The Normal Distribution: Practice

A normal curve is symmetric about its mean, which is also where its peak sits.

$\text{axis of symmetry: } x = \mu = 60$

The value $0$ is the centre of the standard normal curve, the one whose mean has already been subtracted off. This distribution has not been standardized, so its centre is still $60$.

Read the middle number of the rule, which is the one attached to two standard deviations.

$P(\mu - 2\sigma < X < \mu + 2\sigma) \approx 95\%$

The figure $34\%$ is half of the $68\%$ band, the area from the mean out to one standard deviation on a single side.

The table is built to record cumulative area, accumulated from the far left up to the value $z$.

$\Phi(z) = P(Z < z)$

So the table entry is the area to the left of $z$. Everything else is derived from that by subtraction: the area to the right is $1 - \Phi(z)$, and the height of the curve is not an area and so is never a probability at all.

The table gives areas to the left, and this question asks for the area to the left, so read the entry directly with no adjustment.

$P(Z < 1.0) = \Phi(1.0) = 0.8413$

The value $0.1587$ is the leftover area in the right tail, $1 - 0.8413$, and $0.3413$ is the area from $0$ out to $1$, which is what you get after also removing the left half.

A z-score of $2$ means "two standard deviations above the mean", so start at the mean and walk two standard deviations.

$x = \mu + z\sigma = 20 + 2(3) = 26$

The value $23$ is only one standard deviation up, and $40$ comes from doubling the mean, which has nothing to do with the spread.

Subtract the mean first, keeping the sign, then divide by the standard deviation.

$z = \frac{88 - 100}{12} = \frac{-12}{12} = -1$

The value sits below the mean, so its z-score must be negative. Dropping the minus sign gives $1$, which would place the value at $112$ instead, on the wrong side of the curve.

The two events differ only by the single outcome $X = 12$, and for a continuous quantity a single outcome has zero area and therefore zero probability.

$P(X \le 12) = P(X < 12) + P(X = 12) = P(X < 12) + 0$

So the two probabilities are equal, and the endpoint can be included or excluded freely. This is a genuine difference from discrete probability, where the two do differ whenever $12$ is a possible outcome, because there the outcome $X = 12$ carries real weight.

Read the last number of the rule, the one attached to three standard deviations.

$P(\mu - 3\sigma < X < \mu + 3\sigma) \approx 99.7\%$

It is close to everything but not quite everything: about $0.3\%$ of the distribution still lies further out than three standard deviations. The normal curve never actually reaches zero, so $100\%$ is wrong no matter how far out you go.

In the standard form $\varphi(z) = \frac{1}{\sqrt{2\pi}} e^{-z^2/2}$, the exponent $-z^2/2$ is largest when $z^2$ is smallest, and $z^2$ is smallest at $z = 0$, the centre.

$\text{peak at } z = 0 \;\Longleftrightarrow\; \text{peak at } x = \mu$

Since $z = 0$ means $x = \mu$, the peak of any normal curve sits at its mean. It is only at $x = 0$ when the mean happens to be $0$, which is true of the standard curve and of no other.

This asks for the area to the left, which is exactly what the table stores, so read the entry for $z = 2.0$ directly.

$P(Z < 2.0) = \Phi(2.0) = 0.9772$

The value $0.0228$ is the small tail left over on the right, and $0.9544$ is the area of the central band from $-2$ to $2$, which is a different question.

For an area between two values, take the area to the left of the upper value and remove the area to the left of the lower one.

$P(0 < Z < 1) = \Phi(1) - \Phi(0) = 0.8413 - 0.5000 = 0.3413$

The entry $0.8413$ still includes the whole left half of the curve, which this question does not want, and $0.6826$ is the full band from $-1$ to $1$, which is twice as wide as the one asked for.

The z-score is a distance measured in standard deviations, not in the original units, and its sign records the direction.

$z = \frac{x - \mu}{\sigma} = -2 \;\Longleftrightarrow\; x = \mu - 2\sigma$

So the value sits two standard deviations below the mean. Reading it as "$2$ units below" throws away the division by $\sigma$, which is the whole point of standardizing.