Question

The average score for games played in the NFL is 22.1 and the standard deviation is 9.3 points. 46 games are randomly selected. Round all answers to 4 decimal places where possible and assume a normal distribution. a. What is the distribution of 2?2- NO b. What is the distribution of r? -NG C. Plā > 23.8432) - d. Find the 66th percentile for the mean score for this sample size. e. P(20.5432 < 3 < 23.4856) - f. Q3 for the distribution- 8. PO > 1018.5872) - h. For part c) and e), is the assumption of normal necessary? Yes No

Answer 1

a.

$\large \boldsymbol{\bar{x}\sim N\left ( 22.1,1.3712 \right )}$

- 2-122.193 -13712 V46

b.

$\large \boldsymbol{\sum x\sim N\left ( 1016.6,63.0757 \right )}$

- Σι-Νημνισ] Σx-N[1016.6,63.0757]

c.

$\boldsymbol{P(\bar{x}>23.8432)=0.1018}$

The excel function is:

=1-NORMDIST(23.8432,22.1,1.3712,TRUE)

d.

The 66th percentile for the mean score for this sample size is $\boldsymbol{22.6656}$

The excel function is:

=NORMINV(0.66,22.1,1.3712)

e. $\boldsymbol{P(20.5432< \bar{x}<23.4856)=0.7158}$

The excel function is:

=NORMDIST(23.4856,22.1,1.3712,TRUE)-NORMDIST(20.5432,22.1,1.3712,TRUE)

f. Q3 for the $\bar{x}$ distribution is $\boldsymbol{23.0249 }$

The excel function is:

=NORMINV(0.75,22.1,1.3712)

g. $\boldsymbol{P(\sum x>1018.5872)=0.4874}$

The excel function is:

=1-NORMDIST(1018.5872,1016.6,63.0757,TRUE)

h. No because the sample size is greater than 30