91Ó°ÊÓ

Consider the following table of data. The number of games required to complete the randomly selected world series is the random variable Y.

Mean value

$$\begin{gathered} \mathrm{μ}=\underset{}{∑}y·P(Y=y) \\ \mathrm{μ}=4\left(0.2\right)+5(0.229)+6\left(0.229\right)+7\left(0.343\right) \\ \mathrm{μ}=0.8+1.15+1.4+2.4 \\ \mathrm{μ}=5.72 \end{gathered}$$

Standard deviation

$$\begin{gathered} \mathrm{σ}=\sqrt{\underset{}{∑}{x}^{2}P(X=x)-{\mathrm{μ}}^2} \\ \mathrm{σ}=\sqrt{16\left(0.2\right)+25\left(0.229\right)+36\left(0.229\right)+49\left(0.343\right)-{\left(5.72\right)}^2} \\ \mathrm{σ}=\sqrt{1.2576} \\ \mathrm{σ}=1.1214 \end{gathered}$$

The average number of games needed to finish the world series = 5.72

The standard deviation = 1.122

Interpretation

The average number of games needed to complete a world series is 1.122, compared to a mean of 5.72.

Consider the following table of data. The number of games required to complete the world series is the random variable X.

The number of games required to complete the randomly selected world series is the random variable Y. The random variable X represents the number of games needed to complete the world series.

For already completed games, the random variable Y is defined.

As a result, the random variable X has the same estimate as the random variable Y.

As a result, the mean is 5.72.

1.122 is the standard deviation.