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

Let π be a random variable which is normally distributed with mean 63 and variance 144. Determine π(37.56 β€ π β€ 57.36).

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### Video Transcript

Let π be a random variable which is normally distributed with mean 63 and variance 144. Determine the probability that π 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 π and a standard deviation π 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 β thatβs π squared or the standard deviation squared β of 144. Weβre looking to find the probability that π is greater than or equal to 37.56 and less than or equal to 57.36; thatβs represented by the area shaded.

The first step once weβve sketched out our curve with normal distribution questions is to calculate the π§-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.

Weβve already specified our value for π. The variance is the square of the standard deviation π. So we can square root 144 to get the value for the standard deviation. The square root of 144 is 12. So the standard deviation π for our data is 12. Substituting our top most value of π 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 canβt look up a π§-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 π§ is less than 0.47 in our standard normal table is 0.6808. Weβre actually interested in though the probability that π§ 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 π§ is greater than 0.47 is 0.3192. In turn, that also tells us the probability that π§ is less than negative 0.47 is equal to 0.3192. What that also means is the probability that π is less than 57.36 is 0.3192.

Letβs repeat this process for an π-value of 37.56. Substituting 37.56 into the formula for the π§-value and we get 37.56 minus 63 all over 12, which is equal to negative 2.12. Once again, weβll use the symmetry of the curve and instead find the probability that π§ is greater than 2.12. This is found by subtracting the probability that π§ is less than 2.12 from one. Thatβs one minus 0.9830, which is 0.0170.

So we now also know the probability that π is less than 37.56 to be equal to 0.0170. We can now find the probability that π is greater than 37.56 and less than 57.36 by finding the difference between the two probabilities. Thatβ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 π is greater than 37.56 and less than 57.36, whereas in the question weβre being asked to find the probability that π 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 π is greater than or equal to 37.56 and less than or equal to 57.36 is, therefore, equal to 0.3022.