Mean and Standard Deviation for Discrete Probability Distributions

Question

Find the mean and standard deviation.

x P(X=x)

-----------------

15 0.6

22 0.4

Answer 1

The mean is also the expected value. This is done by multiplying each numerical outcome by the probability of its occurrence, and adding all the results.

In this distribution, there aren't many calculations. It will E(X) = 0.6 • 15 + 0.4 • 22 = 17.8. This is, effectively, a weighted average.

To find the standard deviation, we first find the variance. This can be done by the formula Var(X) = E(X²) - [E(X)]²

We have E(X), but we don't have E(X²). We can get this by multiplying the squares of each outcome by their probabilities to give us a weighted average of the squares of the outcomes.

E(X²) = 0.6 • 15² + 0.4 • 22² = 328.6. Now we use the formula Var(X) = E(X²) - [E(X)]²

Var(X) = 328.6 - 17.8² = 11.76

σ_x = √Var(X) = √11.76 ≈ 3.429

Answer 2

Let's start with the mean. To calculate the mean, or expected value of X (i.e., E[X]), we'll multiply each possible value of X by its corresponding probability. So in this problem, the mean of X is 15(0.6) + 22(0.4) = 9 + 8.8 = 17.8.

Now that we know that the mean of X is 17.8, we can use this to calculate the standard deviation. First what we'll want to do is square the difference between each possible value of X and the mean. So (15 - 17.8)² = (-2.8)² = 7.84 and (22 - 17.8)² = (4.2)² = 17.64.

Next we'll want to add the products of these squared differences and the corresponding probabilities of X to get the variance of X. So:

Var(X) = 7.84(0.6) + 17.64(0.4) = 4.704 + 7.056 = 11.76

And the standard deviation is just the (positive) square root of the variance, so the standard deviation of X is the square root of 11.76, or approximately 3.429.