Video Transcript
Let 𝑥 denote a discrete random variable which can take the values zero, two, and five. Given that 𝑥 has probability distribution function 𝑓 of 𝑥 equals 𝑎 over six 𝑥 plus six, find the standard deviation of 𝑥. Give your answer to two decimal places.
The standard deviation of a discrete random variable is a measure of spread of its probability distribution. We’re given the probability distribution function 𝑓 of 𝑥 for this discrete random variable, which is the probability that 𝑥 is equal to each value in its range, but it is in terms of an unknown value 𝑎. Before we can calculate the standard deviation of 𝑥, we must first determine the value of 𝑎.
To do this, we recall that the sum of all probabilities in a probability distribution must be equal to one. In other words, if we find expressions for 𝑓 of zero, 𝑓 of two, and 𝑓 of five, which are the probabilities for each of the values in the range of this discrete random variable, we can then sum these and form an equation, which we can solve to find the value of 𝑎.
Substituting zero first of all into the probability distribution function, we have 𝑓 of zero is equal to 𝑎 over six multiplied by zero plus six, which is 𝑎 over six. 𝑓 of two is 𝑎 over six multiplied by two plus six, which is 𝑎 over 18. And 𝑓 of five is 𝑎 over six multiplied by five plus six, which is 𝑎 over 36. As we’ve already said, the sum of these three values must be equal to one. We can write each term on the left-hand side with a common denominator of 36. And then summing these three terms, we have nine 𝑎 over 36 is equal to one. This simplifies to 𝑎 over four is equal to one. And then multiplying both sides of the equation by four, we find that 𝑎 is equal to four.
We can now find the probability that 𝑥 is equal to each value in its range and write down the probability function explicitly. The probabilities are four-sixths, four eighteenths, and four thirty-sixths, each of which can be simplified. We will write the probability function down as a table. So this is the probability distribution function of 𝑥. We have the values in the range of the discrete random variable in the top row, which are zero, two, and five, and then their corresponding probabilities in the second row, which in simplified form are two-thirds, two-ninths, and one-ninth.
We’re asked to find the standard deviation of 𝑥. We denote this using the Greek letter 𝜎, or sometimes we write this with a subscript of a capital 𝑋 if there are multiple variables in the same problem. The standard deviation is equal to the square root of the variance, which we write either as 𝜎 squared, 𝜎 subscript 𝑋 squared, or var of 𝑋. The formula for calculating the variance of a discrete random variable is as follows: it’s equal to the expectation of 𝑋 squared minus the expectation of 𝑋 squared. We need to be very clear on the difference in notation here as the two parts of the formula look quite similar.
The first term is the expectation of 𝑋 squared. We square the 𝑋-values first and then find their expected or average value, whereas the second term is the expected or average value of 𝑋, which we then square. There’s quite a lot of work to be done here, so we’ll break it down into stages. First, we need to find the expected value of 𝑋, which we do by multiplying each 𝑋-value by its 𝑓 of 𝑥 value and then finding the sum.
We can add an extra row to our table for this. The first value is zero multiplied by two-thirds, which is just zero. The second value is two multiplied by two-ninths; that’s four-ninths. And the third value is five multiplied by one-ninth. The sum of these values is zero plus four-ninths plus five-ninths, which is nine-ninths. And this simplifies to the integer one. So we’ve found the expectation of 𝑋. This is the average value of the discrete random variable.
Next, we need to compute the expected value of 𝑋 squared. The formula for this is the sum of each 𝑋 squared value multiplied by its 𝑓 of 𝑥 value. Now the probability distribution for 𝑋 squared is inherited directly from the probability distribution of 𝑋. If the values in the range of 𝑋 are zero, two, and five, then the values in the range of 𝑥 squared are the squares of these values, which are zero, four, and 25. The probabilities for each of these values are identical to the second row in our table. This is because, for example, the probability that 𝑥 squared is equal to four is the same as the probability that 𝑥 itself is equal to two.
So we then add a row to multiply each 𝑋 squared value by 𝑓 of 𝑥. We have zero multiplied by two-thirds, four multiplied by two-ninths, and 25 multiplied by one-ninth. These values simplify to zero, eight-ninths, and twenty-five ninths. The sum of these three values is 33 over nine. So this is the expected value of 𝑥 squared.
Next we compute the variance of 𝑋, which is the expected value of 𝑋 squared minus the expected value of 𝑋, which we then square. So that’s 33 over nine minus one squared. So 33 over nine minus one squared, that’s 33 over nine minus one or 33 over nine minus nine over nine. That’s 24 over nine, which simplifies to eight-thirds.
We found the variance of 𝑥, but we haven’t quite finished. The last step is we need to take the square root of this value to give the standard deviation of 𝑥. 𝜎 is therefore equal to the square root of eight over three, which as a decimal is 1.6329 continuing. The question specifies that we should give our answer to two decimal places. So as the digit in the third decimal place is a two, we’ll be rounding down. The standard deviation of 𝑥 then to two decimal places is 1.63. This tells us that essentially on average outcomes of the discrete random variable 𝑥 are 1.63 units away from their expected or average value.
