The Normal Distribution: Additional Practice (Set 2)

"Unusual" means far from the mean in standard deviations, so compute every z-score and compare the sizes, ignoring sign.

$\frac{31 - 25}{4} = 1.5, \qquad \frac{118 - 100}{15} = 1.2, \qquad \frac{210 - 180}{25} = 1.2$

$\frac{6.2 - 7.0}{0.5} = \frac{-0.8}{0.5} = -1.6$

The largest distance is $|-1.6| = 1.6$, belonging to $x = 6.2$ with $\mu = 7.0$ and $\sigma = 0.5$. Being below the mean makes an observation no less unusual, and the raw gap of only $0.8$ is misleading because that distribution is so tightly packed.

The area to the left of $1116$ is $0.0668$. The table has no negative entries, so find the positive twin: $1 - 0.0668 = 0.9332 = \Phi(1.5)$, which means the cutoff sits at $z = -1.5$.

$-1.5 = \frac{1116 - 1200}{\sigma} = \frac{-84}{\sigma}$

Solve for the standard deviation.

$\sigma = \frac{-84}{-1.5} = 56 \text{ hours}$

The figure $84$ h is the distance from the mean down to the cutoff, and that distance covers one and a half standard deviations, not one.

First compute the target. The area to the right of $0.9$ is what is left over from the total.

$P(Z > 0.9) = 1 - \Phi(0.9) = 1 - 0.8159 = 0.1841$

By symmetry the right tail beyond $+0.9$ is the mirror image of the left tail below $-0.9$, so $\Phi(-0.9)$ and $P(Z < -0.9)$ are two more names for that same area of $0.1841$. All three of those agree.

The odd one out is the remaining expression.

$\Phi(0.9) - 0.5 = 0.8159 - 0.5 = 0.3159$

That is the area from the centre out to $0.9$, a band inside the curve rather than a tail, and $0.3159 \ne 0.1841$.

"The top $5\%$" means $5\%$ of the area lies above the cutoff, so $95\%$ lies below it.

$\Phi(z) = 0.95 \;\Longrightarrow\; z = 1.645$

The other three are all real cutoffs, just for other questions: $z = 1.28$ leaves $10\%$ above it, $z = 1.96$ leaves $2.5\%$ above it (which is why $\pm 1.96$ encloses the middle $95\%$), and $z = 2.0$ leaves $2.28\%$ above it. Keep the one-tailed and two-tailed versions apart, since $1.96$ and $1.645$ are easy to swap.

The word "exactly" matters here. Going out $2$ standard deviations captures $95.45\%$, which is what the 68-95-99.7 rule rounds to $95\%$, but it is not exactly $95\%$.

For exactly $95\%$ in the middle, each tail must hold $2.5\%$, so the area to the left of the upper cutoff is

$1 - 0.025 = 0.975 = \Phi(1.96)$

$\Longrightarrow\; z = 1.96$

So the band runs from $-1.96$ to $1.96$. The value $1.645$ cuts off a $5\%$ tail on one side only, which leaves $90\%$ in the middle, not $95\%$.

The band is symmetric, so the area left over 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 $c$ is therefore everything but the upper tail.

$\Phi(c) = 1 - 0.0668 = 0.9332 \;\Longrightarrow\; c = 1.5$

A quick sanity check: $c = 1.0$ would enclose only about $68\%$ and $c = 2.0$ about $95\%$, so a band holding $86.64\%$ must sit between them.

Standardize the cutoff.

$z = \frac{87 - 72}{10} = \frac{15}{10} = 1.5$

A distinction means scoring at or above the cutoff, which is the area to the right. (For a continuous distribution the endpoint itself carries no area, so "at least $87$" and "more than $87$" give the same answer.)

$P(Z > 1.5) = 1 - \Phi(1.5) = 1 - 0.9332 = 0.0668 = 6.68\%$

The figure $93.32\%$ is the fraction who fail to earn a distinction, and $15\%$ mistakes the raw gap of $15$ points for a percentage.

Standardize each finishing time against its own race.

$z_P = \frac{195 - 240}{30} = \frac{-45}{30} = -1.5, \qquad z_Q = \frac{46 - 55}{5} = \frac{-9}{5} = -1.8$

For race times, lower is better, so the better performance is the one further BELOW its mean, which means the more negative z-score. Runner Q, at $-1.8$ standard deviations, beat their field by more than runner P did at $-1.5$.

Comparing $195$ minutes with $46$ minutes directly is meaningless, because the two races are not the same distance. That is exactly the situation the z-score exists to fix.

Read the formula as a composition of the graph transformations from chapter 2. Replacing the input $z$ by $\frac{x - \mu}{\sigma}$ does two things at once: subtracting $\mu$ shifts the graph right by $\mu$, and dividing by $\sigma$ multiplies every horizontal distance from the centre by $\sigma$. The outer factor $\frac{1}{\sigma}$ then divides every height by $\sigma$.

$\text{shift right } \mu, \quad \text{widths} \times \sigma, \quad \text{heights} \div \sigma$

A note on vocabulary, since chapter 2 was strict about it. That lesson names a horizontal scaling by the input multiplier, never by its reciprocal, and the input multiplier here is $\frac{1}{\sigma}$, so in its language this is a horizontal compression by $\frac{1}{\sigma}$. It is the same move: dividing the inputs by $\frac{1}{\sigma}$ multiplies them by $\sigma$. Describing it by what happens to distances avoids the argument altogether and stays correct even when $\sigma < 1$.

The vertical shrink is essential: without it, a wide curve would also be a tall one and the total area would grow to $\sigma$ instead of staying at $1$. The two rescalings cancel exactly, which is why every area is preserved and one table serves every normal distribution. And $\mu$ and $\sigma$ play quite different roles, so a description that shifts by $\sigma$ and scales by $\mu$ has swapped them, which is not a small slip.

The rule gives the area for $z = 1$, $z = 2$ and $z = 3$, and nothing else. But a real cutoff lands wherever the data put it.

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

A z-score of $1.5$ has no entry in the rule, and interpolating between $68\%$ and $95\%$ by eye is guesswork. The table supplies $\Phi(1.5) = 0.9332$ directly.

The rule is not wrong, just coarse: it is the numerical answers $68.27\%$, $95.45\%$ and $99.73\%$ rounded off. And it applies perfectly well to every normal curve, because it is stated in standard deviations from the mean rather than in raw units.

The denominator $1024$ is the same for every $k$, so the shape of the eleven probabilities is entirely the shape of the numerators, which are the tenth row of Pascal's triangle.

$1,\; 10,\; 45,\; 120,\; 210,\; 252,\; 210,\; 120,\; 45,\; 10,\; 1$

Those counts are huge in the middle and tiny at the ends, for a concrete reason: there are $252$ different ways to arrange five heads among ten flips, and only one way to get all ten. Divided by $1024$, they trace a hump.

Adding to $1$ is true of every probability distribution, including perfectly flat ones, so it cannot explain the shape. Fairness only makes the hump symmetric; a biased coin still produces a hump, just an off-centre one.

The band is symmetric about the centre, so the leftover area splits equally between the two tails.

$\frac{1 - 0.9544}{2} = \frac{0.0456}{2} = 0.0228 \text{ in each tail}$

The area to the left of $c$ is therefore everything except the upper tail.

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

This is the table's version of the "$95$" in the 68-95-99.7 rule: two standard deviations really enclose $95.44\%$, not a round $95\%$.