Reading Normal Probabilities

Start from the area to the left

For Z ~ N(0, 1), a table gives $\Phi \left(z\right)=P\left(Z<z\right)$Φ(z) = P(Z < z), the area to the left of z. Every other area is built from these.

At z = 0 the area to the left is a half, since the curve is symmetric about 0: P(Z < 0) = 0.5. Further right, the table gives P(Z < 1) = 0.8413.

The area to the left of z = 0 is half of the whole: P(Z < 0) = 0.5.

A right-hand tail

The area to the right of z is everything that is not to its left. The whole area is 1, so $P\left(Z>z\right)=1-\Phi \left(z\right)$P(Z > z) = 1 − Φ(z). For z = 1: P(Z > 1) = 1 − 0.8413 = 0.1587.

Whether the end point is included makes no difference, since a single value has probability 0: $P\left(Z\ge 1\right)$P(Z ≥ 1) is also 0.1587. By symmetry, P(Z > 1) is the same as P(Z < −1).

The tail to the right of z = 1. The table gives the 0.8413 to its left, so the tail is 1 − 0.8413 = 0.1587.

A band between two values

The area between a and b is everything to the left of b, take away everything to the left of a: $P\left(a<Z<b\right)=\Phi \left(b\right)-\Phi \left(a\right)$P(a < Z < b) = Φ(b) − Φ(a). Subtract the smaller area from the larger.

For example, $P\left(1<Z<2\right)=\Phi \left(2\right)-\Phi \left(1\right)=0.9772-0.8413=0.1359$P(1 < Z < 2) = Φ(2) − Φ(1) = 0.9772 − 0.8413 = 0.1359.

A band with a negative edge needs symmetry first. For P(−1.5 < Z < 0.5), the table gives $\Phi \left(0.5\right)=0.6915$Φ(0.5) = 0.6915 and $\Phi \left(1.5\right)=0.9332$Φ(1.5) = 0.9332, so $\Phi \left(-1.5\right)=1-0.9332=0.0668$Φ(−1.5) = 1 − 0.9332 = 0.0668. Then P(−1.5 < Z < 0.5) = 0.6915 − 0.0668 = 0.6247.

On the original scale

The lifetimes L of a brand of battery follow $N\left(500,40^2\right)$N(500, 40²), in hours. What share of the batteries last more than 560 hours? Standardize first: $z=\frac{560-500}{40}=1.5$z = (560 − 500)/40 = 1.5. Then P(L > 560) = P(Z > 1.5) = 1 − 0.9332 = 0.0668, so about 6.7% of the batteries.

What share last between 440 and 520 hours? The two edges standardize to $\frac{440-500}{40}=-1.5$(440 − 500)/40 = −1.5 and $\frac{520-500}{40}=0.5$(520 − 500)/40 = 0.5, so the answer is P(−1.5 < Z < 0.5) = 0.6247, about 62%.

A sketch is a quick check. The band from 440 to 520 holds the peak, so it should hold well over half of the area, and 0.6247 does. A tail that starts 1.5 standard deviations out should be small, and 0.0668 is.

The battery lifetimes, $N\left(500,40^2\right)$N(500, 40²), with the band from 440 to 520 hours shaded. Its edges are 1.5 standard deviations below the mean and half a standard deviation above it, and it holds 0.6247 of the area.

The usual mistakes

Giving the area to the left as the answer for a tail. The table gives 0.8413 for z = 1, and that is P(Z < 1); the tail P(Z > 1) is 1 − 0.8413.

Running a band from the far left. P(0 < Z < 1.5) is not 0.9332, which includes the half below 0; it is 0.9332 − 0.5 = 0.4332.

Subtracting the wrong way round. A probability is never negative, so $\Phi \left(a\right)-\Phi \left(b\right)$Φ(a) − Φ(b) with a below b has the two areas in the wrong order.