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
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 144 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
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
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
The probability that 𝑋 is greater
than or equal to 37.56 and less than or equal to 57.36 is, therefore, equal to