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umka2103 [35]
2 years ago
7

Please help if you can.​

Mathematics
1 answer:
ivanzaharov [21]2 years ago
7 0

Answer:

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Step-by-step explanation:

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The final margin of an NFL football game is approximately normally distributed, with a mean equal to the gambling point spread a
nika2105 [10]

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.

7 0
3 years ago
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