Answered

Scoring Touchdowns: The probability that a football team scores $0, 1, 2, 3, 4$, or $5$ touchdowns in a game is $0.3, 0.3, 0.3, 0.05, 0.025$, and $0.025$, respectively.



Part 1 of 3:

Find the mean. Round your answer to three decimal places, if needed.

Mean: $\mu =$



Part 2 of 3:

Find the variance. Round your answer to at least three decimal places, if needed.

Variance: $\sigma^2 =$

Verified Answer :

  • The mean \(\mu\) is calculated as \(\sum x \cdot P(X=x) = \boxed{1.275}\).
  • The expected square \(E[X^2]\) is \(\sum x^2 \cdot P(X=x) = 2.975\).
  • The variance \(\sigma^2\) is given by \(E[X^2] - \mu^2 = 2.975 - 1.275^2 = \boxed{1.349}\).

Explanation

  1. Analyze the probability distribution of touchdowns

To determine the statistical properties of the touchdown distribution, we first identify the probability mass function for the number of touchdowns \(X\), where \(X \in \{0, 1, 2, 3, 4, 5\}\). The given probabilities are \(P(0)=0.3\), \(P(1)=0.3\), \(P(2)=0.3\), \(P(3)=0.05\), \(P(4)=0.025\), and \(P(5)=0.025\). We first verify that the sum of these probabilities is \(0.3 + 0.3 + 0.3 + 0.05 + 0.025 + 0.025 = 1.0\), confirming it is a valid probability distribution.

  1. Calculate the mean of the distribution

The mean \(\mu\) is calculated as the sum of the products of each outcome and its corresponding probability:
\[\mu = E[X] = \sum_{x=0}^{5} x \cdot P(X=x)\]
Substituting the given values:
\[\mu = (0 \cdot 0.3) + (1 \cdot 0.3) + (2 \cdot 0.3) + (3 \cdot 0.05) + (4 \cdot 0.025) + (5 \cdot 0.025)\]
\[\mu = 0 + 0.3 + 0.6 + 0.15 + 0.1 + 0.125 = 1.275\]
Thus, the mean number of touchdowns is \(1.275\).

  1. Calculate the variance of the distribution

To calculate the variance \(\sigma^2\), we use the formula \(\sigma^2 = E[X^2] - \mu^2\). First, we find the expected value of the square of the touchdowns, \(E[X^2] = \sum_{x=0}^{5} x^2 \cdot P(X=x)\):
\[E[X^2] = (0^2 \cdot 0.3) + (1^2 \cdot 0.3) + (2^2 \cdot 0.3) + (3^2 \cdot 0.05) + (4^2 \cdot 0.025) + (5^2 \cdot 0.025)\]
\[E[X^2] = 0 + 0.3 + 1.2 + 0.45 + 0.4 + 0.625 = 2.975\]
Next, we substitute this into the variance formula:
\[\sigma^2 = 2.975 - (1.275)^2\]
\[\sigma^2 = 2.975 - 1.625625 = 1.349375\]
Rounding to three decimal places, the variance is \(1.349\).

Examples

This calculation is essential in sports analytics for coaches and analysts to evaluate a team's offensive performance. By understanding the mean and variance, they can predict likely game outcomes, set betting odds, or analyze historical performance trends, such as how consistently a team scores touchdowns versus relying on outliers.

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