Problem 24 - Algebra 2 and Trigonometry

Question

In 22-24 , use the normal approximation to estimate each probability. Round your answers to three decimal places. The probability that a basketball team will win any game is .5. What is the probability that the team will win at least 12 out of 25 games?

Step-by-Step Solution

Answer

The probability is approximately 0.655.

Step 1: Identify the Problem as a Binomial Distribution

The question asks for the probability of winning at least 12 out of 25 games, where the probability of winning any single game is 0.5. This is a binomial distribution problem with parameters \( n = 25 \) and \( p = 0.5 \).

Step 2: Use the Normal Approximation of the Binomial Distribution

For large \( n \), the binomial distribution can be approximated by a normal distribution. The normal approximation is appropriate here because \( n = 25 \) is sufficiently large. To use this approximation, calculate the mean (\( \mu \)) and standard deviation (\( \sigma \)) of the binomial distribution using the formulas: \( \mu = np \) and \( \sigma = \sqrt{np(1-p)} \).

Step 3: Calculate Mean and Standard Deviation

Calculate the mean: \( \mu = 25 \times 0.5 = 12.5 \). Calculate the standard deviation: \( \sigma = \sqrt{25 \times 0.5 \times 0.5} = \sqrt{6.25} = 2.5 \).

Step 4: Apply Continuity Correction

To apply the continuity correction, adjust the probability question to encompass the discrete nature of the binomial distribution. Instead of finding \( P(X \geq 12) \), look for \( P(X \geq 11.5) \) in the normal approximation.

Step 5: Convert to Standard Normal Variable (Z-score)

Calculate the Z-score for \( X = 11.5 \) using the formula: \[ Z = \frac{X - \mu}{\sigma} = \frac{11.5 - 12.5}{2.5} = \frac{-1}{2.5} = -0.4 \].

Step 6: Find the Probability Using the Z-table

Use the Z-table to find \( P(Z \geq -0.4) \). Looking up \( -0.4 \) in the Z-table gives a probability of approximately 0.655, which is the probability that the team will win at least 12 out of 25 games.

Key Concepts

Binomial Distribution

A binomial distribution is a probability distribution that summarizes the likelihood of a variable taking one of two independent states and is used when there are a fixed number of trials. Each trial is independent, and the probability of success (winning, in this context) is constant on every trial.

• **Trials**: Individual events, like each basketball game.
• **Success/Failure**: Winning or losing a game.
• **Parameters**: Characterized by two parameters—number of trials () and probability of success per trial ( ). In our exercise, \( n = 25 \) and \( p = 0.5 \).

A classic example of a binomial distribution is flipping a coin: head or tails, win or lose. The goal here is to find the probability of winning a certain number of games (successes) out of numerous attempts.

Continuity Correction

Continuity correction is essential when using the normal approximation to estimate a binomial distribution. The binomial distribution is discrete, representing specific values (like winning exactly 12 games). However, the normal distribution is continuous.

• **Adjustment**: It bridges the gap between continuous and discrete by shifting our target value slightly to better capture the probability.
• For example, instead of calculating the probability of winning at least 12 games (discrete), we adjust our target to 11.5 (continuous), seeking \( P(X \geq 11.5) \) from the normal distribution.

This slight adjustment usually involves adding or subtracting 0.5 to/from the integer in question. It's crucial for improving the approximation's accuracy when transitioning between a discrete scenario and the continuous model.

Z-score

The Z-score converts a particular score in a normal distribution to standard units—making it comparable across different distributions. It essentially tells us how many standard deviations a score is from the mean.

• **Formula**: \( Z = \frac{X - \mu}{\sigma} \)
• **Interpretation**: If the Z-score is negative, the score is below the mean. If positive, it's above.

In our exercise, we used the Z-score to determine where the adjusted value (11.5) falls on the standard normal distribution:

• Calculated \( Z \) as \( \frac{11.5 - 12.5}{2.5} = -0.4 \).
• This score indicates that 11.5 is 0.4 standard deviations below the mean of 12.5.

Z-scores allow us to use standard normal probability tables, or Z-tables, to find probabilities for different scores.

Probability Estimation

Once we have the Z-score, we can estimate the probability using a Z-table, which outlines cumulative probabilities corresponding to each Z-score in a standard normal distribution (mean of 0 and standard deviation of 1).

• Z-scores help convert our real-world problem into one that fits a standard, well-understood framework.
• Using the Z-table, we find \( P(Z \geq -0.4) \), which yields approximately 0.655.

Therefore, this probability estimation suggests there is a 65.5% chance the basketball team will win at least 12 out of 25 games. Z-tables simplify this vast sea of probabilities into an easily navigable format, offering valuable insights into our questions of interest. By increasing your understanding of such conversion techniques, you'll master probability estimation in various contexts.