Question

The average score for games played in the NFL is 20.8 and the standard deviation is 9 points. 46 games are randomly selected. Round all probabilities to 4 decimal places where possible, percentiles to 2 decimal places and assume a normal distribution. What is the distribution of xˉ?

Answer 1

The distribution of the average score of NFL games for a sample of 46 games follows a normal distribution with a mean of 20.8 and a standard error of approximately 1.3263.

Step-by-step explanation:

The distribution of ׈ for the average score of NFL games is a sampling distribution. Given that the population mean (μ) is 20.8 and the population standard deviation (σ) is 9 points, for a sample size (n) of 46, the sampling distribution of the sample mean will have a mean equal to the population mean (׈ = μ = 20.8) and a standard deviation equal to the population standard deviation divided by the square root of the sample size (σˆ = σ/√n = 9/√46). This standard deviation is known as the standard error of the mean (SEM).

Using the information, the standard error of the mean (SEM) is SEM = 9/√46 ≈ 1.3263 (rounded to four decimal places). Now, the distribution of ׈ can be represented as Xˆ ~ N(20.8, 1.3263).