91Ó°ÊÓ

On September \(7,2008,\) the Pittsburgh Pirates lost their 82 nd game of the 2008 season and tied the \(1933-1948\) Philadelphia Phillies major sport record (baseball, football, basketball, and hockey) for most consecutive losing seasons at 16. One year later on September \(7,2009,\) they lost their 82 nd game of the 2009 season, and the record became theirs alone. The only way things could get much worse for the Pirates was to lose their 82 nd game earlier in the season. Sure enough, on August \(21,2010,\) they lost their 82 nd game of the 2010 season, extending their streak to 18 consecutive seasons. A major league baseball season consists of 162 games, so for the Pirates to end their streak, they will eventually need to win at least 81 games in a season. a. Over the course of the streak, the Pirates have won approximately \(42 \%\) of their games. For simplicity, assume the number of games they win in a given season follows a binomial distribution with \(n=162\) and \(p=0.42 .\) What is their expected number of wins in a season? b. What is the probability that the Pirates will win at least 81 games in a given season? (You may use technology to find the exact binomial probability or use the normal distribution to approximate the probability by finding a \(z\) -score for 81 and then evaluating the appropriate area under the normal curve.)

Step by step solution

01

Interpreting the Binomial Distribution

The problem states that the number of wins follows a binomial distribution with parameters \(n = 162\) (total games) and \(p = 0.42\) (probability of winning a single game). We need these parameters for our calculations.

02

Calculate Expected Number of Wins

The expected value \(E(X)\) of a binomial distribution \(X\) with parameters \(n\) and \(p\) is given by the formula \(E(X) = np\). Substitute \(n = 162\) and \(p = 0.42\) to find the expected number of wins: \[ E(X) = 162 \times 0.42 = 68.04 \]Therefore, the expected number of wins in a season is approximately 68.

03

Probability Calculation Using Normal Approximation

Since calculating the exact binomial probability may be complex, we'll use the normal distribution to approximate. The normal approximation is valid because \(np\) and \(n(1-p)\) are large. Calculate the mean \(\mu = np = 68.04\) and the standard deviation \(\sigma = \sqrt{np(1-p)} = \sqrt{162 \times 0.42 \times 0.58} \approx 6.24\).

04

Calculate the Z-Score for 81 Wins

The z-score is calculated using the formula \[ z = \frac{X - \mu}{\sigma} \]where \(X = 81\), \(\mu = 68.04\), and \(\sigma = 6.24\). Substituting these values in, we get: \[ z = \frac{81 - 68.04}{6.24} \approx 2.07 \]

05

Find Probability from Z-Score

Using standard normal distribution tables or technology, we find the probability \(P(Z \geq 2.07)\). This corresponds to the area to the right of \(z = 2.07\) under the standard normal curve. The probability \(P(Z \geq 2.07)\) is approximately 0.0192.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expected Value

In probability and statistics, the "Expected Value" is the average number of expected events, calculated using the probabilities of all possible outcomes. This concept is crucial when dealing with random variables. Simply put, it provides an estimate of the center of a probability distribution.
For the Pittsburgh Pirates' win-loss record, we use the formula for the expected value, specifically in a binomial distribution:

• The formula is given by \(E(X) = np\), where \(n\) is the number of trials (games) and \(p\) is the probability of success (winning a game).

Using the provided values:

• \(n = 162\) games in a season.
• \(p = 0.42\), the odds of winning any single game.

By substituting these into the formula: \[ E(X) = 162 \times 0.42 = 68.04 \]This calculates the expected number of games the Pirates are likely to win in a season --> approximately 68 games. This helps set expectations for performance over the season.

Normal Approximation

When a binomial distribution has many trials, calculating exact probabilities directly can be cumbersome. Here, "Normal Approximation" comes to the rescue. Given that a binomial distribution can be closely approximated by a normal distribution, especially when the product of \(n\) and \(p\) is large enough, it provides a simpler approach.
For the Pirates:

• The mean \(\mu\) (expected number of wins) is given by \(np = 68.04\).
• The standard deviation \(\sigma\) is calculated as \(\sigma = \sqrt{np(1-p)}\).

Substituting into the formula for standard deviation: \[ \sigma = \sqrt{162 \times 0.42 \times 0.58} \approx 6.24 \]This normal approximation allows us to use the standard normal distribution, a far simpler model, to estimate probabilities for a large number of trials like a baseball season.

Probability Calculation

Probability calculation in this context refers to determining the likelihood of achieving a certain number of wins. Using the normal approximation, we simplify this calculation.
First, identify the desired outcome: winning at least 81 games. Then, use the normal approximation to determine how this compares with the expected performance:

• Estimate the probability of the Pirates winning 81 or more games using the normal distribution.

Rather than calculate directly through complicated binomial probabilities, we use the approximation with the calculated mean and standard deviation, simplifying the practicality of evaluating long sequences of trials.

• This complexity is reduced to finding an area under the normal curve, and it's manageable with statistical tables or technology.

Z-Score

"Z-Score" is a statistical measurement that describes a value's relation to the mean of a group of values, in terms of standard deviations. It helps in finding out how far and in what direction an observation deviates from the mean.
For the Pirates' probability question, calculating the z-score is essential to use the normal distribution for probability estimation:The Z-Score formula is: \[ z = \frac{X - \mu}{\sigma} \]Where:

• \(X\) is the target number of wins (81),
• \(\mu\) is the expected average (68.04),
• \(\sigma\) is the standard deviation (6.24).

Plug these values into the formula to calculate: \[ z = \frac{81 - 68.04}{6.24} \approx 2.07 \]This z-score indicates how many standard deviations 81 is above the average win count.

• A positive z-score shows a value greater than the mean.
• Using this z-score, we determine the tail area in the standard normal distribution table, which shows how likely such an outcome is.

This translates the probability into an intuitive measure of the "unlikeliness" or rarity of winning 81 or more games in a season.