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Baseball hitting Suppose a baseball player has a 0.200 probability of getting a hit in each time at-bat. a. Describe the shape, mean, and standard deviation of the sampling distribution of the proportion of times the player gets a hit after 36 at- bats. b. Explain why it would not be surprising if the player has a 0.300 batting average after 36 at-bats.

Short Answer

Expert verified
The distribution is approximately normal with mean 0.200 and standard deviation 0.047. A 0.300 average after 36 at-bats is possible due to variability.

Step by step solution

01

Understand Distribution Type

With a probability of success (hit) of 0.200 and a sample size of 36, we're dealing with a binomial distribution that can be approximated by a normal distribution when the conditions are met.
02

Conditions for Approximation

We check if the sample size is large enough to approximate a binomial distribution with a normal distribution. This requires both np and n(1-p) to be greater than 5, where n is the number of trials and p is the probability of success: \[ np = 36 \times 0.2 = 7.2 \] \[ n(1-p) = 36 \times 0.8 = 28.8 \] Both conditions are satisfied, allowing for a normal approximation.
03

Compute the Mean of the Sampling Distribution

The mean \(\mu\) of the sampling distribution of the proportion of hits is given by \(\mu = p = 0.200\).
04

Calculate the Standard Deviation of the Sampling Distribution

The standard deviation \(\sigma\) of the sampling distribution of the proportion is calculated using the formula: \[ \sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.2 \times 0.8}{36}} \approx 0.047 \]
05

Analyze a 0.300 Batting Average

A 0.300 batting average translates to 10.8 hits in 36 at-bats. Given the initial mean of 7.2 hits (from Step 2), this deviation from the mean lies within the range of possible outcomes because of the observed standard deviation and normal variability.
06

Determine Likelihood of Observed Outcome

The probability of observing such an outcome (0.300 average) can be analyzed by calculating the z-score, but without exact z-table values, understanding standard deviation behavior and the ability of stocking batter averages gives reasons why a 0.300 average within early attempts is possible.

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Key Concepts

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

Binomial Distribution
The binomial distribution is a discrete probability distribution that models scenarios where there are two possible outcomes: success or failure. In the context of our baseball player, getting a hit is considered a success, while not getting a hit is a failure. The player has a probability of success, denoted as \( p = 0.200 \), for each time at-bat. When we perform a set number of trials, each with the same probability of success, the distribution of these outcomes follows a binomial distribution.

Key characteristics of a binomial distribution include having a fixed number of trials, each independent of the others, and a constant probability of success. Thus, the scenario of a player going to bat 36 times satisfies these conditions since each at-bat is an independent trial with a consistent probability of achieving a hit. This setup allows us to subsequently model and analyze the player's performance effectively using this distribution.
Normal Approximation
Normal approximation is a method used to estimate the binomial distribution with a normal distribution when certain conditions are met. This is particularly useful because normal distributions are easier to work with and well-understood. For a binomial distribution, we can switch to a normal approximation if the sample size is sufficiently large and certain conditions regarding the probability of success are met.

In our problem, the player goes to bat 36 times with a success probability of 0.200. The conditions require that both \( np \) and \( n(1-p) \) be greater than 5. Here, \( np = 36 \times 0.2 = 7.2 \) and \( n(1-p) = 36 \times 0.8 = 28.8 \). Both values satisfy the condition, validating the use of a normal approximation. This facilitates more straightforward calculations and allows us to use z-scores to understand potential outcomes.
Standard Deviation
The standard deviation is a measure of variability or dispersion of a set of values. In the context of a sampling distribution, it tells us how much the sample proportions are expected to vary from the overall population proportion. This metric is essential as it provides insight into the reliability and variability of our binomial distribution estimates.

For the player's at-bats, the formula for the standard deviation of a sampling distribution derived from a binomial process is \( \sigma = \sqrt{\frac{p(1-p)}{n}} \). Plugging our values into this formula gives \( \sigma = \sqrt{\frac{0.2 \times 0.8}{36}} \approx 0.047 \). This small value indicates that while the player might have varying outcomes in each at-bat, the overall performance closely centers around the mean proportion of hits.
Mean of Sampling Distribution
The mean of a sampling distribution offers a central value or expected outcome we anticipate from our sampled trials. It is the average number we expect over numerous trials, and for a binomial distribution, it equals the probability of success \( p \) multiplied by the number of trials \( n \).

Thus, for the player, the mean of the sampling distribution of the proportion of hits is simply \( \mu = p = 0.200 \). It signifies that on average, the player will get a hit 20% of the time. In a set of 36 at-bats, we multiply the mean proportion by the number of attempts to get a total expected value, yielding \( 36 \times 0.2 = 7.2 \) hits. This calculation provides us with a baseline to evaluate the player's performance and compare it against observed batting averages like the 0.300 scenario mentioned in the exercise.

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Most popular questions from this chapter

Other scenario for exit poll Refer to Examples 1 and 2 about the exit poll, for which the sample size was \(3889 . \mathrm{In}\) that election, \(40.9 \%\) voted for Whitman. a. Define a binary random variable \(X\) taking values 0 and 1 that represents the vote for a particular voter \((1=\) vote for Whitman and \(0=\) another candidate \()\) State its probability distribution, which is the same as the population distribution for \(X\). b. Find the mean and standard deviation of the sampling distribution of the proportion of the 3889 people in the sample who voted for Whitman.

Multiple choice: Standard deviation Which of the following is not correct? The standard deviation of a statistic describes a. The standard deviation of the sampling distribution of that statistic. b. The standard deviation of the sample data measurements. c. How close that statistic falls to the parameter that it estimates. d. The variability in the values of the statistic for repeated random samples of size \(n\).

Exam performance An exam consists of 50 multiplechoice questions. Based on how much you studied, for any given question you think you have a probability of \(p=0.70\) of getting the correct answer. Consider the sampling distribution of the sample proportion of the 50 questions on which you get the correct answer. a. Find the mean and standard deviation of the sampling distribution of this proportion. b. What do you expect for the shape of the sampling distribution? Why? c. If truly \(p=0.70,\) would it be very surprising if you got correct answers on only \(60 \%\) of the questions? Justify your answer by using the normal distribution to approximate the probability of a sample proportion of 0.60 or less.

PDI The scores on the Psychomotor Development Index (PDI), a scale of infant development, have a normal population distribution with mean 100 and standard deviation 15. An infant is selected at random. a. Find the \(z\) -score for a PDI value of 90 . b. A study uses a random sample of 225 infants. Using the sampling distribution of the sample mean PDI, find the \(z\) -score corresponding to a sample mean of 90 . c. Explain why a PDI value of 90 is not surprising, but a sample mean PDI score of 90 for 225 infants would be surprising.

Finite populations The formula \(\sigma / \sqrt{n}\) for the standard deviation of \(\bar{x}\) actually is an approximation that treats the population size as infinitely large relative to the sample size \(n\). The exact formula for a finite population size \(N\) is Standard deviation \(=\sqrt{\frac{N-n}{N-1}} \frac{\sigma}{\sqrt{n}}\) The term \(\sqrt{(N-n) /(N-1)}\) is called the finite population correction. a. When \(n=300\) students are selected from a college student body of size \(\mathrm{N}=30,000\), show that the standard deviation equals \(0.995 \sigma / \sqrt{n}\). (When \(n\) is small compared to the population size \(\mathrm{N},\) the approximate formula works very well.) b. If \(n=\mathrm{N}\) (that is, we sample the entire population), show that the standard deviation equals \(0 .\) In other words, no sampling error occurs, since \(\bar{x}=\mu\) in that case.

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