# 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# ?

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

What is the standard deviation of the random variable

##### 1 Answer

Mean:

Variance:

Standard deviation:

#### Explanation:

We are given that

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 = 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

The variance is a measure of the "spread" of

That's because the variance *squared* distance between each possible value and

#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

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 *square* of the units of

That's where standard deviation comes in. The standard deviation

#sigma= sqrt (sigma^2)#

For this problem, that works out to be

#sigma = sqrt(0.64)=0.8#

So every time we pick an *(See: confidence intervals.)*