Question

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

Answer 1

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