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A softball player with a batting "hit" average of 0.250 can expect 5 official at-bats in an upcoming two-game series. What is the mean number of hits expected in these 5 official at-bats? What is the variance and standard deviation of the number of hits expected? For the information above, calculate a binomial probability distribution and its cumulative probability distribution. Determine the probabilities of the following various events occurring:

What is the probability of getting 5 hits?
What is the probability of getting less than 2 hits?
What is the probability of getting no hits?
What is the probability of getting 3 or more hits?
What is the probability of being thrown out at second base after hitting a single?

User Pldg
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1 Answer

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Final answer:

The mean number of hits expected in 5 official at-bats with a batting average of 0.250 is 1.25. The variance is 0.9375 and the standard deviation is 0.9682. Using the binomial probability distribution, the probabilities of various events occurring can be calculated.

Step-by-step explanation:

The mean number of hits expected in the 5 official at-bats can be calculated by multiplying the batting average (0.250) by the number of at-bats (5). So, the mean number of hits expected is 0.250 * 5 = 1.25.

The variance of the number of hits expected can be calculated using the formula: variance = np(1-p), where n is the number of at-bats and p is the batting average. So, the variance is 5 * 0.250 * (1-0.250) = 0.9375.

The standard deviation of the number of hits expected is the square root of the variance. So, the standard deviation is sqrt(0.9375) = 0.9682.

The binomial probability distribution can be calculated using the formula: P(X=k) = (n choose k) * p^k * (1-p)^(n-k), where X is the number of hits, n is the number of at-bats, k is the number of hits, and p is the batting average.

The cumulative probability distribution can be calculated by summing the individual probabilities for each event up to and including the event of interest.

The probabilities of the various events occurring are as follows:

Probability of getting 5 hits: P(X=5) = (5 choose 5) * 0.250^5 * (1-0.250)^(5-5) = 0.0009765625.

Probability of getting less than 2 hits: P(X<2) = P(X=0) + P(X=1) = (5 choose 0) * 0.250^0 * (1-0.250)^(5-0) + (5 choose 1) * 0.250^1 * (1-0.250)^(5-1) = 0.57421875.

Probability of getting no hits: P(X=0) = (5 choose 0) * 0.250^0 * (1-0.250)^(5-0) = 0.2373046875.

Probability of getting 3 or more hits: P(X>=3) = P(X=3) + P(X=4) + P(X=5) = (5 choose 3) * 0.250^3 * (1-0.250)^(5-3) + (5 choose 4) * 0.250^4 * (1-0.250)^(5-4) + (5 choose 5) * 0.250^5 * (1-0.250)^(5-5) = 0.11328125.

Probability of being thrown out at second base after hitting a single: This probability is not provided in the given information and cannot be calculated with the given data.

User Regestea
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