Video Transcript
In a school, the weights of
students are normally distributed with mean 61 kilograms and standard deviation
eight kilograms. What percentage of students weigh
between 50.6 kilograms and 61.64 kilograms?
Recall that the graph of a normal
distribution is bell shaped and symmetric about the mean and the total area under
the curve is one. It can be very useful when dealing
with normal distribution questions to sketch the curve using the information given
to us.
Well, we have been given that the
mean weight, 𝜇, is 61 kilograms. Meanwhile, the standard deviation,
𝜎, is eight kilograms. We have been asked to calculate the
percentage of students weighing between 50.6 and 61.64 kilograms, which are values
lying on either side of the mean, indicated by the shaded area shown. Written mathematically, this is the
probability that 𝑋 is greater than or equal to 50.6 and less than or equal to
61.64.
Now the first step with most normal
distribution calculations is to normalize the data, which allows us to calculate it
using a standard normal table. We can do this by subtracting the
mean and dividing by the standard deviation inside the parentheses. This means we will get 𝑍 in the
middle of the inequalities. And we can substitute 𝜇 equals 61
and 𝜎 equals eight for the values on the left and right. So we have the probability that 𝑍
is between 50.6 minus 61 over eight and 61.64 minus 61 over eight. And this simplifies to the
probability that 𝑍 is between negative 1.3 and 0.08.
Now the standard normal table that
we are using gives the probabilities of big 𝑍 being greater than or equal to zero
and less than or equal to little 𝑧. Therefore, we want to be able to
express our probability in this form so that we can evaluate it.
Sketching the curve, we note that
it’s much the same as before, but now it’s centered around zero. In order to calculate this area
using the tables, we can split it into two regions, where one is to the right of
zero and one to the left, as shown. That is, we have the region where
𝑍 is between negative 1.3 and zero and the region where 𝑍 is between zero and
0.08. Having done this, we can see that
the region on the right is of the correct form for us to be able to look it up in
the tables. However, for the region on the
left, we will have to use the symmetry of the graph to find the area.
Let us clear some space so we can
write all this out. We start off by rewriting the
calculation in terms of the two separate regions. We then want to use the symmetric
property of the normal distribution, which tells us that the probability of a
variable being between negative 𝑥 and zero is the same as the probability of it
being between positive 𝑥 and zero. So this first probability is equal
to the probability of 𝑍 being between zero and 1.3. Finally, we can now look up each of
these 𝑧-values in the standard normal distribution tables. And this gives us 0.4032 plus
0.0319, which is 0.4351.
Thus, we can conclude that the
percentage of students that weigh between 50.6 kilograms and 61.64 kilograms is
43.51 percent.
