Question Video: Determining the Variance for a Discrete Random Variable Mathematics

The function in the given table is a probability function of a discrete random variable 𝑋. Find the variance of 𝑋. If necessary, give your answer to two decimal places.

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Video Transcript

The function in the given table is a probability function of a discrete random variable 𝑋. Find the variance of 𝑋. If necessary, give your answer to two decimal places.

In the table, we see that 𝑋 can take four values: three, five, seven, and eight. The probability of our random variable taking each of these values is two 𝐴, five 𝐴 squared, five 𝐴 squared, and 𝐴, respectively. We know that the sum of these probabilities must equal one. And each of the individual probabilities must be greater than or equal to zero and less than or equal to one. This means that two 𝐴 plus five 𝐴 squared plus five 𝐴 squared plus 𝐴 is equal to one. Collecting like terms, we have 10𝐴 squared plus three 𝐴 is equal to one. We can then subtract one from both sides of our equation so we have a quadratic equation equal to zero. 10𝐴 squared plus three 𝐴 minus one equals zero.

This quadratic can be factored into two sets of parentheses, five 𝐴 minus one and two 𝐴 plus one. As the product of these equals zero, either five 𝐴 minus one equals zero or two 𝐴 plus one equals zero. Solving the first equation, we get 𝐴 is equal to one-fifth. And solving the second equation, 𝐴 is equal to negative one-half. As all the probability values must be greater than or equal to zero and less than or equal to one, 𝐴 cannot equal negative one-half. We can now substitute 𝐴 equals one-fifth into the bottom row of our table. Two 𝐴 is equal to two-fifths, and five 𝐴 squared is equal to one-fifth. Therefore, the probability of 𝑋 being equal to three is two-fifths, and the probability of 𝑋 being equal to five, seven, or eight are all equal to one-fifth.

At this stage, it is worth checking once again that our four values sum to one. Two-fifths plus one-fifth plus one-fifth plus one-fifth is equal to five-fifths, which equals one. We can now move on to calculate the variance of 𝑋 using the fact that the variance of 𝑋 is equal to 𝐸 of 𝑋 squared minus the 𝐸 of 𝑋 all squared, where 𝐸 of 𝑋 is the expected value or mean. This formula is also sometimes referred to as the mean of the squares minus the square of the mean. We can calculate the mean by summing the products of π‘₯ sub 𝑖 and 𝑓 of π‘₯ sub 𝑖. 𝐸 of 𝑋 is equal to three multiplied by two-fifths plus five multiplied by one-fifth plus seven multiplied by one-fifth plus eight multiplied by one-fifth. This is equal to 26 over five or twenty-six fifths. The mean 𝐸 of 𝑋 of the discrete random variable 𝑋 is twenty-six fifths.

We can calculate the 𝐸 of 𝑋 squared in a similar way. This is because if we consider a different random variable π‘Œ, which takes values which are the square of our random variable 𝑋, then the probability of each value of our new variable will be the same as the probability of each 𝑋-value for the original variable. 𝐸 of 𝑋 squared is therefore equal to three squared multiplied by two-fifths plus five squared multiplied by one-fifth plus seven squared multiplied by one-fifth plus eight squared multiplied by one-fifth. This is equal to 156 over five or one hundred and fifty-six fifths.

We now have values for both the mean and the mean of the squares. The variance of 𝑋 is therefore equal to 156 over five minus 26 over five squared. This is equal to 104 over 25, which as a decimal is 4.16. As our answer is exact to two decimal places, we don’t need to do any rounding. The variance of the discrete random variable 𝑋 is 4.16.

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