Question

In the accompanying table, the random variable x represents the number of televisions in a household in a certain country. Determine whether or not the table is a probability distribution. If it is a probability distribution, find its mean and standard deviation.xP(x)00.0310.1220.2430.3140.1850.12

Answer 1

For the table to be a complete probability distribution we need to check that the addition of the probabilities (P(x)) is equal to 1.

$0.03+0.12+0.24+0.31+0.18+0.12=1$

As it is equal to one, we can be certain that no household has more than 6 TVs. The function P(x) is indeed a probability distribution

Now, we need to find its mean and standard deviation

$\begin{gathered} \operatorname{mean}=\mu=\sum ^{}_{}xP(x) \\ \Rightarrow\mu=0\cdot0.03+1\cdot0.12+2\cdot0.24+3\cdot0.31+4\cdot0.18+5\cdot0.12 \\ \Rightarrow\mu=2.85 \end{gathered}$

And the Standard Deviation is:

$SD=\sigma=\sqrt[]{\sum^{}_{}}P(x)(x-\mu)^2$

Then, in our case:

$\begin{gathered} \sigma=\sqrt[]{0.03(-2.85)^2+0.12(-1.85)^2+0.24(-0.85)^2+0.31(0.15)^2+0.18(1.15)^2+0.12(2.15)^2} \\ \Rightarrow\sigma=1.2757\ldots \end{gathered}$