The Normal Distribution: Practice

The table stores the area to the left, so the area to the right is everything that is left over out of the total area $1$.

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

The entry $0.9332$ answers $P(Z < 1.5)$ instead. Subtracting from $0.5$ rather than $1$ would give $0.4332$, which forgets that the whole left half of the curve also lies below $1.5$.

Standardize the cutoff before touching the table.

$z = \frac{488 - 500}{8} = \frac{-12}{8} = -1.5$

Now use symmetry, since the table lists only positive z-scores.

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

So about $6.68\%$ of bags are underweight. The value $0.9332$ is the fraction weighing more than $488$ g, which is the complement of what was asked.

Convert the cutoff to a z-score.

$z = \frac{46 - 40}{5} = \frac{6}{5} = 1.2$

The question asks for the area to the right, so subtract the table entry from $1$.

$P(Z > 1.2) = 1 - \Phi(1.2) = 1 - 0.8849 = 0.1151$

About $11.51\%$ of the batteries outlast $46$ hours. The entry $0.8849$ is the fraction that last less than $46$ hours.

Raw scores from different tests are measured on different scales, so they cannot be compared directly. Standardize each one.

$z_{\text{Alex}} = \frac{78 - 70}{4} = 2.0, \qquad z_{\text{Bo}} = \frac{88 - 80}{10} = 0.8$

Alex is $2.0$ standard deviations above the class mean while Bo is only $0.8$ above, so Alex performed better relative to the class.

Both students beat their mean by $8$ points, which is exactly why the raw gap is misleading: on Alex's test $8$ points is a large distance, and on Bo's it is a small one, because the spreads differ.

If $15.87\%$ of the area lies above the cutoff, then the area below it is the rest.

$1 - 0.1587 = 0.8413$

Search the table for the entry $0.8413$. It appears exactly at $z = 1.0$, so the cutoff sits one standard deviation above the mean.

$x = \mu + z\sigma = 500 + 1.0(100) = 600$

A score of $600$ or better earns the award.

Subtract the area to the left of the lower endpoint from the area to the left of the upper one, using symmetry for the negative endpoint.

$\Phi(-2) = 1 - \Phi(2) = 1 - 0.9772 = 0.0228$

$P(-2 < Z < 2) = 0.9772 - 0.0228 = 0.9544$

This is the exact table version of the "$95$" in the 68-95-99.7 rule, and it shows that the rule is rounded: the true area out to two standard deviations is a little over $95\%$. The entry $0.9772$ alone would also count the whole left tail below $-2$, which the band excludes.

The area to the right of a point is everything that is not to its left, and the total area is $1$.

$P(Z > 1.7) = 1 - \Phi(1.7) = 1 - 0.9554 = 0.0446$

The bare entry $\Phi(1.7) = 0.9554$ is the area to the left, the opposite region. Subtracting from $0.5$ instead of $1$ gives $0.5 - 0.9554 = -0.4554$, a negative number, which cannot be a probability at all, and $\Phi(1.7) - 0.5 = 0.4554$ is the area from the centre out to $1.7$.

Locate both endpoints in standard deviations. Since $162 = 170 - 8$, it sits at $z = -1$; since $186 = 170 + 2(8)$, it sits at $z = +2$.

Now add the bands of the rule between those two points. From $-1$ to $0$ is $34\%$, from $0$ to $+1$ is another $34\%$, and from $+1$ to $+2$ is $13.5\%$.

$34\% + 34\% + 13.5\% = 81.5\%$

The answer $95\%$ would need the band to be symmetric, running from $-2$ to $+2$, but this one reaches only one standard deviation below the mean.

Evaluate the density formula at $z = 0$ and see what kind of quantity comes out.

$\varphi(0) = \frac{1}{\sqrt{2\pi}} e^{0} = \frac{1}{\sqrt{2\pi}} \approx 0.3989$

This is the value of the function, that is, the height of the curve at its peak, and a height is never a probability. The probability that $Z$ equals exactly $0$ is $0$, the probability that $Z < 0$ is $0.5$ by symmetry, and the area from $0$ to $1$ is $0.3413$.

Handle the negative endpoint with symmetry. The area to the right of $-1.1$ mirrors the area to the left of $+1.1$.

$P(Z > -1.1) = P(Z < 1.1) = \Phi(1.1) = 0.8643$

Most of the curve lies to the right of a point that sits below the mean, so the answer must be larger than $0.5$. The value $0.1357$ is the small area to the left of $-1.1$, the complement of what was asked.

Convert both endpoints with the symmetry rule, then subtract.

$\Phi(-1.0) = 1 - 0.8413 = 0.1587, \qquad \Phi(-2.0) = 1 - 0.9772 = 0.0228$

$P(-2.0 < Z < -1.0) = 0.1587 - 0.0228 = 0.1359$

This band is the mirror image of the band from $1$ to $2$, so it has exactly the same area, which is a good check on the arithmetic. The value $0.0228$ alone is the whole tail below $-2$, with no lower cutoff at $-1$.

The result is the Central Limit Theorem, and it is the reason the bell turns up in so many unrelated measurements: heights, measurement errors, and the slots of a Galton board are all totals of many small independent nudges.

$\text{many small independent effects, none dominating} \;\longrightarrow\; \text{normal}$

The proviso earns its keep. When one ingredient is allowed to swamp the sum, the bell never forms, which is why heavy-tailed quantities such as incomes and stock returns are not normal despite also being built out of many effects. The theorem's proof is far beyond this course, which is why the lesson names it rather than deriving it. The Binomial Theorem expands $(a+b)^n$, and the other two are results about polynomial roots.