This note covers the normal distribution: its bell-shaped curve, how to find probabilities as areas under it, and how to work backwards from a probability to a value (inverse normal). It models many measurements, such as masses, heights, reaction times and manufacturing errors.
Key ideas
A continuous random variable \(X\) that follows a normal distribution is written \(X \sim N(\mu, \sigma^2)\). The two parameters are the mean \(\mu\) and the variance \(\sigma^2\), so the standard deviation is \(\sigma\). Notice that the second number in the brackets is the variance. If a question says \(X \sim N(500, 12^2)\), then \(\sigma = 12\).
Shape of the curve.
- It is bell-shaped and symmetrical about the vertical line \(x = \mu\).
- The mean, median and mode are all equal to \(\mu\).
- The total area under the curve is 1, because the total probability is 1.
- The curve never touches the horizontal axis, but it gets very close beyond about 3 standard deviations from the mean.
- A larger \(\sigma\) gives a wider, flatter curve. A smaller \(\sigma\) gives a narrower, taller curve.
As a rough guide, about 68% of values lie within one standard deviation of the mean, about 95% within two, and about 99.7% within three.
Probability is area. For a continuous variable, \(P(X = a) = 0\) for any single value, so \(P(X \lt a)\) and \(P(X \le a)\) are the same. Only intervals have probability. This means you do not need to worry about "less than" versus "less than or equal to".
Finding probabilities. Use the normal cumulative distribution function on your GDC. Enter the lower bound, the upper bound, the mean and the standard deviation. For an unbounded side, enter a very large number such as \(10^{99}\) (or \(-10^{99}\)). Always sketch the curve and shade the required region first. The sketch lets you check that your answer is sensible.
Using symmetry. Because the curve is symmetrical, \(P(X \gt \mu + a) = P(X \lt \mu - a)\). Also \(P(X \lt \mu) = 0.5\). The complement rule still applies: \(P(X \gt a) = 1 - P(X \lt a)\).
Inverse normal. Sometimes the probability is given and you must find the value. Use the inverse normal function on your GDC. It takes the area to the left of the value, the mean and the standard deviation, and returns the value. If the question gives an area to the right, subtract it from 1 first.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(X \sim N(\mu, \sigma^2)\) | \(X\) is normally distributed with mean \(\mu\) and standard deviation \(\sigma\). |
| \(P(a \lt X \lt b)\) | Area under the curve between \(a\) and \(b\). Use the normal cumulative distribution function. |
| \(P(X \gt a) = 1 - P(X \lt a)\) | Complement rule, or use the upper bound \(10^{99}\) on your GDC. |
| \(P(X \lt \mu) = 0.5\) | The curve is symmetrical, so half the area is each side of the mean. |
| \(P(X \gt \mu + a) = P(X \lt \mu - a)\) | Symmetry about the mean. |
| Inverse normal | Given \(P(X \lt k) = q\), the GDC returns \(k\). Always use the area to the left. |
Worked examples
Example 1 A machine fills sugar bags. The mass of sugar, \(X\) grams, is normally distributed with mean 500 and standard deviation 12. Find (a) the probability a bag has less than 485 g, (b) the probability a bag has between 480 g and 510 g.
\(X \sim N(500, 12^2)\). Sketch a bell curve centred at 500.
(a) Shade the area to the left of 485. Use the normal cumulative distribution function with lower bound \(-10^{99}\), upper bound 485, mean 500 and standard deviation 12: \(P(X \lt 485) = 0.1056\dots \approx 0.106\).
(b) Shade the area between 480 and 510. Use the same function with lower bound 480 and upper bound 510: \(P(480 \lt X \lt 510) = 0.7499\dots \approx 0.750\).
Answer: (a) 0.106; (b) 0.750.
Example 2 For the sugar bags in Example 1, the company labels the heaviest 10% of bags as "extra full". Find the minimum mass of a bag that gets this label.
We need \(k\) such that \(P(X \gt k) = 0.1\). The inverse normal function uses the area to the left, so \(P(X \lt k) = 1 - 0.1 = 0.9\).
Use the inverse normal function with area 0.9, mean 500 and standard deviation 12: \(k = 515.378\dots\).
Sense check: the answer is above the mean, as it should be for the top 10%.
Answer: about 515 g.
Example 3 The battery life \(T\) hours of a tablet is normally distributed with mean 9.0 and standard deviation 0.8. It is known that \(P(T \gt 10.2) = 0.0668\) to 3 s.f. (a) Without a calculator, write down \(P(T \lt 7.8)\). (b) Hence find \(P(7.8 \lt T \lt 10.2)\).
The mean is 9.0. The value 10.2 is \(1.2\) above the mean, and 7.8 is \(1.2\) below the mean. Symmetry about the mean applies.
(a) \(P(T \lt 7.8) = P(T \gt 10.2) = 0.0668\).
(b) The two tails together have probability \(2 \times 0.0668 = 0.1336\). The area between is the rest: \(1 - 0.1336 = 0.8664\).
Answer: (a) 0.0668; (b) 0.866.
Common errors
- Common error: entering the variance instead of the standard deviation into the calculator. For \(N(50, 16)\), the standard deviation is 4, not 16.
- Common error: giving the inverse normal function an area to the right. The function needs the area to the left.
- Common error: skipping the sketch. Without it, probabilities greater than 0.5 for a value below the mean (or the reverse) go unnoticed.
- Common error: thinking \(P(X \lt a)\) and \(P(X \le a)\) differ for a normal variable. They are equal.
- Common error: rounding values before they go into the next step. Keep full calculator values and round only the final answer.
Exam tip: Draw and label a quick sketch showing the mean, the value or values given, and the shaded area. State the distribution, such as \(X \sim N(500, 12^2)\), before using the calculator.
Check yourself
- \(X \sim N(50, 4^2)\). Find \(P(X \gt 56)\).
Answer
Use the normal cumulative distribution function with lower bound 56 and upper bound \(10^{99}\): \(P(X \gt 56) = 0.0668\). - \(X \sim N(50, 16)\). Find \(k\) such that \(P(X \lt k) = 0.2\).
Answer
The variance is 16, so \(\sigma = 4\). Inverse normal with area 0.2, mean 50 and standard deviation 4 gives \(k = 46.6\). - \(X \sim N(100, 15^2)\) and \(P(X \gt 115) = 0.1587\). Find \(P(X \lt 85)\) and \(P(85 \lt X \lt 115)\).
Answer
85 is 15 below the mean and 115 is 15 above, so \(P(X \lt 85) = 0.1587\). Then \(P(85 \lt X \lt 115) = 1 - 2(0.1587) = 0.683\). - \(X \sim N(8, 25)\). State the mean and standard deviation, and write down \(P(X \lt 8)\).
Answer
The mean is 8 and the standard deviation is \(\sqrt{25} = 5\). By symmetry \(P(X \lt 8) = 0.5\).