Question

The probability distribution of the number of students absent on Mondays, is as follows: X 0 1 2 3 4 5 6 7 f(x) 0.02 0.03 0.26 0.34 0.22 0.08 0.04 0.01 (a) What is the probability that more than 3 students are absent. (b) Compute the expected value of the random variable X . Interpret this expected value. (c) Compute the variance and standard deviation of the random variable X . (d) Compute the expected value and variance of Y=7X+3 . (e) Compute the covariance Cov(X,Y) (f) What is the value of the ratio Cov(X,Y)/Var(X) ? (g) What is the value of Cov(X,Y)/Var(X) for Y=β0+β1X ?

Answer 1

a) Add up all the probabilities $f(x)$ where $x>3$:

$f(4)+f(5)+f(6)+f(7)=0.35$

b) The expected value is

$E[X]=\displaystyle\sum_xx\,f(x)=3.16$

Since X is the number of absent students on Monday, the expectation E[X] is the number of students you can expect to be absent on average on any given Monday. According to the distribution, you can expect around 3 students to be consistently absent.

c) The variance is

$V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2$

where

$E[X^2]=\displaystyle\sum_xx^2\,f(x)=11.58$

So the variance is

$V[X]=11.58-3.16^2\approx1.59$

The standard deviation is the square root of the variance:

$\sqrt{V[X]}\approx1.26$

d) Since $Y=7X+3$ is a linear combination of $X$, computing the expectation and variance of Y is easy:

$E[Y]=E[7X+3]=7E[X]+3=25.12$

$V[Y]=V[7X+3]=7^2V[X]\approx78.13$

e) The covariance of X and Y is

$\mathrm{Cov}[X,Y]=E[(X-E[X])(Y-E[Y])]=E[XY]-E[X]E[Y]$

We have

$XY=X(7X+3)=7X^2+3X$

so

$E[XY]=E[7X^2+3X]=7E[X^2]+3E[X]=90.54$

Then the covariance is

$\mathrm{Cov}[X,Y]=90.54-3.16\cdot25.12\approx11.16$

f) Dividing the covariance by the variance of X gives

$\dfrac{\mathrm{Cov}[X,Y]}{V[X]}\approx\dfrac{11.16}{1.59}\approx0.9638$