The Normal Distribution Examples

A wild pack of Chihuahuas terrorizing the countryside has a mean height of 7.5 inches, with a standard deviation of 1.5 inches. We feel sorry for the person who had to measure that. What proportion of these Chihuahuas are between 6 and 9 inches tall?

You can use this table to answer the question.

When we want to know something about probabilities or proportions of normal distributions, we need to work with Z-scores. We use them to convert a value into the number of standard deviations it is from the mean. The formula is:

μ is another fancy code name for the mean of the normal distribution, while σ is its standard deviation. We can find the Z-scores for 6 and 9 inches now.

How much of the normal distribution falls within 1 standard deviation above or below the mean? According to the Empirical Rule, that's 68% of the distribution. Problem solved, no table needed. The table will tell us the same thing, though.

People's monthly electric bills in Shmoopsville are normally distributed with a mean of $225 and a standard deviation of $55. Those Shmoopers spend a lot of time online. In a group of 500 customers, how many would we expect to have a bill that is $100 or less?

You can use this table to answer the question.

Compared to what we've worked on before, this problem only has one extra step at the end. We'll work out the proportion of the distribution that is below $100, using the Z-score and a standard normal table, and then multiply by how many individuals we actually have.

A journey to the answer begins with a single step, and that's finding the Z-score.

= -2.27

What probability are we looking for again? Oh yeah, that someone spends less than $100 on their bill. Let's draw that:

Our table doesn't give us probabilities below Z, only above Z. It also doesn't include negative Z-scores. Both of these problems fade away when we remember that a normal distribution is symmetric around the mean.

So, we have Pr(x < -Z) = Pr(x > Z). We also have Z = 2.27. We've got a great idea here: let's find 2.27 on the standard normal table to find the probability. Pretty genius, right?

Zoom and enhance, right on the spot in the table we need. The probability of someone paying $100 or less is 0.011604. Out of 500 people, that comes out to (0.011604)(500) = 5.802 people. We'll say "around 6 people" to keep things from getting messy.

On a recent English test, the scores were normally distributed with a mean of 74 and a standard deviation of 7. What proportion of the class would be expected to score between 60 and 80 points?

You can use this table to answer the question.

We can't work with English scores, we need Z-scores.

Much better. Now let's graph this.

We need to find the probability in the middle, Pr(-2 < x < 0.86). That's equal to Pr(x > -2) – Pr(x > 0.86). Both of these are "greater than" probabilities, because that's all the standard normal table we're using gives us.

Let's start with Pr(x > -2). But we also need to start with what's available in the table, which is Pr(x > 2). The normal distribution is symmetric, so we can flip this around to Pr(x < -2). Notice the change in the inequality.

This is the opposite of what we want. The total probability under a normal curve is 1, though, so we can take 1 – Pr(x < -2) to find Pr(x is NOT less than -2), or Pr(x > -2).

If you're even a little confused, check out what we just said, but in picture form.

The upshot is that we need to find Pr(x > 2) from the standard normal table and subtract it from 1. This will give us Pr(x > -2).

Pr(x > -2) = 1 – 0.022750 = 0.97725

We could have also used the Empirical Rule to approximate the probability of x being greater than 2. About 95% of the distribution is between -2 and 2, which means that about 5% is less than -2 or greater than 2. Half of that, or 2.5%, or 0.025, would be on the high end. It wouldn't have been exactly right, but it would have been close, and a little easier to find to boot.

Anyway, all that's left now is to subtract the probability of the other Z-score, Z = 0.86. From the table that is:

Pr(x > 0.86) = 0.194894

Pr(-2 < x < 0.86) = Pr(x > -2) – Pr(x > 0.86)

= 0.97725 – 0.194894

= 0.782356

So, about 78% of the class is expected to score between 60 and 80 points on the test.