### Video Transcript

Let π denote the discrete random
variable that can take values zero, two, four, and six. Given that the probability π
equals zero is one-seventh, the probability π equals two is two-sevenths, and the
probability π equals four is two-sevenths, find the standard deviation of π,
giving your answer to two decimal places.

Weβve been told that this discrete
random variable can take the values zero, two, four, and six. But weβve only been given the
probabilities for three of them. Before we can proceed with
calculating the standard deviation, we need to find the probability that π is equal
to the fourth value, which is six. We can work this out because the
sum of all the values in a probability distribution π of π must be equal to
one. So one-seventh plus two-sevenths
plus two-sevenths plus the probability π is equal to six is equal to one. This simplifies to five-sevenths
plus the probability π equals six equals one. And then we can subtract
five-sevenths from each side to find that the probability π is equal to six is
two-sevenths.

We may now find it helpful to
express this probability distribution in a table, with the values in the range of
the discrete random variable in the top row and their associated probabilities in
the second row. So the probability distribution of
π can be represented as shown. We are asked to find the standard
deviation of π, which is a measure of spread of its probability distribution. We use the Greek letter π, or
sometimes π subscript π, to denote the standard deviation. And itβs equal to the square root
of the variance.

The variance of a discrete random
variable π can be calculated using the formula shown. Itβs equal to the expected value of
π squared minus the expected value of π squared. The difference in notation is
really important here. In the second term, we find the
expected or average value of π and then we square it, whereas in the first term, we
square the π-values first and then find their expected or average value. Thereβs quite a lot of work to be
done here, so weβll break it down into stages. And weβll begin by calculating the
expectation of π. This is the sum of each π-value in
the probability distribution multiplied by its associated probability. And we can add a row to our table
to work out these values.

First, we have zero multiplied by
one-seventh, which is zero; then two multiplied by two-sevenths, which is
four-sevenths; four multiplied by two-sevenths, which is eight-sevenths; and finally
six multiplied by two-sevenths, which is twelve-sevenths. The expected value of π is then
the sum of these four values, which is 24 over seven. Next, we need to calculate the
expected value of π squared. The formula for this is the sum of
each π squared value multiplied by its corresponding probability. And the probabilities for π
squared are inherited directly from the probability distribution of π.

We can add another row to our table
for the π squared values, which are zero, four, 16, and 36. And then weβll add another row in
which we multiply these values by the probabilities. Zero multiplied by one-seventh is
zero. Four multiplied by two-sevenths is
eight-sevenths. 16 multiplied by two-sevenths is
thirty-two sevenths. And finally, 36 multiplied by
two-sevenths is seventy-two sevenths. The expectation of π squared is
the sum of these four values, which is 112 over seven.

So having calculated both the
expectation of π and the expectation of π squared, weβre now able to calculate the
variance of π. Itβs equal to 112 over seven for
the expectation of π squared minus 24 over seven squared for the expectation of π
squared. And this is equal to 208 over
49. The final step in calculating the
standard deviation is to take the square root of this value. π then is the square root of 208
over 49, which in simplified surd form is four root 13 over seven. We can then evaluate this as a
decimal, and it is 2.0603 continuing.

The question specifies that we
should give our answer to two decimal places. So we round down to 2.06. The standard deviation of π to two
decimal places is 2.06, which means that, on average, observations of this discrete
random variable π are 2.06 units away from their mean.