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#?

1 Answer
Feb 5, 2017

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.)