Question

How do you find the mean of the random variable `x`?

X= 0,1,2,3
P(x) = 0.15, 0.35, 0.45, 0.05

What is the variance and standard deviation of the random variable `x`?
What is the standard deviation of the random variable `x`?

Answer 1

Mean: `mu=1.4`
Variance: `sigma^2=0.64`
Standard deviation: `sigma=0.8`

Explanation:

We are given that `X` could take on the values `{0,1,2,3}` with respective probabilities `{0.15, 0.35, 0.45, 0.05}`. Since `X` is discrete, we can imagine `X` as a 4-sided die that's been weighted so that it lands on "0" 15% of the time, "1" 35% of the time, etc.

The question is, when we roll this die once, what value should we expect to get? Or perhaps, if we roll the die a huge number of times, what should the average value of all those rolls be?

Well, of the 100% of the rolls, 15% should be "0", 35% should be "1", 45% should be "2", and 5% should be "3". If we add all these together, we'll have what's known as a weighted average.

In fact, if we placed these relative weights at their matching points on a number line, the point that would "balance the scale" is the mean that we seek.

This is a good way to interpret the mean of a discrete random variable. Mathematically, the mean `mu` is the sum of all the possible values, weighted by their probabilities. As a formula, this is:

`mu = E[X] = sum_("all " x)[x * P(X=x)]`

In our case, this works out to be:

`mu = [0*P(0)]+[1*P(1)]+[2*P(2)]+[3*P(3)]`
`color(white)mu=(0)(0.15)+(1)(0.35)+(2)(0.45)+(3)(0.05)`
`color(white)mu="       "0"       "+"    "0.35"    "+"     "0.9"     "+"    "0.15`
`color(white)mu=1.4`

So, over a large number of rolls, we would expect the average roll value to be `mu=1.4`.

The variance is a measure of the "spread" of `X`. Going back to our "balanced number line" idea, if we moved our weights out from our "centre of gravity" `mu` so that they are twice as far away, `mu` itself wouldn't change, but the variance would increase, by a factor of 4.

That's because the variance `sigma^2` of a random variable is the average squared distance between each possible value and `mu`. (We square the distances so that they're all positive.) As a formula, this is:

`sigma^2="Var"(X)=E[(X-mu)^2]`

Using a bit of algebra and probability theory, this becomes

`sigma^2=E[X^2]-mu^2`
`color(white)(sigma^2)=sum_("all x")x^2P(X=x)" "-" "mu^2`

For this problem, we get

`sigma^2=[0^2*P(0)]+[1^2*P(1)]+[2^2*P(2)]`
`color(white)(sigma^2=)+[3^2*P(3)]" "-" "1.4^2`
`color(white)(sigma^2)=(0)(0.15)+(1)(0.35)+(4)(0.45)+(9)(0.05)`
`color(white)(sigma^2=)-1.96`
`color(white)(sigma^2)=0.64`

So the average squared distance between each possible `X` value and `mu` is `sigma^2=0.64`.

Standard deviation is easy—it's just the square root of the variance. But, why bother with it if it's pretty much the same? Because the units of `sigma^2` are the square of the units of `X`. If `X` measures time, for example, its variance is in units of `"(time)"^2`, which really doesn't help us if we're trying to establish a "margin of error".

That's where standard deviation comes in. The standard deviation `sigma` of `X` is a measure of how far from `mu` we should expect `X` to be. It's simply

`sigma= sqrt (sigma^2)`

For this problem, that works out to be

`sigma = sqrt(0.64)=0.8`

So every time we pick an `X`, the expected distance between `mu` and that `X` is `sigma=0.8`. And since `sigma` is in the same "units" as `X`, it's much more easy to use to help us construct a margin of error. (See: confidence intervals.)