Answer :
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
