QUESTION IMAGE
Question
this table gives approximate probabilities for scoring 0, 1, 2, 3, or 4 runs in one inning of major league baseball. (although it is possible to score more than 4 runs in one inning, the probability is very small, so it is ignored in this question.)
| x | p(x) |
| 0 | 0.76 |
| 1 | 0.12 |
| 2 | 0.05 |
| 3 | 0.04 |
| 4 | 0.03 |
round all answers to 2 decimal places.
compute the sum of these probabilities:
compute the mean number of runs/inning, using this probability distribution:
compute the standard deviation of this probability distribution:
question help: message instructor post to forum
Step1: Calculate sum of probabilities
$0.76 + 0.12+0.05 + 0.04+0.03=1.00$
Step2: Calculate the mean $\mu$
The formula for the mean of a discrete - probability distribution is $\mu=\sum_{i}x_{i}P(x_{i})$.
$\mu=(0\times0.76)+(1\times0.12)+(2\times0.05)+(3\times0.04)+(4\times0.03)$
$=0 + 0.12+0.10 + 0.12+0.12$
$=0.46$
Step3: Calculate the variance $\sigma^{2}$
The formula for the variance of a discrete - probability distribution is $\sigma^{2}=\sum_{i}(x_{i}-\mu)^{2}P(x_{i})$.
$(0 - 0.46)^{2}\times0.76+(1 - 0.46)^{2}\times0.12+(2 - 0.46)^{2}\times0.05+(3 - 0.46)^{2}\times0.04+(4 - 0.46)^{2}\times0.03$
$=( - 0.46)^{2}\times0.76+(0.54)^{2}\times0.12+(1.54)^{2}\times0.05+(2.54)^{2}\times0.04+(3.54)^{2}\times0.03$
$=0.2116\times0.76 + 0.2916\times0.12+2.3716\times0.05+6.4516\times0.04+12.5316\times0.03$
$=0.160816+0.0350+0.1186+0.2581+0.3759$
$=0.9484$
Step4: Calculate the standard deviation $\sigma$
$\sigma=\sqrt{\sigma^{2}}=\sqrt{0.9484}\approx0.97$
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Sum of probabilities: $1.00$
Mean number of runs/inning: $0.46$
Standard deviation: $0.97$