Question

The table below gave approximately probabilities for scoring 0123 or 4 runs in one inning of major leagues baseball although it is possible to score more than 4 runs in 1 inning the probability is very small so it is ignored in this question

Answer 1

The sum of the probabilities is always equal to 1. Summing up the probabilities given on the data set, we have

`\sum P(x)=0.7+0.15+0.09+0.04+0.02=1.00`

Given the probability distribution table, the mean number per inning can be computed as

`\begin{gathered} \mu=(0*0.7)+(1*0.15)+(2*0.09)+(3*0.04)+(4*0.02) \\ \mu=0.53 \end{gathered}`

For the computation of the standard deviation, we need to solve first for the square of the difference of the number of innings on the mean. We have the following

`\begin{gathered} (0-0.53)^2=0.2809 \\ (1-0.53)^2=0.2209 \\ (2-0.53)^2=2.1609 \\ (3-0.53)^2=6.1009 \\ (4-0.53)^2=12.0409 \end{gathered}`

We then multiply these values on the corresponding probability of each inning given in the probability distribution table. We have

`\begin{gathered} 0.2809*0.7=0.19663 \\ 0.2209*0.15=0.033135 \\ 2.1609*0.09=0.194481 \\ 6.1009*0.04=0.244036 \\ 12.0409*0.02=0.240818 \end{gathered}`

We now get the sum of the values computed above and get the square of it.

`\begin{gathered} 0.19663+0.033135+0.194481+0.244036+0.240818 \\ \sigma=\text{ }\sqrt[]{0.9091}=0.953 \end{gathered}`