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Ch. 6 - Normal Probability Distributions
Triola - Elementary Statistics 14th Edition
Triola14th EditionElementary StatisticsISBN: 9780137366446Not the one you use?Change textbook
All textbooksTriola 14th EditionCh. 6 - Normal Probability DistributionsProblem 6.6.13
Chapter 6, Problem 6.6.13

Tennis Replay In a recent year, there were 879 challenges made to referee calls in professional tennis singles play. Among those challenges, 231 challenges were upheld with the call overturned. Assume that in general, 25% of the challenges are successfully upheld with the call overturned.


a. If the 25% rate is correct, find the probability that among the 879 challenges, the number of overturned calls is exactly 231.

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1
Step 1: Recognize that this is a binomial probability problem. The problem involves a fixed number of trials (n = 879), two possible outcomes (success = overturned call, failure = call not overturned), and a constant probability of success (p = 0.25). The goal is to find the probability of exactly 231 successes (k = 231).
Step 2: Write the formula for the binomial probability distribution: P(X = k) = (n choose k) * p^k * (1 - p)^(n - k). Here, (n choose k) is the binomial coefficient, which can be calculated as (n! / (k! * (n - k)!)).
Step 3: Substitute the given values into the formula. Use n = 879, k = 231, and p = 0.25. The formula becomes: P(X = 231) = (879 choose 231) * (0.25)^231 * (0.75)^(879 - 231).
Step 4: Calculate the binomial coefficient (879 choose 231). This is done using the formula: (879! / (231! * (879 - 231)!)). This step involves factorials, which can be computed using a calculator or software.
Step 5: Multiply the binomial coefficient by the probabilities raised to their respective powers. Specifically, compute (0.25)^231 and (0.75)^(879 - 231), then multiply these values by the binomial coefficient to find the final probability.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Binomial Distribution

The binomial distribution models the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success. In this context, the number of challenges (879) represents the trials, while the probability of a challenge being upheld (25%) is the success probability. This distribution is essential for calculating the probability of observing exactly 231 overturned calls.
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Probability Mass Function (PMF)

The probability mass function gives the probability of a discrete random variable taking on a specific value. For a binomial distribution, the PMF can be calculated using the formula P(X = k) = (n choose k) * p^k * (1-p)^(n-k), where n is the number of trials, k is the number of successes, and p is the probability of success. This function is crucial for determining the likelihood of exactly 231 challenges being upheld.
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Normal Approximation to the Binomial

For large sample sizes, the binomial distribution can be approximated by a normal distribution, which simplifies calculations. This approximation is valid when both np and n(1-p) are greater than 5. In this scenario, with 879 challenges and a 25% success rate, the normal approximation can be used to estimate probabilities and assess the likelihood of observing 231 overturned calls.
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Related Practice
Textbook Question

Small Sample Weights of M&M plain candies are normally distributed. Twelve M&M plain candies are randomly selected and weighed, and then the mean of this sample is calculated. Is it correct to conclude that the resulting sample mean cannot be considered to be a value from a normally distributed population because the sample size of 12 is too small? Explain.

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Textbook Question

Basis for the Range Rule of Thumb and the Empirical Rule. In Exercises 45–48, find the indicated area under the curve of the standard normal distribution; then convert it to a percentage and fill in the blank. The results form the basis for the range rule of thumb and the empirical rule introduced in Section 3-2.


About __ % of the area is between z = -1 and z = 1 (or within 1 standard deviation of the mean).

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Textbook Question

Finding Bone Density Scores. In Exercises 37–40 assume that a randomly selected subject is given a bone density test. Bone density test scores are normally distributed with a mean of 0 and a standard deviation of 1. In each case, draw a graph, then find the bone density test score corresponding to the given information. Round results to two decimal places.


Find the bone density scores that are the quartiles Q1, Q2, and Q3.

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Textbook Question

IQ Scores. In Exercises 5–8, find the area of the shaded region. The graphs depict IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15 (as on the Wechsler IQ test).

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