edges, because there are many ways to balance rights and lefts and only one way to bounce right every single time. More routes means more balls, so the pile takes the shape of the coefficients, which is the bell.

Gravity is uniform across the board and the balls do not interact; the shape comes entirely from counting.

Standardize both endpoints.

$z_1 = \frac{157 - 164}{7} = -1.0, \qquad z_2 = \frac{178 - 164}{7} = 2.0$

Find the area of the band, converting the negative endpoint by symmetry.

$P(-1.0 < Z < 2.0) = \Phi(2.0) - \Phi(-1.0) = 0.9772 - 0.1587 = 0.8185$

Apply that fraction to the group.

$800 \times 0.8185 = 654.8$

So about $655$ women fall in that range. Using $\Phi(2.0)$ alone would give about $782$, but that would wrongly include every woman shorter than $157$ cm.

The constant $\frac{1}{\sqrt{2\pi}}$ appears in both heights, so it cancels in the ratio, leaving only the exponential.

$\frac{\varphi(2)}{\varphi(0)} = \frac{e^{-2^2/2}}{e^{-0^2/2}} = \frac{e^{-2}}{e^{0}} = e^{-2}$

Now evaluate, using $e \approx 2.718$.

$e^{-2} = \frac{1}{e^2} \approx \frac{1}{7.389} \approx 0.135$

So the curve at two standard deviations out stands at only about $0.14$ of its peak height, roughly a seventh. The value $0.37$ is $e^{-1}$, which is the ratio at $z = \sqrt{2}$, not at $z = 2$. The squared term in the exponent is what makes the curve collapse so quickly.

The event $|Z| < 0.7$ is the band running from $-0.7$ to $0.7$, so subtract the two areas to the left.

$\Phi(-0.7) = 1 - \Phi(0.7) = 1 - 0.7580 = 0.2420$

$P(-0.7 < Z < 0.7) = 0.7580 - 0.2420 = 0.5160$

So just over half the distribution lies within $0.7$ standard deviations of the mean, which is believable given that a full standard deviation each way captures about $68\%$. The value $0.4840$ is the complement, the two tails outside the band.

The 68-95-99.7 rule is a statement about the normal curve, and it inherits every one of that curve's assumptions, symmetry above all.

$\text{rule assumes: symmetric, bell-shaped, thin tails}$

Income data has none of those. The long right tail drags the mean above the median, and the bands $\mu \pm \sigma$ no longer sit sensibly around the bulk of the data. The rule could easily place a band boundary below zero income, which is meaningless.

The mean and the standard deviation can still be computed for any data set (and a standard deviation is never negative, being a square root of an average of squares). What fails is not the arithmetic but the model: the numbers no longer mean what the rule assumes they mean. That is exactly why "normal" is such a misleading name for this curve.