Teams in Major League Baseball play 162 games in a season. The all-time record f

Question

Teams in Major League Baseball play 162 games in a season. The all-time record for wins in a season is 116. Let's do some computations on the chance of beating this record. (NOTE that you will definitely want to use a spreadsheet on these computations.) a) Let's start with a simplistic model that a team has a fifty percent chance of winning a game. If that's the case, what's the probability of a team winning 117 or more games in a season? b) Does the model in Part A seem realistic? Why? c) Now suppose the team has a sixty percent chance of winning a game. What's the probability of a team winning 117 or more games in a season? d) Ditto, with a seventy percent chance. e) Clearly, some teams are better than others in the league. So a model that assigns the same chance of victory to each game of the season may be too simplistic. How might we construct a more accurate model (using binomial distributions) for the number of games a team wins in a season?

Explanation / Answer

Solution:-

a) The probability of team winning 117 or more games in a season is less than 0.000001.

n = 162, p = 0.50, x = 117

By applying binomial distribution:-

P(x, n, p) = nCx*p x *(1 - p)(n - x)

P(x > 117) = less than 0.000001

b)  Model in Part A does not seem realistic beacus the probability of team winning 117 or more games in a season is less than 0.000001.

c) The probability of team winning 117 or more games in a season is 0.000792.

n = 162, p = 0.60, x = 117

By applying binomial distribution:-

P(x, n, p) = nCx*p x *(1 - p)(n - x)

P(x > 117) = 0.000792

d)

The probability of team winning 117 or more games in a season is 0.3005.

n = 162, p = 0.70, x = 117

By applying binomial distribution:-

P(x, n, p) = nCx*p x *(1 - p)(n - x)

P(x > 117) = 0.3005.

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