Revision: Statistics and Probability

For \(X \sim \mathcal{N}(\mu, \sigma^2)\), the probability of an interval is an area under the normal curve, an integral of the density whose antiderivative cannot be written with the usual functions (polynomials, exponentials, logarithms): the normal cumulative distribution function of a graphing calculator computes it. Its name depends on the model: Normal C.D, normalcdf, normCdf.

1. State the model: \(X \sim \mathcal{N}(\mu, \sigma^2)\), with its mean \(\mu\) and its standard deviation \(\sigma\) — not the variance \(\sigma^2\): for \(X \sim \mathcal{N}(50, 16)\), \(\sigma = 4\).
2. Sketch and shade first: draw a bell curve centred at \(\mu\), mark the bound or bounds, and shade the region of the event. The sketch predicts the size of the answer: a region that contains the mean and most of the curve has a probability close to \(1\), a region that lies on one side of \(\mu\) a probability less than \(0.5\), and a region far in one tail a probability close to \(0\).
3. Enter the bounds of the region, as in the table below, then \(\mu\) and \(\sigma\), each in the field that carries its name: the order of the fields depends on the model, and the command normalcdf takes the lower bound, the upper bound, \(\mu\), then \(\sigma\). A region with no left end takes as its lower bound a negative number so large that it lies very many standard deviations below the mean, such as \(-10^{99}\), and a region with no right end a positive number as large, such as \(10^{99}\), as its upper bound.
4. Compare the screen with the sketch, and report the probability to \(3\) significant figures, unless the question asks otherwise.

Event	Lower bound	Upper bound
\(a < X < b\)	\(a\)	\(b\)
\(X < b\)	\(-10^{99}\)	\(b\)
\(X > a\)	\(a\)	\(10^{99}\)

Strict or not, an inequality gives the same probability: \(X\) is continuous, so \(P(X = a) = 0\) and \(P(X \leq a) = P(X < a)\).
The sketch and the mock-up show the computation of \(P(60 < X < 72)\) for \(X \sim \mathcal{N}(64, 5^2)\): the region contains the mean and most of the curve, and the screen gives \(P(60 < X < 72) \approx 0.733\).

The reverse question gives a probability and asks for a value of \(X\). The inverse normal function of the calculator (Inverse Normal, invNorm) returns the value \(k\) such that \(P(X < k) = p\), for \(0 < p < 1\), from the area \(p\) to the left of \(k\), \(\sigma\) and \(\mu\).

1. State the model \(X \sim \mathcal{N}(\mu, \sigma^2)\) and name the unknown value \(k\).
2. Sketch and shade: place \(k\) on the side of \(\mu\) that the probability indicates — a left area less than \(0.5\) puts \(k\) below the mean — and shade the given area.
3. Turn the condition into a left area. The condition \(P(X < k) = p\) is entered as it is; \(P(X > k) = q\) becomes \(P(X < k) = 1 - q\). An interval centred on \(\mu\) that contains the proportion \(p\) of the values leaves \(\frac{1 - p}{2}\) in each tail: its ends are the values with left areas \(\frac{1 - p}{2}\) and \(\frac{1 + p}{2}\).
4. Enter the area, \(\mu\) and \(\sigma\), each in its field (the command invNorm takes the area, \(\mu\), then \(\sigma\); the tail is set to the left, on the models that ask for one), then check the position of \(k\) against the sketch and report it to \(3\) significant figures, or to the precision of the data.

The mock-up finds \(k\) such that \(P(X < k) = 0.9\) for \(X \sim \mathcal{N}(64, 5^2)\): a left area of \(0.9\) puts \(k\) above the mean, and \(k \approx 70.4\).