Question

In: Statistics and Probability

The average score for games played in the NFL is 22.2 and the standard deviation is...

The average score for games played in the NFL is 22.2 and the standard deviation is 8.9 points. 12 games are randomly selected. Round all answers to 4 decimal places where possible and assume a normal distribution. What is the distribution of ¯ x x¯ ? ¯ x x¯ ~ N(,) What is the distribution of ∑ x ∑x ? ∑ x ∑x ~ N(,) P( ¯ x x¯ > 19.7462) = Find the 65th percentile for the mean score for this sample size. P(21.9462 < ¯ x x¯ < 24.8846) = Q1 for the ¯ x x¯ distribution = P( ∑ x ∑x < 236.9544) = For part c) and e), is the assumption of normal necessary? YesNo

Solutions

Expert Solution

µ = 22.2, σ = 8.9, n = 12

a) Standard deviation = 8.9/SQRT(12) = 2.5692

Distribution of x̅ ~ N(8.9, 2.5692)

b) Mean = 22.2*12 = 266.4

standard deviation = 2.5692*12 = 30.8305

Distribution of ΣX ~ N(266.4, 30.8305)

c)

P(X̅ > 19.7462) =

= P( (X̅-μ)/(σ/√n) > (19.7462-22.2)/(8.9/√12) )

= P(z > -0.9551)

= 1 - P(z < -0.9551)

Using excel function:

= 1 - NORM.S.DIST(-0.9551, 1)

= 0.8302

d)

65th percentile for the mean score for this sample size:

P(x < a) = 0.65

Z score at p = 0.35 using excel = NORM.S.INV(0.35) =0.3853

Value of X = µ + z*(σ/√n) = 22.2 + (0.3853)*8.9/√12 = 23.1900

e)

P(21.9462 < X̅ < 24.8846) =

= P( (21.9462-22.2)/(8.9/√12) < (X-µ)/(σ/√n) < (24.8846-22.2)/(8.9/√12) )

= P(-0.0988 < z < 1.0449)

= P(z < 1.0449) - P(z < -0.0988)

Using excel function:

= NORM.S.DIST(1.0449, 1) - NORM.S.DIST(-0.0988, 1)

= 0.3913

f)

First quartile, Q1 = 25%

P(x < a) = 0.25

Z score at p = 0.75 using excel = NORM.S.INV(0.75) =-0.6745

Value of X = µ + z*(σ/√n) = 22.2 + (-0.6745)*8.9/√12 = 20.4671

g)

P(X > 236.9544) =

= P( (X-µ)/σ > (236.9544-266.4)/30.8305)

= P(z > -0.9551)

= 1 - P(z < -0.9551)

Using excel function:

= 1 - NORM.S.DIST(-0.9551, 1)

= 0.8302


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