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

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}`