If 𝑥 is a random variable with mean 150 and coefficient of variation 2.5 percent, find the approximate variance of 𝑥.
Let’s first remind ourselves what is meant by a coefficient of variation. This is a relative measure of variation. It compares the standard deviation 𝜎 with the mean 𝜇 of a random variable. The coefficient of variation is defined as 𝜎, the standard deviation, over 𝜇, the mean. And if it’s given as a percentage rather than a decimal, then this fraction will be multiplied by 100.
In this question, we’re told that the coefficient of variation is 2.5 percent, which means that the standard deviation 𝜎 is 2.5 or two and a half percent of the mean 𝜇. Now, we’re asked to use the information given to find the approximate variance of 𝑥, which will be the value 𝜎 squared. But before we can do this, we’ll first need to work out the value of 𝜎, the standard deviation. We can substitute the information we know into the formula we’ve written down. We don’t know the standard deviation. So we’ll leave this as 𝜎. We do know the value of 𝜇, the mean is 150. We multiply this by 100 to convert to a percentage. And we know that the coefficient of variation is 2.5 percent. So we have an equation that we can solve. 𝜎 over 150 multiplied by 100 is equal to 2.5.
Now, we can actually simplify this equation quite a lot. Both 150 and 100 are multiples of 50. 100 divided by 50 is equal to two, and 150 divided by 50 is equal to three. So our equation simplifies to 𝜎 over three multiplied by two equals 2.5. That’s two 𝜎 over three equals 2.5. To solve this equation, we need to do two things. And we can do them in either order. We need to multiply both sides by three, which will give two 𝜎 is equal to 7.5, and then divide both sides by two, giving 𝜎 equals 3.75.
So we’ve found the standard deviation of our random variable 𝑥. It’s equal to 3.75. But we need to work out the variance. We therefore need to square this value of 3.75 which gives 14.0625. Now, the question asks us to find the approximate variance of 𝑥. So we can round this value to a sensible degree of accuracy. Let’s round to three significant figures. The fourth significant figure, the six, is our deciding digit. And as this is greater than five, it tells us that we’re rounding up. So the zero in the first decimal place will become a one. The approximate variance of this random variable 𝑥 then, to three significant figures, is 14.1.