Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
To find [tex]\(E(X)\)[/tex], the expected number of bases for Albert Pujols in a typical at bat in 2009, we use the formula for the expected value of a discrete random variable. The expected value [tex]\(E(X)\)[/tex] is given by the sum of each outcome multiplied by its probability.
First, let's summarize the data:
- [tex]\(0\)[/tex] bases: [tex]\(382\)[/tex] times
- [tex]\(1\)[/tex] base: [tex]\(93\)[/tex] times
- [tex]\(2\)[/tex] bases: [tex]\(45\)[/tex] times
- [tex]\(3\)[/tex] bases: [tex]\(1\)[/tex] time
- [tex]\(4\)[/tex] bases: [tex]\(47\)[/tex] times
The total number of at bats is [tex]\(568\)[/tex]. The probability of each outcome is the number of times that outcome occurred divided by the total number of at bats.
We can calculate the probabilities as follows:
- Probability of [tex]\(0\)[/tex] bases: [tex]\(\frac{382}{568}\)[/tex]
- Probability of [tex]\(1\)[/tex] base: [tex]\(\frac{93}{568}\)[/tex]
- Probability of [tex]\(2\)[/tex] bases: [tex]\(\frac{45}{568}\)[/tex]
- Probability of [tex]\(3\)[/tex] bases: [tex]\(\frac{1}{568}\)[/tex]
- Probability of [tex]\(4\)[/tex] bases: [tex]\(\frac{47}{568}\)[/tex]
Next, we multiply each outcome by its probability and sum the results to find [tex]\(E(X)\)[/tex]:
[tex]\[ E(X) = 0 \cdot \frac{382}{568} + 1 \cdot \frac{93}{568} + 2 \cdot \frac{45}{568} + 3 \cdot \frac{1}{568} + 4 \cdot \frac{47}{568} \][/tex]
Breaking this into parts:
[tex]\[ 0 \cdot \frac{382}{568} = 0 \][/tex]
[tex]\[ 1 \cdot \frac{93}{568} = \frac{93}{568} \][/tex]
[tex]\[ 2 \cdot \frac{45}{568} = \frac{90}{568} \][/tex]
[tex]\[ 3 \cdot \frac{1}{568} = \frac{3}{568} \][/tex]
[tex]\[ 4 \cdot \frac{47}{568} = \frac{188}{568} \][/tex]
Now, summing these values:
[tex]\[ E(X) = 0 + \frac{93}{568} + \frac{90}{568} + \frac{3}{568} + \frac{188}{568} \][/tex]
Combine the fractions:
[tex]\[ E(X) = \frac{93 + 90 + 3 + 188}{568} = \frac{374}{568} \][/tex]
Using division, we get:
[tex]\[ E(X) \approx 0.6584507042253521 \][/tex]
To the nearest thousandth, [tex]\(E(X)\)[/tex] is:
[tex]\[ 0.658 \][/tex]
Therefore, the correct answer is:
B. 0.658
First, let's summarize the data:
- [tex]\(0\)[/tex] bases: [tex]\(382\)[/tex] times
- [tex]\(1\)[/tex] base: [tex]\(93\)[/tex] times
- [tex]\(2\)[/tex] bases: [tex]\(45\)[/tex] times
- [tex]\(3\)[/tex] bases: [tex]\(1\)[/tex] time
- [tex]\(4\)[/tex] bases: [tex]\(47\)[/tex] times
The total number of at bats is [tex]\(568\)[/tex]. The probability of each outcome is the number of times that outcome occurred divided by the total number of at bats.
We can calculate the probabilities as follows:
- Probability of [tex]\(0\)[/tex] bases: [tex]\(\frac{382}{568}\)[/tex]
- Probability of [tex]\(1\)[/tex] base: [tex]\(\frac{93}{568}\)[/tex]
- Probability of [tex]\(2\)[/tex] bases: [tex]\(\frac{45}{568}\)[/tex]
- Probability of [tex]\(3\)[/tex] bases: [tex]\(\frac{1}{568}\)[/tex]
- Probability of [tex]\(4\)[/tex] bases: [tex]\(\frac{47}{568}\)[/tex]
Next, we multiply each outcome by its probability and sum the results to find [tex]\(E(X)\)[/tex]:
[tex]\[ E(X) = 0 \cdot \frac{382}{568} + 1 \cdot \frac{93}{568} + 2 \cdot \frac{45}{568} + 3 \cdot \frac{1}{568} + 4 \cdot \frac{47}{568} \][/tex]
Breaking this into parts:
[tex]\[ 0 \cdot \frac{382}{568} = 0 \][/tex]
[tex]\[ 1 \cdot \frac{93}{568} = \frac{93}{568} \][/tex]
[tex]\[ 2 \cdot \frac{45}{568} = \frac{90}{568} \][/tex]
[tex]\[ 3 \cdot \frac{1}{568} = \frac{3}{568} \][/tex]
[tex]\[ 4 \cdot \frac{47}{568} = \frac{188}{568} \][/tex]
Now, summing these values:
[tex]\[ E(X) = 0 + \frac{93}{568} + \frac{90}{568} + \frac{3}{568} + \frac{188}{568} \][/tex]
Combine the fractions:
[tex]\[ E(X) = \frac{93 + 90 + 3 + 188}{568} = \frac{374}{568} \][/tex]
Using division, we get:
[tex]\[ E(X) \approx 0.6584507042253521 \][/tex]
To the nearest thousandth, [tex]\(E(X)\)[/tex] is:
[tex]\[ 0.658 \][/tex]
Therefore, the correct answer is:
B. 0.658
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.