The Normal Distribution: Free Response

For a continuous quantity, probability is defined as area under the density curve, never as the height of the curve at a point. The engineer read off a height, $0.15$, which is a different kind of number entirely: it can even exceed $1$ on some curves, which a probability never can.

The actual probability of filling in exactly the average time is the area above that single value, and a single point has zero width, so

$P(X = \mu) = 0.$

What the engineer should say instead is that the curve is tallest at the mean, which tells you fill times cluster there, but that fact is read from the shape of the curve, not turned into a probability by itself.

The area between two points is the area up to the larger one with the area up to the smaller one removed, since the smaller region is counted inside the larger one.

$P(42 < X < 58) = 0.82 - 0.35 = 0.47.$

So about $47\%$ of fill times fall in that window. Nothing here needed the shape of the curve or its mean and standard deviation, only the two given areas.

The model is not broken. "Probability $0$" and "impossible" coincide for a finite, discrete list of outcomes, but a continuous quantity has infinitely many possible values packed into every interval, and the areas above them still have to add up to $1$. The only way that can happen is if each individual point contributes nothing on its own, exactly the way an infinitely thin slice of area is genuinely zero even though the curve still passes through it:

$P(X = 50) = 0, \qquad \text{yet} \qquad P(49.5 < X < 50.5) > 0.$

Asking for "exactly $50$ seconds" is really asking for $50.000\ldots$ with infinitely many trailing zeros, a claim about infinitely many decimal places that no stopwatch and no real bottling event could ever confirm or refute. Real questions about this machine are always about intervals, such as between $49.5$ and $50.5$ seconds, and those intervals do have positive area and positive probability, as shown above. So the model is behaving exactly as a continuous model should; the engineer's objection mistakes a feature of continuous probability for a flaw in it.

Locate both endpoints in standard deviations from the mean. Since $58.5 = 60 - 3(0.5)$, that endpoint sits $3$ standard deviations below the mean. Since $60.5 = 60 + 1(0.5)$, that endpoint sits $1$ standard deviation above the mean. So the question asks for the area from $-3$ to $+1$.

Add the rule's own band figures across that stretch: from $-3$ to $-2$ is $2.35\%$, from $-2$ to $-1$ is $13.5\%$, from $-1$ to $0$ is $34\%$, and from $0$ to $+1$ is another $34\%$:

$2.35\% + 13.5\% + 34\% + 34\% = 83.85\%.$

So about $83.85\%$ of bolts fall in that range.

Locate $59$ mm in standard deviations: $59 = 60 - 2(0.5)$, so it sits exactly $2$ standard deviations below the mean. By the rule, $95\%$ of bolts lie within $2$ standard deviations of the mean, so $5\%$ lie outside that band, split evenly between the two tails by symmetry:

$\frac{100\% - 95\%}{2} = \frac{5\%}{2} = 2.5\%.$

About $2.5\%$ of bolts fall below $59$ mm and are rejected as undersized.

The $95\%$ figure is not the true computed area, it is a rounded value chosen because it is easy to remember and close enough for a quick estimate. The more precisely computed area under the normal curve within $2$ standard deviations of the mean, found numerically rather than rounded for convenience, is about $95.45\%$:

$95.45\% - 95\% = 0.45\text{ percentage points}.$

The rule's number and the computed number agree to within about half a percentage point, which is all the rule ever promised.

A very large batch of bolts is exactly the situation where that small gap becomes visible: with enough bolts measured, the observed fraction settles toward the true, more precisely computed area, about $95.45\%$, rather than toward the rounded $95\%$ used for quick mental estimates. So the supervisor's batch is not misbehaving; it is revealing the rounding the rule always carried.

Subtract the mean to find the distance from the centre, then divide by the standard deviation to convert that distance into standard deviations.

$z = \frac{50 - 38}{6} = \frac{12}{6} = 2.0.$

The delivery took $2$ standard deviations longer than average.

The table entry $\Phi(z)$ gives the area to the left of $z$, but this question wants the area to the right, so subtract from the total area of $1$:

$P(X > 50) = P(Z > 2.0) = 1 - \Phi(2.0) = 1 - 0.9772 = 0.0228.$

About $2.28\%$ of deliveries take longer than $50$ minutes.

