Normal probabilities with the cumulative table

◈ 6 cards

Tail, upper tail and interval for a normal X: standardise, round z to two decimals, read Φ(z) from the cumulative table, then subtract from 1 or subtract two cells.

The table is cumulative

The z-table on the paper gives $Φ (z) = P (Z \leq z)$ — the area to the left of $z$ — for $z$ from $- 3.4$ to $3.4$ in steps of 0.01. The row gives the first two digits, the column the third: $Φ (1.20)$ is the row 1.2, column .00 cell, 0.8849. Two properties do the rest of the work:

Complement: $P (Z > z) = 1 - Φ (z)$ , because the total area is 1.
Symmetry: $Φ (- z) = 1 - Φ (z)$ , because the bell is symmetric about 0. The table prints the negative rows anyway; the identity is the check.

IBS1's printed table is a different table — the area from 0 to $z$ , so it reads 0.3849 where the paper's reads 0.8849. Never transplant a textbook lookup; re-derive it on the cumulative table.

Worked example — four questions about one distribution

Northfield Credit's invoices are $X \sim N (120, 1 5^{2})$ . Each question is: standardise, read, arrange.

Below. $P (X < 138)$ : $z = 1.20$ , so $P (X < 138) = Φ (1.20) = 0.8849$ . The table answers a "less than" question directly.

Above. $P (X > 99)$ : $z = (99 - 120) /15 = - 1.40$ . The table gives $Φ (- 1.40) = 0.0808$ , the area below 99. The area above is the complement:

$P (X > 99) = 1 - Φ (- 1.40) = 1 - 0.0808 = 0.9192$

Equivalently, by symmetry, $P (Z > - 1.40) = P (Z < 1.40) = Φ (1.40) = 0.9192$ — the same cell either way.

Between. $P (105 < X < 135)$ : the two z-scores are $- 1.00$ and $1.00$ . The area between is the area below the upper limit minus the area below the lower:

$P (105 < X < 135) = Φ (1.00) - Φ (- 1.00) = 0.8413 - 0.1587 = 0.6826$

— the empirical rule's 68 %, now to four decimals.

Far tail. $P (X > 150)$ : $z = 2.00$ , $P = 1 - Φ (2.00) = 1 - 0.9772 = 0.0228$ . About one invoice in 44 exceeds \$150.

Rounding z

The table has two decimals. Round $z$ to two decimals before the lookup: an invoice of \$100 has $z = - 20/15 = - 1.333 \dots$ , read at $- 1.33$ as 0.0918. Do not interpolate between cells, and do not round the probability to fewer decimals than the table gives. When a question's $z$ needs rounding, the marking scheme allows the small drift that rounding causes — the course's numeric items widen the tolerance to ±0.002 on exactly those questions and keep ±0.0005 when $z$ lands on a cell.

The three shapes, in one line each

Question	Arrangement
$P(X < a)$	$Φ (z_{a})$
$P(X > a)$	$1 - Φ (z_{a})$
$P(a < X < b)$	$Φ (z_{b}) - Φ (z_{a})$

Strict and non-strict inequalities give the same number (L7.1). A sketch of the bell with the region shaded is the fastest guard against the most common error — reporting $Φ (z)$ when the question asked for the area above.

The negative rows of the cumulative table. Φ(−1.40) = 0.0808 gives P(X > 99) = 0.9192; Φ(−1.00) = 0.1587 is the lower end of the 105–135 interval; Φ(−1.33) = 0.0918 is the cell the L7.6 exam question reads.