Question

The average score for games played in the NFL is 22.4 and the standard deviation is 9.1 points. 43 games are randomly selected. Round all answers to 4 decimal places where possible and assume a normal distribution.

A. What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)

B. What is the distribution of ∑x∑x? ∑x∑x ~ N(,)

C. P(¯xx¯ > 22.3185) =

D. Find the 60th percentile for the mean score for this sample size.

E. P(21.0185 < ¯xx¯ < 23.6939) =

F. Q1 for the ¯xx¯ distribution =

G. P(∑x∑x > 951.0955) =

H. For part c) and e), is the assumption of normal necessary? NoYes

Please answer for e, f, g, h

Answer 1

The average score for games played in the NFL is 22.4 and the standard deviation is 9.1 points. 43 games are randomly selected.

= 22.4    = 9.1 n =43

z-score =

Central limit theorem states that if the sample size is large ( n > 30) then the distribution of the means of similar sample size will approximately follow normal distribution.

A. What is the distribution of

B. What is the distribution of ∑x∑x? ∑x∑x ~ N(,)

Central limit theorem states that if the sample size is large ( n > 30) then the distribution of the sum of similar sample size will approximately follow normal distribution.

C. P(x¯ > 22.3185)

z-score = -0.06

P(Z > -0.06) = P(Z < 0.06)

Ans: 0.52392   .......usiing normal distribution tables

We use normal percentage tables for this. The tables provide value for greater than 'z' so our sign is proper and for 'p < 50%' so that is appropriate as well. If they weren't we would have to rearrange.

D. Find the 60th percentile for the mean score for this sample size.

60th percentile means a value below which 60% data lies.

P( < a) = 0.6

P( >  a) = 0.4 ...................subtracting from 1 on both sides

P( > )= 0.4

= 0.2534 ................using normal distribution percentage tables

a =22.7516

E. P(21.0185 < < 23.6939) = P( < 23.6939) - P( < 21.0185)

=P( Z < 0.93) - P( Z < -1)

=P(Z < 0.93) - [1 - P( Z < 1)]

= 0.82381 - (1 -0.84135)

Ans: 0.66516

F. Q1 for the distribution

Q1 is the 1st quartile where v=below this value 25% of data lies

P( < b) = 0.25

P( >- b) = 0.25  

P( > )= 0.25

= 0.67449 ................using normal distribution percentage tables

b = 21.464

G. P(> 951.0955)

P(Z > -0.20) = P( Z < 0.2)

Ans: 0.57926

H. For part c) and e), is the assumption of normal necessary? NoYes

Since for CLT can be used if the data is normal or if n > 30, here n = 43 > 30, so the assumption is not neeeded.

No