13.3 The Normal distribution as a model — Learn

Prerequisites: the mean and standard deviation from Section 11.5; the idea of area under a curve from Section 8.2; the binomial model from Section 13.1.

By the end of this section you can: describe the Normal curve and its two parameters; explain why a probability is an area for a continuous variable; standardise a value; read a probability from a standard Normal table; and find a value when you are given a probability.

1. Continuous random variables

A continuous random variable can take any value in a range. A height, a mass and a time are continuous. You cannot list every possible value, so you cannot give each one a probability.

For a continuous variable, a probability is an area under a curve. The curve is called a probability density function. The total area under the curve is 1.

One consequence matters for every calculation. The area under a single point is zero, so

For a continuous variable, "less than" and "less than or equal to" give the same probability. This is different from the discrete binomial variable, where $P(X = r)$ can be large.

2. The Normal model

The Normal distribution is a continuous model with a symmetric, bell-shaped curve. It has two parameters.

• The mean $\mu$ sets the centre of the curve.
• The standard deviation $\sigma$ sets the spread. A larger $\sigma$ gives a wider, flatter curve.

We write

The second number is the variance, $\sigma^2$, not the standard deviation. So $X \sim N(20, 25)$ has mean 20 and standard deviation $\sqrt{25} = 5$. This notation is a common trap. Read the second number as the variance every time.

The curve is symmetric about $\mu$. About 68% of the area lies within one standard deviation of the mean, about 95% within two, and about 99.7% within three. You will check these numbers in Section 13.4.

3. Standardising

Every Normal variable can be turned into the standard Normal variable $Z \sim N(0, 1)$, which has mean 0 and standard deviation 1. The change of variable is called standardising:

The value $z$ counts how many standard deviations $x$ lies from the mean. A negative $z$ means $x$ is below the mean.

Standardising does not change a probability. It moves the question onto the curve in every standard Normal table:

The symbol $\Phi$ (a capital Greek phi) is the area to the left of $z$ under the standard Normal curve.

Example 1. The mass of a bag of flour is $X \sim N(20, 5^2)$ grams. Find $P(X < 26)$.

Standardise $x = 26$:

From the table, $\Phi(1.2) = 0.8849$. So

4. A standard Normal table

The table gives $\Phi(z) = P(Z \le z)$ to four decimal places.

For a negative $z$, use the symmetry of the curve:

This table is an extract. In an examination, your calculator or the table you are given provides more values. The method is the same.

Example 2. $Z \sim N(0, 1)$. Find $P(Z > 2)$.

The table gives the area to the left. The area to the right is the complement:

Example 3. $X \sim N(20, 5^2)$. Find $P(X > 12)$.

Standardise $x = 12$:

The area to the right of $-1.6$ equals the area to the left of $1.6$, by symmetry:

Example 4. $X \sim N(20, 5^2)$. Find $P(15 < X < 27)$.

Standardise both ends:

Subtract the smaller area from the larger:

5. Finding a value from a probability

Sometimes the probability is given and the value of $X$ is wanted. Work backwards through the same steps.

Example 5. $X \sim N(50, 6^2)$. Find the value $a$ such that $P(X < a) = 0.90$.

Read the table backwards. Find the value $z$ with $\Phi(z) = 0.90$. A full table gives $z = 1.2816$. Then unstandardise:

So 90% of the values lie below 57.69.

This small table gives the $z$-value for the probabilities used most often.

6. Progress check

Return to the chapter problem. The supplier claims that the height of a mature sunflower is normally distributed with mean 45 cm and standard deviation 6 cm. Let $X$ be the height of a mature plant. Find the probability that a plant is taller than 52 cm.

7. Try it: area under the Normal curve

Predict first. The figure shows a Normal curve with sliders for the mean $\mu$ and the standard deviation $\sigma$, and a value $x$. The shaded area is $P(X < x)$.

Before you move anything, predict what happens to the shaded area when you increase $\sigma$ while keeping $\mu$ and $x$ fixed. Write down whether the area rises, falls or stays the same.

Test. Move the $\sigma$ slider. Watch the shaded area and the readout. Then move $\mu$ and watch the shaded area. Note the two points marked at $\mu - \sigma$ and $\mu + \sigma$.

Explain. Write two sentences. The first explains why increasing $\sigma$ changes the shaded area even though $\mu$ and $x$ are fixed. The second explains what is special about the two marked points, where the curve is steepest.

Watch out. Four mistakes are common.

Reading the variance as the standard deviation. In $X \sim N(20, 25)$, the standard deviation is 5, not 25. Standardise with $z = \dfrac{x - 20}{5}$.

Using the wrong tail. The table gives the area to the left. For "greater than", subtract from 1. Sketch the curve and shade the region before you read the table.

Forgetting to standardise. A value of 52 is not a $z$-value. Divide by $\sigma$ first.

Treating a continuous variable like a discrete one. For a continuous variable, $P(X = 52) = 0$. Only ranges have a positive probability.

Re-check: write the probability as an area, standardise, sketch, then read the table. Check that your answer lies between 0 and 1.

You should now be able to

• describe the Normal curve and name its two parameters;
• explain why "less than" and "less than or equal to" agree for a continuous variable;
• standardise a value using $z = \dfrac{x - \mu}{\sigma}$;
• read $P(Z \le z)$ from a standard Normal table and use $\Phi(-z) = 1 - \Phi(z)$;
• find a probability for a general Normal variable;
• find a value when you are given a probability.