The table gives Q(z), the area to the right of z for positive z. Every other probability is built from it: a right tail is Q(z), and a left tail is 1 − Q(z) or Q(a), depending on the sign.
This lesson is part of probability distributions. It assumes you can already standardise a normal random variable.
How do I match each tail to an expression?
Sketch a bell curve, mark the mean, and shade what the question asks for. Then use the table below.
| Shaded region | Condition | Expression |
|---|---|---|
| Right of a positive z | P(Z > z) | Q(z) |
| Left of a negative z | P(Z < −a) | Q(a) |
| Left of a positive z | P(Z < z) | 1 − Q(z) |
| Right of a negative z | P(Z > −a) | 1 − Q(a) |
| Between two values | P(a < Z < b) | Subtract the tails from 1, or subtract probabilities |
The key habit is to decide from the sketch, not from the sign alone. The sign tells you which side of the mean, and the shading tells you which tail.
Worked example: three probabilities, one distribution
Here is an original set. The mass X of a fictional mango is normally distributed as X ~ N(400, 30²), in grams. Find three probabilities.
(a) P(X > 445). The region is the right tail. z = (445 − 400) ÷ 30 = 1.5. So P = Q(1.5) = 0.0668.
(b) P(X < 370). The region is the left tail. z = (370 − 400) ÷ 30 = −1. Left of a negative z, so P = Q(1) = 0.1587.
(c) P(370 < X < 445). This is the middle strip. Remove the two tails from the whole curve.
P = 1 − 0.1587 − 0.0668 = 0.7745.
Size check: the strip runs from 1 standard deviation below the mean to 1.5 above, so most of the curve is shaded, and 0.77 fits.
The mistake that costs marks
The common slip is to read the table at the positive value of a negative z and then subtract it from 1 as if it were an upper tail. For part (b) that gives 1 − 0.1587 = 0.8413, which is the region to the right of 370.
| Step | Wrong | Right |
|---|---|---|
| Sketch | (skipped) | Shade left of 370 |
| z | −1 | −1 |
| Expression | 1 − Q(1) | Q(1) |
| P(X < 370) | 0.8413 | 0.1587 |
| Size check | A thin left tail cannot be 84% | 0.16, small, as expected |
The size check is quick: 370 sits one standard deviation below the mean, so the area below it is small, not large.
A second example: everything below a positive z
Use the same mangoes. Find P(X < 430).
z = (430 − 400) ÷ 30 = 1. The region is everything to the left of a positive z, so P = 1 − Q(1) = 1 − 0.1587 = 0.8413.
Here the answer is large because the shading includes the whole left half of the curve.
Check yourself
X ~ N(20, 2²). Find P(X < 17) and P(X > 17).
Answer
z = (17 − 20) ÷ 2 = −1.5.
P(X < 17) is a left tail below a negative z, so it equals Q(1.5) = 0.0668.
P(X > 17) is the rest of the curve, so it equals 1 − 0.0668 = 0.9332.
The two answers add to 1, as they must.
What to study next
The next step is using tail probabilities to find unknown values. Continue with finding unknown normal-distribution parameters, or sharpen your diagram checks with reconciling a calculation with the shaded diagram.
If you want a teacher to watch you sketch and shade on new questions, see online one-to-one Additional Mathematics tuition.