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The World Series in baseball is won by the first team to win four games (ignoring the 1903 and 1919-1921 World Series, when it was a best of nine). Thus it takes at least four games and no more than seven games to establish a winner. As found on the Major League Baseball Web site in World Series Overview, historically, the lengths of the World Series are as given in the following table.

 Number  of games  Frequency  Relative  frequency 4200.205230.236220.227350.35\begin{array}{c|c|c} \hline \begin{array}{c} \text { Number } \\ \text { of games } \end{array} & \text { Frequency } & \begin{array}{c} \text { Relative } \\ \text { frequency } \end{array} \\ \hline 4 & 20 & 0.20 \\ 5 & 23 & 0.23 \\ 6 & 22 & 0.22 \\ 7 & 35 & 0.35 \\ \hline \end{array}

a. If X denotes the number of games that it takes to complete a World Series, identify the possible values of the random variable X. b. Do the first and third columns of the table provide a probability distribution for X? Explain your answer. c. Historically, what is the most likely number of games it takes to complete a series? d. Historically, for a randomly chosen series, what is the probability that it ends in five games? e. Historically, for a randomly chosen series, what is the probability that it ends in five or more games? f. The data in the table exhibit a statistical oddity. If the two teams in a series are evenly matched and one team is ahead three games to two, either team has the same chance of winning game number six. Thus there should be about an equal number of six- and seven-game series. If the teams are not evenly matched, the series should tend to be shorter, ending in six or fewer games, not seven games. Can you explain why the series tend to last longer than expected?

Question

How much should a healthy kitten weigh? A healthy 10-week-old (domestic) kitten should weigh an average of μ=24.5\mu=24.5 ounces with a (95% of data) range from 14 to 35 ounces. Let x be a random variable that represents the weight (in ounces) of a healthy 10-week-old kitten. Assume that x has a distribution that is approximately normal.

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Given:

μ=24.5\mu=24.5

σ5.25\sigma\approx 5.25

The standardized score is the value xx decreased by the mean and then divided by the standard deviation.

z=xμσ=3324.55.251.62z=\dfrac{x-\mu}{\sigma}=\dfrac{33-24.5}{5.25}\approx 1.62

Look in the row containing the integer and tenth of the zz-score, while looking in the column that contains the corresponding hundredth in table 5, then you find the probability to the left of the zz-score.

P(x>33)=P(z>1.62)=1P(z<1.62)=10.9474=0.0526P(x>33)=P(z>1.62)=1-P(z<1.62)=1-0.9474=0.0526

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