91Ó°ÊÓ

A baseball player in the major leagues who plays regularly will have about 500 at-bats (that is, about 500 times he can be the hitter in a game) during a season. Suppose a player has a 0.300 probability of getting a hit in an at- bat. His batting average at the end of the season is the number of hits divided by the number of at-bats. When we consider the 500 at-bats as a random sample of all possible at-bats for this player, this batting average is a sample proportion, so it has a sampling distribution describing where it is likely to fall. a. Describe the shape, mean, and standard deviation of the sampling distribution of the player's batting average. b. Explain why a batting average of 0.320 or of 0.280 would not be especially unusual for this player's year-end batting average. (That is, you should not conclude that someone with a batting average of 0.320 is necessarily a better hitter than a player with a batting average of \(0.280 .\) Both players could have a probability of 0.300 of getting a hit.)

Step by step solution

01

Identify the Problem Type

The problem asks about the sampling distribution of a sample proportion, which relates to a binomial distribution converted to a normal distribution due to a large number of trials.

02

Determine the Shape of the Distribution

For large sample sizes like 500, the sampling distribution of a proportion is approximately normal due to the Central Limit Theorem, because both np and n(1-p) are greater than 5 for this situation.

03

Calculate the Mean of the Sampling Distribution

The mean of the sample proportion sampling distribution is equal to the probability of success, which in this case is the probability of getting a hit: \( \mu = p = 0.300 \).

04

Calculate the Standard Deviation of the Sampling Distribution

The standard deviation of the sampling distribution of a sample proportion is given by \( \sigma = \sqrt{\frac{p(1-p)}{n}} \), where \( p = 0.300 \) and \( n = 500 \). This evaluates to \( \sigma = \sqrt{\frac{0.300 \times 0.700}{500}} \approx 0.020 \).

05

Use the Standard Normal Distribution for Probabilities

To find if hitting 0.320 or 0.280 is unusual, convert these into z-scores using \( z = \frac{x - \mu}{\sigma} \) for each value. For 0.320, \( z = \frac{0.320 - 0.300}{0.020} = 1 \). For 0.280, \( z = \frac{0.280 - 0.300}{0.020} = -1 \).

06

Interpret Normal Distribution Z-Scores

A z-score of 1 or -1 indicates the score is one standard deviation away from the mean. For a normal distribution, data within one standard deviation (approximately 68% of the data) is considered common, suggesting neither 0.320 nor 0.280 is unusual.

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Key Concepts

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

Central Limit Theorem

The central limit theorem is a fantastic tool statisticians use to understand data better. It tells us that, given a large enough sample size, the distribution of sample means or sample proportions will approach a normal distribution. This happens no matter what the shape of the original population distribution is. It's like the universe's way of balancing everything out.
For instance, in our baseball player's example, he has 500 at-bats, which is quite a large sample size. This is crucial because it means the sampling distribution of his batting average can be viewed as a normal distribution, thanks to the central limit theorem. This is true as long as the sample size is large enough for both conditions, \(np\) and \(n(1-p)\), to be greater than 5. This balances out the chances and assures us we can rely on the normal curve to make predictions about his batting average.

Binomial Distribution

The concept of a binomial distribution is at the heart of understanding how likely different outcomes are over a set number of trials. For instance, each at-bat of our baseball player is considered a trial. In a binomial setting:

• Each trial is independent.
• There are only two possible outcomes - a hit or a miss.
• The probability of success is the same on each trial.

In our exercise, the player having an at-bat and hitting out of 500 times is perfectly modeled by a binomial distribution. However, because our sample size is pretty large, we conveniently use the central limit theorem to approximate this binomial distribution to a normal distribution. But originally, it begins with the understanding that each swing of the bat is another chance to succeed or fail.

Z-Score

A z-score is a statistical measure that tells us how many standard deviations a particular value is from the mean. It's like a map that helps locate where we stand in the distribution. Calculating a z-score helps us understand whether a data point is typical or unusual compared to the overall distribution.
In this scenario, for example, the player's batting averages of 0.320 and 0.280 are converted to z-scores. Using the formula \( z = \frac{x - \mu}{\sigma} \), where \( \mu = 0.300 \), \( \sigma \approx 0.020 \), we find z-scores of 1 and -1 respectively. This means both averages lie within one standard deviation from the mean. According to how normal distributions work, this range is somewhat ordinary since about 68% of data in a normal distribution falls within one standard deviation. Hence, neither average is considered unusual.

Sample Proportion

The sample proportion is essentially the fraction of successful outcomes in our sample. It is calculated as the number of successes divided by the total number of trials. In our baseball player's case, it represents his batting average.
For this problem, the sample proportion is the player's batting average over the course of 500 at-bats during a season. It’s the real-world reflection of how our player is doing statistically. We examine this proportion to draw insights. What's interesting and vital is that as the sample size (number of at-bats) grows, the sampling distribution of the sample proportion becomes approximately normal, allowing us to apply normal probability theories to analyze it easily.