Question

lesson 6.4 - combining random variables
check your understanding
most sports fans believe in home field advantage - the idea that home teams have a slight edge over the away teams because they have wild fans cheering for them. lets investigate by looking at results from all 285 games of a recent nfl football season. let ( x ) = the number of points scored by the home team, which can be approximated by a normal distribution with ( mu = 22.98 ) and ( sigma = 9.30 ) and ( y ) = the number of points scored by the away team, which can be approximated by a normal distribution with ( mu = 20.90 ) and ( sigma = 9.41 ).
define ( d = x - y ). assume that ( x ) and ( y ) are independent.
a. calculate and interpret ( mu_d ).
b. calculate ( sigma_d ).
c. suppose we randomly choose a game and record the number of points from the home team, and then independently randomly choose another game and record the number of points from the away team. what is the probability that the home team has more points?

Explanation:

Step1: Calculate mean of D

$\mu_D = \mu_X - \mu_Y = 22.98 - 20.90 = 2.08$
Interpretation: On average, home teams score 2.08 more points than away teams.

Step2: Calculate variance of D

$\sigma_D^2 = \sigma_X^2 + \sigma_Y^2 = 9.30^2 + 9.41^2 = 86.49 + 88.5481 = 175.0381$

Step3: Calculate standard deviation of D

$\sigma_D = \sqrt{175.0381} \approx 13.23$

Step4: Find z-score for D>0

$z = \frac{0 - \mu_D}{\sigma_D} = \frac{0 - 2.08}{13.23} \approx -0.157$

Step5: Calculate probability P(D>0)

$P(Z > -0.157) \approx 1 - 0.437 = 0.563$

Answer:

a. $\mu_D = 2.08$
b. $\sigma_D \approx 13.23$
c. $\approx 0.563$