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In the National Basketball Association, the top free throw shooters usually have probability of about 0.90 of making any given free throw. a. During a game, one such player (Dolph Schayes) shot 10 free throws. Let \(X=\) number of free throws made. What must you assume in order for \(X\) to have a binomial distribution? (Studies have shown that such assumptions are well satisfied for this sport.) b. Specify the values of \(n\) and \(p\) for the binomial distribution of \(X\) in part a. c. Find the probability that he made (i) all 10 free throws and (ii) 9 free throws.

Short Answer

Expert verified
Assume trials are independent with a constant probability of success. Parameters: \(n=10\), \(p=0.90\). Probabilities: all 10 free throws \(=0.3487\), nine free throws \(=0.3874\).

Step by step solution

01

Understanding the Binomial Distribution

For a random variable to have a binomial distribution, the following assumptions must be satisfied: there are a fixed number of trials, each trial has only two possible outcomes (success or failure), the probability of success is the same for each trial, and the trials are independent of each other. In this context, each free throw is a trial, making a free throw is a success, and the probability of success is constant.
02

Identifying Parameters for the Binomial Distribution

In this problem, we are dealing with the free throws of a basketball player. The player takes a fixed number of trials, specifically 10 free throws, which gives us the parameter \(n = 10\). The probability of making any given free throw is the probability of success, which is \(p = 0.90\). Hence, the random variable \(X\) is binomially distributed with parameters \(n = 10\) and \(p = 0.90\).
03

Calculating the Probability of Making All 10 Free Throws

To find the probability of making all 10 free throws (\(X=10\)), we use the binomial probability formula: \[P(X=k) = \binom{n}{k} p^k (1-p)^{n-k},\]where \(\binom{n}{k}\) is the number of combinations. For \(k=10\):\[P(X=10) = \binom{10}{10} (0.90)^{10} (1-0.90)^{0} = 1 \times 0.90^{10}\]Calculating this gives approximately 0.3487.
04

Calculating the Probability of Making 9 Free Throws

To find the probability of making exactly 9 free throws (\(X=9\)), we use the binomial probability formula again for \(k=9\):\[P(\text{X=9}) = \binom{10}{9} (0.90)^{9} (1-0.90)^{1} = 10 \times 0.90^{9} \times 0.10\]This simplifies to approximately 0.3874.

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Key Concepts

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

Probability Theory
Probability theory is a branch of mathematics that deals with calculating the likelihood of various outcomes. It plays a crucial role in understanding events that occur randomly, such as the success of a basketball player's free throws.

In our exercise, we apply probability to determine how often we expect certain numbers of successful free throws by investigating the chances of different outcomes. Probability theory allows us to use mathematical formulas to predict the chances of a player making all, some, or none of their free throws.

Key factors include:
  • Determining the probability of success (e.g., making a successful free throw)
  • Assessing the nature of the trials (fixed number, two possible outcomes)
  • Understanding how different outcomes can be calculated using mathematical concepts
These ideas use the concept of binomial distribution, which helps us model the likelihood of an event occurring a specific number of times under defined circumstances.
Random Variable
In statistics, a random variable represents a variable whose outcomes are determined by a random phenomenon. It's a way to numerically express possible outcomes of an experiment in a probabilistic framework.

In our basketball scenario, we use the random variable \(X\) to denote the number of successful free throws made by a player. Random variables can be discrete, as in this case where \(X\) can take on values from 0 to 10, indicating the number of throws made successfully.

Understanding random variables is essential because:
  • They allow us to quantify the outcomes of an event.
  • They provide a means to apply the binomial probability formula to calculate specific probabilities.
  • They help in interpreting the statistical distribution of outcomes, letting us see the probabilities associated with different success rates.
These concepts are fundamental in translating real-world random occurrences into mathematical analysis.
Free Throw Statistics
Free throw statistics in basketball reflect a player's likelihood of making a basket when shooting from the free throw line. They are crucial for evaluating player performance and developing effective strategies.

The probability of a successful free throw, denoted by \(p\), is key to calculating basketball player's performance. In professional scenarios, top players often have a very high probability of success, such as the 0.90 probability mentioned in our exercise.

