The average score for games played in the NFL is 21.1 and the standard deviation is 8.9 points. 46 games are randomly selected. Round all answers to 4 decimal places where possible and assume a normal distribution. A. P(21.4317 < ¯x < 22.7561) = B. Q1 for the ¯x distribution = The line over should be over the x.

A) P(21.4317 < ¯x < 22.7561) = 0.2975

B) Q1 for the ¯x distribution = 21.9844

Explanation:

The Central Limit theorem allows us to say that

Sample mean = Population mean = 21.1 points

Mean of sampling distribution = σₓ = (σ/√n)

σ = population standard deviation = 8.9 points

n = sample size = 46

σₓ = (8.9/√46) = 1.3122334098 = 1.3122

A) P(21.4317 < ¯x < 22.7561) =

This is a normal distribution problem

To find this probability, we will use the normal probability tables

We first normalize/standardize 21.4317 and 22.7561.

The standardized score of any value is that value minus the mean divided by the standard deviation.

For 21.4317

z = (x - μ)/σ = (21.4317 - 21.1)/1.3122 = 0.25

For 22.7561

z = (x - μ)/σ = (22.7561 - 21.1)/1.3122 = 1.26

The required probability

P(21.4317 < ¯x < 22.7561) = P(0.25 < z < 1.26)

Checking the tables

P(21.4317 < ¯x < 22.7561) = P(0.25 < z < 1.26)

= P(z < 1.26) - P(z < 0.25)

= 0.89617 - 0.59871

= 0.29746 = 0.2975 to 4 d.p.

B) Q1 for the distribution is the first quartile. The first quartile is greater than 25% of the distribution.

P(x > Q1) = 0.25

Let the z-score that corresponds to Q1 be z'

P(x > Q1) = P(z > z') = 0.25

But P(z > z') = 1 - P(z ≤ z') = 0.25

P(z ≤ z') = 1 - 0.25 = 0.75

From the normal distribution tables,

z' = 0.674

z' = (Q1 - μ)/σ

0.674 = (Q1 - 21.1)/1.3122

Q1 = 0.674×1.3122 + 21.1 = 21.9844228 = 21.9844 to 4 d.p.

Hope this Helps!!!

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