487,771 views
11 votes
11 votes
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

User Garrett Berg
by
2.8k points

1 Answer

16 votes
16 votes

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}

User Xin
by
3.0k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.