Key aspects include:
  • Comparing individual performance against team averages or historical data.
  • Using statistical norms and benchmarks to set player goals and improvement areas.
  • Applying statistical analysis to predict future performance and outcomes.
By understanding and analyzing these statistics, coaches and players can improve their methods and gain strategic advantages.
Independent Trials
Independent trials are a fundamental concept in probability, describing situations where the outcome of one trial does not affect another. This is crucial when applying the binomial distribution, as seen in calculating the probabilities of successful free throws.

In the context of basketball free throws:
  • Each throw is considered an independent trial.
  • The player's likelihood of making any one throw is unaffected by previous throws.
  • The same probability of success applies consistently across all trials.
This independence allows us to use the binomial formula confidently, ensuring that each throw is statistically identical in terms of its success probability. It's a powerful assumption, giving us insights into performance consistency and reliability over multiple attempts.

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Most popular questions from this chapter

The juror pool for the upcoming murder trial of a celebrity actor contains the names of 100,000 individuals in the population who may be called for jury duty. The proportion of the available jurors on the population list who are Hispanic is 0.40 . A jury of size 12 is selected at random from the population list of available jurors. Let \(X=\) the number of Hispanics selected to be jurors for this jury. a. Is it reasonable to assume that \(X\) has a binomial distribution? If so, identify the values of \(n\) and \(p\). If not, explain why not. b. Find the probability that no Hispanic is selected. c. If no Hispanic is selected out of a sample of size 12 , does this cast doubt on whether the sampling was truly random? Explain.

For a normal distribution, use Table A, software, or a calculator to find the probability that an observation is a. at least 1 standard deviation above the mean. b. at least 1 standard deviation below the mean. c. In each case, sketch a curve and show the tail probability.

A quiz in a statistics course has four multiple-choice questions, each with five possible answers. A passing grade is three or more correct answers to the four questions. Allison has not studied for the quiz. She has no idea of the correct answer to any of the questions and decides to guess at random for each. a. Find the probability she lucks out and answers all four questions correctly. b. Find the probability that she passes the quiz.

The state of Ohio has several statewide lottery options. One is the Pick 3 game in which you pick one of the 1000 three-digit numbers between 000 and 999\. The lottery selects a three-digit number at random. With a bet of \(\$ 1,\) you win \(\$ 500\) if your number is selected and nothing (\$0) otherwise. (Many states have a very similar type of lottery.) (Source: Background information from www.ohiolottery.com.) a. With a single \(\$ 1\) bet, what is the probability that you win \(\$ 500 ?\) b. Let \(X\) denote your winnings for a \(\$ 1\) bet, so \(x=\$ 0\) or \(x=\$ 500\). Construct the probability distribution for \(X\). c. Show that the mean of the distribution equals 0.50 , corresponding to an expected return of 50 cents for the dollar paid to play. Interpret the mean. d. In Ohio's Pick 4 lottery, you pick one of the 10,000 four-digit numbers between 0000 and 9999 and (with a \(\$ 1\) bet ) win \(\$ 5000\) if you get it correct. In terms of your expected winnings, with which game are you better off - playing Pick 4 , or playing Pick 3 in which you win \(\$ 500\) for a correct choice of a three-digit number? Justify your answer.

Consider a game of poker being played with a standard 52 -card deck (four suits, each of which has 13 different denominations of cards). At a certain point in the game, six cards have been exposed. Of the six, four are diamonds. Your opponent makes a bet of \(\$ 20\), and you must decide whether to call the bet. If you do call the bet, you will receive one more card. If that final card turns out to be another diamond, you will win \(\$ 100\). If not, you will lose the hand as well as the \(\$ 20\) you called in order to receive the final card. On the other hand, if you do not call the bet, the hand ends immediately, your opponent wins, and you neither win nor lose any more money. a. Specific the probability distribution for \(X=\) expected winnings. b. Find the expected value of \(X\). Based on the expected value, should you call the \(\$ 20\) bet and receive one more card or not call the bet?

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