This is a random sample of four homes representing the probability of the homes of having dust mite level greater than 2μg/g.
x:0,1,2,3,4
$P(x=0)=0.09, P(x=1)=0.30, p(x=2)=0.37, p(x=3)=0.20, p(x=4)=0.04$
If that dust mite level exceeds 2μg/g each home will spend $ 2000 for an allergen air purification system. Find the mean and variance of the total amount spend by the four sampled homes.
First what I did was that I calculated the mean of the distribution:
$0.09 \cdot (0) + 0.30 \cdot (1) + 0.37 \cdot (2) + 0.20 \cdot (3) + 0.04 \cdot (4) = 1.8$.
Then I multiplied the mean with 2000$ to get the mean of the total amount.
For the variance I calculated variance = $E(x^2)- mean^2$ and I multiplied it by $2000$ to get the variance.
I'm having doubts about my logic, can someone help me?
I'm not entirely sure whether my understanding is correct, but if $T$ is a random variable representing the amount of money paid, then $T=0$ when $x=0,1,2$ and $T=2000$ when $x=3,4$.
That is $T$ is a sort of "scaled" Bernoulli random variable.
Hence the mean is $\mathbb{E}[T]=2000\cdot0.24=480$
and the variance is $\text{var}(T)=2000^2\cdot0.24\cdot0.76=729600$