Standardize both endpoints first.

$z_1 = \frac{29 - 38}{6} = \frac{-9}{6} = -1.5, \qquad z_2 = \frac{44 - 38}{6} = \frac{6}{6} = 1.0.$

The table has no negative entries, so use symmetry for the lower one:

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

For the area between two values, take the area left of the upper one and remove the area left of the lower one:

$P(29 < X < 44) = \Phi(1.0) - \Phi(-1.5) = 0.8413 - 0.0668 = 0.7745.$

About $77.45\%$ of deliveries fall in that window. Since "three-quarters" means $75\%$, and the computed figure $77.45\%$ sits a little above that, the ad's claim holds up; the true share is slightly higher than what is being advertised, not lower.

The table records area to the left, but the organizers described the cutoff in terms of the area to the right (the top slice). Convert by subtracting from the total area of $1$:

$1 - 0.0228 = 0.9772.$

So the cutoff score is the value with $0.9772$ of the distribution below it.

Search the table for the left-area found in part A. It matches $\Phi(2.0) = 0.9772$ exactly, so the cutoff sits $2.0$ standard deviations above the mean. Walk from the mean by that many standard deviations:

$x = \mu + z\sigma = 640 + 2.0(18) = 640 + 36 = 676.$

A score of $676$ or better earns the commendation.

The score $604 = 640 - 2.0(18)$ sits exactly $2.0$ standard deviations below the mean, so its z-score is $-2.0$. By symmetry,

$\Phi(-2.0) = 1 - \Phi(2.0) = 1 - 0.9772 = 0.0228.$

That matches the $2.28\%$ figure the whole problem started from, but now it names the bottom $2.28\%$ of scores instead of the top. This is exactly what the curve's symmetry guarantees: the tail below $2.0$ standard deviations under the mean is a mirror image of the tail above $2.0$ standard deviations over it, so the two tails carry the same area, $2.28\%$ each. It confirms that the arithmetic in part B is internally consistent, since the same $0.0228$ figure appears whether it is read directly off the table on the positive side or reconstructed by symmetry on the negative side.

$z = \frac{235 - 250}{15} = \frac{-15}{15} = -1.0.$

The value $235$ ms sits $1$ standard deviation below the mean, so its z-score is negative.

The student's arithmetic for $\Phi(-1.0) = 1 - \Phi(1.0) = 0.1587$ is correct on its own, but $\Phi(-1.0)$ is the area to the LEFT of $-1.0$, which answers $P(X < 235)$, not the question that was asked. The student stopped one step early and reported a left-tail area as if it were the requested right-tail area.

Since $235$ ms is below the mean, most reaction times should actually be ABOVE it, so the correct probability should be well over $0.5$, and $0.1587$ was never plausible on those grounds alone. Finish the calculation correctly:

$P(X > 235) = 1 - P(X < 235) = 1 - \Phi(-1.0) = 1 - 0.1587 = 0.8413.$

So about $84.13\%$ of reaction times exceed $235$ ms, not $15.87\%$.

Work with an unspecified normal distribution with mean $\mu$ and an unspecified cutoff $c$ satisfying only $c < \mu$, without plugging in any numbers.

The curve is symmetric about $\mu$, so exactly half of the total area lies above $\mu$ and half lies below it:

$P(X > \mu) = 0.5.$

Because $c < \mu$, every value greater than $\mu$ is automatically also greater than $c$, so the region $\{X > \mu\}$ sits entirely inside the region $\{X > c\}$. That means $P(X > c)$ includes all of $P(X > \mu)$, plus the extra sliver of area between $c$ and $\mu$ itself:

$P(X > c) = P(X > \mu) + P(c < X < \mu) = 0.5 + P(c < X < \mu).$

The density curve is strictly positive between $c$ and $\mu$ (they are not equal, since $c < \mu$), so that extra sliver has strictly positive area:

$P(c < X < \mu) > 0.$

Adding a strictly positive amount to $0.5$ gives $P(X > c) > 0.5$, for any normal distribution and any $c$ below its mean, which is exactly the pattern the student's problem, and the corrected answer $0.8413$, both happened to show.