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Problem

The National Football League playoffs are just about to begin. Because of their great record in the regular season, the Steelers get a bye in the first week of the playoffs. In the second week, they will play the winner of the game between the Ravens and the Patriots. A football expert estimates that the Ravens will beat the Patriots with probability 0.45. This same expert estimates that if the Steelers play the Ravens, the mean and standard deviation of the point spread (Steelers points minus Ravens points) will be 6.5 and 10.5, whereas if the Steelers play the Patriots, the mean and standard deviation of the point spread (Steelers points minus Patriots points) will be 3.5 and 12.5. Find the mean and standard deviation of the point spread (Steelers points minus their opponent’s points) in the Steelers game.

Step-by-step solution

Let ‘SR’ denotes that the Steelers point minus ravens point. For this, the Ravens need to beat the Patriots, whose probability has been given to be 0.45 and “SP” denotes the Steelers point minus patriots point, thus,

Similarly, if ‘CP’ denotes the event that the Colts will play the Patriots, then:

Let and denote the mean and standard deviation of the point spread of the

event X. Then, from the information provided in the question,

Since the variance of any variable X is simply the square of its standard deviation. The associated variances in this case are,

For any random X, the expected value or mean value is given by,

Then,

Here, n denotes the total number of observations of X; denotes the value associated with the

random variable X and is the probability associated with event .

Let and denote the mean and standard deviation of the point spread of the

Colts. Then,

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Comment

The variance of any variable X is given by,

Here denotes the mean or expected value of X, n denotes the number of values of X and

denotes the probability of . The standard deviation of X is simply the square root of its

variance.

Therefore,

The standard deviation of C is given by,

Thus the mean and standard deviation of the point spread in the Colt’s game is 4.85 and 1.49 respectively.

Comment

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