Video Transcript
Let 𝑋 denote a discrete random
variable that can take the values one, 𝑎, and three. Given that 𝑋 has a probability
distribution function 𝑓 of 𝑥 equals 𝑥 over six, find the variance of 𝑋. Give your answer to two decimal
places.
The variance of a discrete random
variable is a measure of the extent to which the values of the variable differ from
their mean, or expected value, which we denote as 𝜇. We calculate the variance of a
discrete random variable 𝑋 using the formula variance of 𝑋 equals the expected
value of 𝑋 squared minus the square of the expected value of 𝑋. We need to be really clear on the
difference in notation here. The expected value of 𝑋 squared
means we square the variable first and then find its expected value, whereas the
square of the expected value of 𝑋 means we find the expected value of the discrete
random variable 𝑋 first and then square it. We can replace the expected value
of 𝑋 with the letter 𝜇 if it helps with the distinction.
The formulae for calculating each
of these statistics are as follows. The expected value of 𝑋 is equal
to the sum of each 𝑥-value in the range of the discrete random variable multiplied
by its corresponding probability. The expected value of 𝑋 squared is
the sum of each 𝑥-value squared multiplied by the corresponding probability. These probabilities are the 𝑓 of
𝑥 values in the probability distribution of 𝑥. We’re given this distribution in
the question. We’re told that 𝑓 of 𝑥 is equal
to 𝑥 over six, for the 𝑥-values one, 𝑎, and three.
Before we can calculate the
variance of 𝑋, we need to determine the value of this unknown 𝑎. To do so, we recall that the sum of
all the probabilities in a probability distribution is always equal to one. So, substituting each value of 𝑥
into the probability distribution function and then summing the three probabilities
gives the equation one over six plus 𝑎 over six plus three over six is equal to
one. Multiplying through by six and
simplifying the expression on the left-hand side gives 𝑎 plus four is equal to
six. And then subtracting four from each
side of the equation, we find that 𝑎 is equal to two.
We can now write the probability
distribution of 𝑥 out in full. The three values in the range of
the discrete random variable are one, two, and three, with corresponding
probabilities one over six, two over six, and three over six. To calculate the expected value of
𝑋, we add a new row to the table in which we multiply each 𝑥-value by its
probability, giving one over six, four over six, and nine over six. The expected value of 𝑋 is the sum
of these three values; that’s 14 over six, which simplifies to seven over three.
Next, we want to calculate the
expected value of 𝑋 squared. So, we add a row to the table for
the 𝑥 squared values, which are one, four, and nine. We then add one final row to the
table in which we multiply each 𝑥 squared value by the corresponding 𝑓 of 𝑥
value, giving one over six, eight over six, and 27 over six. The expected value of 𝑋 squared is
the sum of these three values, which is 36 over six, or simply six.
Finally, we substitute the two
values we’ve calculated into the variance formula, giving six minus seven over three
squared. That’s six minus 49 over nine. Using a common denominator of nine,
this is equal to 54 over nine minus 49 over nine, which is five over nine or, as a
decimal, 0.5 recurring. We’re asked to give the variance to
two decimal places. So rounding the answer as required,
we have found that the variance of this discrete random variable 𝑋, to two decimal
places, is 0.56.
