# Video: Estimating Population Percentages from a Normal Distribution in Context

The marks from a statistics exam are normally distributed with mean 𝜇 and standard deviation 𝜎. What percentage of students got a mark between (𝜇 − 2.27 σ) and (𝜇 + 1.73 σ)?

04:09

### Video Transcript

The marks from a statistics exam are normally distributed with mean 𝜇 and standard deviation 𝜎. What percentage of students got a mark between 𝜇 minus 2.27𝜎 and 𝜇 plus 1.73𝜎.

Remember, the graph of a normally distributed data set is this bell-shaped curve. It’s completely symmetrical about the mean and the area underneath the curve is one or 100 percent. In this question, we’re looking to find the percentage of students who got a mark between 𝜇 minus 2.27𝜎 and 𝜇 plus 1.73𝜎. And to do this, we need to find the associated 𝑧-scores. This is essentially a way of standardizing our data. And it allows us to read results of a standard normal table with a mean of zero and a standard deviation of one.

Now at this point, we shouldn’t worry too much that we don’t have numerical values in the question. Instead, we’re going to go ahead and substitute what we know about our data set into the formula for the 𝑧-score. The lower end of our interval is 𝜇 minus 2.27𝜎. So, we’re going to make this our first 𝑥-value. And this means our first 𝑧-value is found by 𝜇 minus 2.27𝜎 minus 𝜇 all over 𝜎. 𝜇 minus 𝜇 is zero. And we can also divide through by 𝜎. And we can see that our lower 𝑧-score is negative 2.27.

Let’s repeat this process with the 𝑥-value at the upper end of our interval. It’s 𝜇 plus 1.73𝜎. So, our associated 𝑧-value here is 𝜇 plus 1.73𝜎 minus 𝜇 all over 𝜎. Once again, 𝜇 minus 𝜇 is zero. And we divide through by 𝜎. And we can see that this 𝑧-score is 1.73. So, we can see that to find the percentage of students who got a mark between 𝜇 minus 2.27𝜎 and 𝜇 plus 1.73𝜎, we need to find the probability that 𝑧 is greater than negative 2.27 and less than 1.73.

And in fact, with normal distribution, the probabilities are cumulative. So, if we find the probability that 𝑧 is less than 1.73, that’s the same as the probability that 𝑥 is less than 𝜇 plus 1.73𝜎. That’s everything to the left-hand side.

To find the shaded area, we’re going to need to subtract the probability that 𝜇 is less than 2.27𝜎, or the probability that 𝑧 is less than negative 2.27. And we can find the probability that 𝑧 is less than 1.73 by looking the value of 1.73 up in the standard normal table.

But you’ll notice that we can’t find negative 2.27 in that table. Instead, we need to use the symmetry of the curve. And we can see that the probability that 𝑧 is less than negative 2.27 is equal to the probability that 𝑧 is greater than 2.27. So, how do we find the probability that 𝑧 is greater than 2.27?

Well, remember we said that the probabilities are cumulative. And, earlier, we said that they also added to one. So, we can subtract the probability that 𝑧 is less than 2.27 from one. And that will give us the probability that 𝑧 is greater than 2.27 and, in turn, the probability that 𝑧 is less than negative 2.27.

If we find a 𝑧-score of 2.27, we see it has an associated probability of 0.9884. And 0.9582 minus one minus 0.9884 is equal to 0.9466. We’re being told to find the percentage of students that got a mark between 𝜇 minus 2.27 and 1.73𝜎. To change from a decimal to a percentage, we multiply our number by 100. 0.9466 multiplied by 100 is 94.66. And our answer is 94.66 percent.