Question

The final margin of an NFL football game is approximately normally distributed, with a mean equal to the gambling point spread and a standard deviation of 14 points. Recall the great 2008 Super Bowl in which the Giants upset the Patriots. The Patriots had been favored by 12 points. What does that fact imply about the Giants' initial (i.e., pre-game) chances of winning

Answer 1

Answer:

The Giants had 0.1949 = 19.49% probability of winning initially.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Mean equal to the gambling point spread and a standard deviation of 14 points. The Patriots had been favored by 12 points.

This means that $\mu = 12, \sigma = 14$

What does that fact imply about the Giants' initial (i.e., pre-game) chances of winning

P(X > 0): Probability of Patriots winning.

P(X < 0): Probability of Giants winning.

So we want to find P(X < 0), which is the pvalue of Z when X = 0. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0 - 12}{14}$

$Z = -0.86$

$Z = -0.86$ has a pvalue of 0.1949

The Giants had 0.1949 = 19.49% probability of winning initially.