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
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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](https://tex.z-dn.net/?f=%5Cmu%20%3D%2012%2C%20%5Csigma%20%3D%2014)
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}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{0 - 12}{14}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B0%20-%2012%7D%7B14%7D)
![Z = -0.86](https://tex.z-dn.net/?f=Z%20%3D%20-0.86)
has a pvalue of 0.1949
The Giants had 0.1949 = 19.49% probability of winning initially.