Suppose That The New England Colonials Baseball Team Is Equa

Suppose That The New England Colonials Baseball Team Is Equally Likely

Suppose that the New England Colonials baseball team is equally likely to win any particular game as not to win it. Suppose also that we choose a random sample of 20 Colonials games.

a. Estimate the number of games in the sample that the Colonials win by giving the mean of the relevant distribution (that is, the expectation of the relevant random variable). Do not round your response.

b. Quantify the uncertainty of your estimate by giving the standard deviation of the distribution. Round your response to at least three decimal places.

Paper For Above instruction

The problem described involves the application of probability distribution concepts to a real-world scenario—specifically, baseball game outcomes. Since the team is equally likely to win or lose each game, the problem can be modeled using a binomial distribution, which is appropriate for binary, independent events occurring at a constant probability. The key parameters in this case are the number of trials (20 games) and the probability of success on each trial (winning a game), which is 0.5 due to the team being equally likely to win or lose.

Part (a) asks for the expected number of wins in the sample of 20 games. The expected value (mean) of a binomial distribution is calculated using the formula:

E(X) = n * p

where n is the number of trials, and p is the probability of success. Substituting the known values:

E(X) = 20 * 0.5 = 10

Therefore, the expected number of wins in the 20-game sample is 10. This value represents the most probable outcome if we were to repeat the sampling process multiple times, and it is the mean of the binomial distribution describing the number of games won.

Part (b) requires the calculation of the standard deviation for the binomial distribution. The standard deviation (σ) measures the variability or spread of the outcomes around the mean. The formula for the standard deviation of a binomial distribution is:

σ = √(n p (1 - p))

Substituting known values yields:

σ = √(20 0.5 0.5) = √(20 * 0.25) = √5 ≈ 2.236

Rounded to three decimal places, the standard deviation is approximately 2.236.

In summary, the expected number of wins in a random sample of 20 games is 10, and the standard deviation of the number of wins is approximately 2.236. These measures provide a statistical understanding of the likely outcomes and the variability inherent in the game results, assuming each game is independent and the probability of winning remains constant at 0.5.

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