Statistics in math?

Question

An investment will be worth $1,000, $2,000, or $5,000 at the end of the year. The probabilities of these values are .25, .60, and .15, respectively. Determine the mean and variance of the worth of the investment.

Answer 1

for any discrete random variable the expectation is:

E(X) = μ = ∑x * P(X = x)

the variance of a discrete random variable is:

Var(X) = σ² = ∑(x - μ)² * P(X = x) = {∑x² * P(X = x) }- μ²

the standard deviation is just the square root of the variance.

Let X be the investment worth.  The mean is:

E(X) = 1000 * 0.25 + 2000 * 0.60 + 5000 * 0.15 = 2200

the variance is:

Var(X) = 1000^2 * 0.25 + 2000^2 * 0.60 + 5000^2 * 0.15 - 2200^2

Var(X) = 1560000

Answer 2

Let X be a random variable denoting the value of the investment at the end of the year.

E[X]=1000(.25) + 2000(.60) + 5000(.15)

=2,200

E[X^2]=(1000^2)(.25) + (2000^2)(.60) + (5000^2)(.15)

=6,400,000

Var[X]=E[X^2]-E[X]^2

=6,400,000 -2,200^2

=1,560,000

Thus the mean is $2200 and the variance is 1,560,000