/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 On September \(7,2008,\) the Pit... [FREE SOLUTION] | 91Ó°ÊÓ

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On September \(7,2008,\) the Pittsburgh Pirates lost their 8 2nd 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 \(16 .\) In fact, their losing streak continued until 2012 with 20 consecutive losing seasons. A Major League Baseball season consists of 162 games, so for the Pirates to end their streak, they need to win at least 81 games in a season (which they did in 2013\()\) 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.) c. Can you think of any factors that might make the binomial distribution an inappropriate model for the number of games won in a season?

Short Answer

Expert verified
a. Expected wins: \(68.04\). b. Probability of at least 81 wins: \(0.0197\). c. Factors: player performance variations, non-random events.

Step by step solution

01

Understand the Problem

We are given that the number of games won by the Pirates in a given season follows a binomial distribution with parameters \(n = 162\) and \(p = 0.42\). Our tasks are to calculate the expected number of wins per season, find the probability of winning at least \(81\) games, and identify any factors that could affect the use of a binomial model.
02

Calculate Expected Number of Wins

The expected number of wins, \(E(X)\), for a binomial distribution is found using the formula \(E(X) = n \cdot p\). Substituting the given values, we get \(E(X) = 162 \cdot 0.42 = 68.04\). Thus, the expected number of wins per season is \(68.04\).
03

Find Probability Using Normal Approximation

To find the probability of winning at least 81 games, we approximate the binomial distribution using a normal distribution. The mean (\(\mu\)) is \(68.04\) and the standard deviation (\(\sigma\)) is \(\sqrt{162 \cdot 0.42 \cdot (1 - 0.42)} \approx 6.27\). Evaluate the \(z\)-score: \(z = \frac{81 - 68.04}{6.27} \approx 2.06\). Find the probability \(P(X \geq 81)\) which corresponds to \(1 - P(Z < 2.06)\). Using standard normal distribution tables or technology, \(P(Z < 2.06) \approx 0.9803\), so \(P(X \geq 81) = 1 - 0.9803 = 0.0197\).
04

Consider Factors for Binomial Model Appropriateness

The binomial distribution assumes each game is an independent event with the same probability of winning (success). Factors such as player injuries, changes in team management, opponent strengths, or any non-random event affecting each game's outcome could make the binomial model inappropriate. Variations in player performance or team morale throughout the season are additional factors.

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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 a crucial concept that gives us an idea of the average outcome of a random event if it could be repeated multiple times. For a binomial distribution, such as with the Pittsburgh Pirates' season, the expected value is calculated using a simple formula: \[ E(X) = n \cdot p \] where \( n \) is the number of trials (in this example, the season games totaling 162) and \( p \) is the probability of a win (0.42). By multiplying these values, the expected number of games won in a season is \( 162 \cdot 0.42 = 68.04 \). Thus, over time, the Pirates are expected to win approximately 68 games per season. This expected value does not guarantee outcomes but gives a central tendency towards which data points might converge if enough data exists.
Normal Approximation
When dealing with large samples in a binomial distribution, it is often helpful to use the normal distribution as an approximation for simplifying calculations. This is possible due to the Central Limit Theorem, which states that the binomial distribution approaches a normal distribution as the number of trials increases. For the Pirates, we can approximate the probability of winning at least 81 games by using the normal distribution. First, we calculate the mean \( \mu = 68.04 \) and the standard deviation \( \sigma \) using: \[ \sigma = \sqrt{n \cdot p \cdot (1-p)} \approx 6.27 \] Next, we determine the \( z \)-score for winning 81 games: \[ z = \frac{81 - 68.04}{6.27} \approx 2.06 \] Using this \( z \)-score, we can find the area under the normal curve that corresponds to probabilities. Calculating \( P(X \geq 81) = 1 - P(Z < 2.06) \), gives \( 0.0197 \), indicating around a 1.97% chance of the Pirates winning at least 81 games.
Probability Calculation
Calculating probabilities in a binomial setting involves determining the likelihood of a particular number of successes over a series of independent trials. In simpler terms, how likely it is for the Pirates to win a certain number of games in a season based on historical performance. Typically, this would mean calculating \( P(X = x) \) for wins, where \( X \) follows a binomial distribution. However, for large \( n \), normal approximation can be more appropriate. Key steps involve:
  • Identifying parameters \( n \) and \( p \).
  • Calculating mean \( \mu = n \cdot p \).
  • Estimating variance \( \sigma^2 = n \cdot p \cdot (1-p) \).
  • Applying normal distribution to approximate binomial probabilities.
Finally, technology or tables are used to find critical probabilities, assisting in making inferences on the Pirates' winning likelihood with precision and ease.
Model Assumptions
The binomial distribution, while extensively used, relies on crucial assumptions that must be evaluated for appropriateness in practical settings. Firstly, it assumes each game is an independent event where the outcome does not affect the other games. It also assumes that the probability of winning each game remains constant over time. However, several real-world factors can challenge these assumptions:
  • Team dynamics like player injuries or changes in team management can alter probabilities from game to game.
  • Opponent strength variations through the season can affect game outcomes.
  • Seasonal morale shifts or external conditions influencing performance.
These factors can introduce dependencies or change probabilities, making the binomial model less appropriate. It’s critical to assess these elements before strictly applying the model to predict outcomes such as the Pirates’ season performance.

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Most popular questions from this chapter

Basketball shots To win a basketball game, two competitors play three rounds of one three-point shot each. The series ends if one of them scores in a round but the other misses his shot or if both get the same result in each of the three rounds. Assume competitors \(\mathrm{A}\) and \(\mathrm{B}\) have \(30 \%\) and \(20 \%\) of successful attempts, respectively, in three-point shots and that the outcomes of the shots are independent events. a. Verify the probability that the series ends in the second round is \(23.56 \%\). (Hint: Sketch a tree diagram and write out the sample space of all possible sequences of wins and losses in the three rounds of the series, find the probability for each sequence and then add up those for which the series ends within the second round). b. Find the probability distribution of \(X=\) number of rounds played to end the series. c. Find the expected number of rounds to be played in the series.

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