# Video: Determining Probabilities for Normal Distribution given the Mean and the Variance

Let 𝑋 be a random variable which is normally distributed with mean 63 and variance 144. Determine 𝑃(37.56 ≤ 𝑋 ≤ 57.36).

04:45

### Video Transcript

Let 𝑋 be a random variable which is normally distributed with mean 63 and variance 144. Determine the probability that 𝑋 is greater than or equal to 37.56 and less than or equal to 57.36.

Remember the graph of the curve representing the normal distribution with a mean of 𝜇 and a standard deviation 𝜎 is bell-shaped and symmetric about the mean. And the total area under the curve is 100 percent or one. A sketch of the curve can be a really useful way to decide how to answer a problem about normally distributed data.

Our data is normally distributed with a mean of 63 and a variance — that’s 𝜎 squared or the standard deviation squared — of 144. We’re looking to find the probability that 𝑋 is greater than or equal to 37.56 and less than or equal to 57.36; that’s represented by the area shaded.

The first step once we’ve sketched out our curve with normal distribution questions is to calculate the 𝑧-value. This is essentially a way of scaling our data or standardising it in what becomes a standard normal distribution. Once we complete this step, we can work from a single standard normal table.

We’ve already specified our value for 𝜇. The variance is the square of the standard deviation 𝜎. So we can square root 144 to get the value for the standard deviation. The square root of 104  is 12. So the standard deviation 𝜎 for our data is 12. Substituting our top most value of 𝑋 of 57.36 and we get 57.36 minus 63 all over 12. And that gives us our first set value of negative 0.47.

Now, we can’t look up a 𝑧-value of negative 0.47 in the standardised normal table. So instead, we consider the fact that the curve is symmetrical. And we look up a value of 0.47 in the standard normal table. The probability that 𝑧 is less than 0.47 in our standard normal table is 0.6808. We’re actually interested in though the probability that 𝑧 is greater than 0.47.

Since the area under the curve sums to one whole or 100 percent, we can subtract 0.6808 from one. And that tells us the probability that 𝑧 is greater than 0.47 is 0.3192. In turn, that also tells us the probability that 𝑧 is less than negative 0.47 is equal to 0.3192. What that also means is the probability that 𝑋 is less than 57.36 is 0.3192.

Let’s repeat this process for an 𝑋-value of 37.56. Substituting 37.56 into the formula for the 𝑧-value and we get 37.56 minus 63 all over 12, which is equal to negative 2.12. Once again, we’ll use the symmetry of the curve and instead find the probability that 𝑧 is greater than 2.12. This is found by subtracting the probability that 𝑧 is less than 2.12 from one. That’s one minus 0.9830, which is 0.0170.

So we now also know the probability that 𝑋 is less than 37.56 to be equal to 0.0170. We can now find the probability that 𝑋 is greater than 37.56 and less than 57.36 by finding the difference between the two probabilities. That’s 0.3192 minus 0.0170. And that gives us a probability of 0.3022.

Now, you might have noticed that we found the probability that 𝑋 is greater than 37.56 and less than 57.36, whereas in the question we’re being asked to find the probability that 𝑋 is greater than or equal to 37.56 and less than or equal to 57.36.

In fact, normal is a continuous distribution. So the difference between less than and less than or equal to is negligible. And we can assume that they are the same.

The probability that 𝑋 is greater than or equal to 37.56 and less than or equal to 57.36 is, therefore, equal to 0.3022.