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Chapter 5: Problem 112

There are 750 players on the active rosters of the 30 Major League Baseball teams. A random sample of 15 players is to be selected and tested for use of illegal drugs. a. If \(5 \%\) of all the players are using illegal drugs at the time of the test, what is the probability that 1 or more players test positive and fail the test? b. If \(10 \%\) of all the players are using illegal drugs at the time of the test, what is the probability that 1 or more players test positive and fail the test? c. If \(20 \%\) of all the players are using illegal drugs at the time of the test, what is the probability that 1 or more players test positive and fail the test?

Short Answer

Expert verified
The probabilities that 1 or more players test positive are calculated using the complementary probabilities of none of the players testing positive. The answers for 5%, 10% and 20% drug use are obtained by: \( 1 - P(0; 15, 0.05) \), \( 1 - P(0; 15, 0.10) \) and \( 1 - P(0; 15, 0.20) \), respectively. The exact probabilities would depend on the exact value of the calculated probabilities.

Step by step solution

01

Calculate the probability for 5% drug use

For the case when 5% of players are using drugs, the probability equation is \( P(0; 15, 0.05) = C(15, 0) * (0.05)^0 * (0.95)^{15} \)\nThen, to find the complementary probability where one or more players test positive, it is \( 1 - P(0; 15, 0.05) \)
02

Calculate the probability for 10% drug use

Same procedure is applied when 10% of players are using drugs: \( P(0; 15, 0.10) = C(15, 0) * (0.10)^0 * (0.90)^{15} \)\nAnd then the complementary probability is \( 1 - P(0; 15, 0.10) \)
03

Calculate the probability for 20% drug use

For 20% of players using drugs: \( P(0; 15, 0.20) = C(15, 0) * (0.20)^0 * (0.80)^{15} \)\nThe complementary probability is \( 1 - P(0; 15, 0.20) \)

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Key Concepts

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

Complementary Probability
Complementary probability is a fundamental concept in probability theory that is used to simplify the calculation of the likelihood of an event not happening. For instance, if you want to find out the probability that at least one event occurs, it is often easier to calculate the probability that no such events occur and then subtract this value from one. This principle is based on the fact that the sum of the probabilities of all possible outcomes of a trial must equal one.

In the context of the baseball exercise provided, the calculation of the complementary probability helps us determine the chances that one or more players out of 15 will test positive for illegal drugs. To calculate this, we first determine the probability that none of the 15 players test positive, which is given by the binomial distribution formula for zero positive tests. Then we subtract this probability from one to find the probability of at least one positive test. That is, the complementary probability is computed as \(1 - P(0; 15, p)\), where \(p\) represents the prevalence of drug use among the players (5%, 10%, or 20%).
Binomial Distribution
The binomial distribution is used when there are two possible outcomes for multiple trials, often represented as 'success' and 'failure.' In statistics, this type of distribution is key to understanding the probability of a specific number of successes in a given number of trials, each with the same probability of success.

In the exercise given, determining how likely it is for a certain number of players to test positive for drugs out of a randomly selected sample of 15 from a larger population is a perfect example of a binomial distribution. The probability of finding exactly 'k' players testing positive is calculated using the formula:
\[ P(k; n, p) = C(n, k) * p^k * (1 - p)^{n-k} \]
where
  • \(C(n, k)\) is the combination of 'n' items taken 'k' at a time,
  • \(p\) is the probability of a single trial resulting in a success (in this case, a player testing positive for drugs), and
  • \(n\) is the number of trials, or sample size.

Understanding this concept is critical in calculating the likelihood of various outcomes based on the binomial distribution.
Statistical Significance
Statistical significance is a measure of whether the results of a study or experiment can reliably indicate a real effect or relationship, rather than being due to random chance. In most research fields, a result is considered statistically significant if the probability of the observed outcome occurring by chance is below a pre-determined threshold, usually set at a 5% level (p-value < 0.05).

When we apply this concept to the baseball players' problem, we're not just interested in the calculated probabilities of one or more players testing positive. We're also concerned with how convincing these probabilities are in indicating a true pattern of drug usage among the population of players. For example, if a high probability is calculated even with a low prevalence rate, it would suggest a significant issue. Conversely, a low calculated probability corresponding to a high prevalence would imply an unlikely event, potentially indicating inaccurate testing or a skewed sample. Statistical significance in such a scenario is vital for making informed decisions based on the data.

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Most popular questions from this chapter

Show that each of the following is true for any values of \(n\) and \(k\). Use two specific sets of values for \(n\) and \(k\) to show that each is true.. $$\text { a. } \quad\left(\begin{array}{l}n \\\0\end{array}\right)=1 \text { and }\left(\begin{array}{l}n \\\n\end{array}\right)=1$$ $$\text { b. }\left(\begin{array}{l}n \\\1\end{array}\right)=n \text { and }\left(\begin{array}{c}n \\\n-1 \end{array}\right)=n \quad \text { c. }\left(\begin{array}{c}n \\\k\end{array}\right)=\left(\begin{array}{c}n \\\n-k\end{array}\right)$$

In a germination trial, 50 seeds were planted in each of 40 rows. The number of seeds germinating in each row was recorded as listed in the following table. $$\begin{array}{cc|cc}\begin{array}{c}\text { Number } \\\\\text { Germinated }\end{array} & \begin{array}{c}\text { Number } \\\\\text { of Rows }\end{array} & \begin{array}{c}\text { Number } \\\\\text { Germinated }\end{array} & \begin{array}{c}\text { Number } \\\\\text { of Rows }\end{array} \\\\\hline 39 & 1 & 45 & 8 \\\40 & 2 & 46 & 4 \\\41 & 3 & 47 & 3 \\\42 & 4 & 48 & 1 \\\43 & 6 & 49 & 1 \\\44 & 7 & & \\\\\hline\end{array}$$ a. Use the preceding frequency distribution table to determine the observed rate of germination for these seeds. b. The binomial probability experiment with its corresponding probability distribution can be used with the variable "number of seeds germinating per row" when 50 seeds are planted in every row. Identify the specific binomial function and list its distribution using the germination rate found in part a. Justify your answer. c. Suppose you are planning to repeat this experiment by planting 40 rows of these seeds, with 50 seeds in each row. Use your probability model from part b to find the frequency distribution for \(x\) that you would expect to result from your planned experiment. d. Compare your answer in part c with the results that were given in the preceding table. Describe any similarities and differences.

Of all the trees planted by a landscaping firm, \(90 \%\) survive. What is the probability that 8 or more of the 10 trees they just planted will survive? (Find the answer by using a table.)

Verify whether or not each of the following is a probability function. State your conclusion and explain. a. \(f(x)=\frac{\frac{3}{4}}{x !(3-x) !},\) for \(x=0,1,2,3\) b. \(f(x)=0.25,\) for \(x=9,10,11,12\) c. \(f(x)=(3-x) / 2,\) for \(x=1,2,3,4\) d. \(f(x)=\left(x^{2}+x+1\right) / 25,\) for \(x=0,1,2,3\)

In a USA Today Snapshot (June 1,2009), the following statistics were reported on the number of hours of sleep that adults get. $$\begin{aligned}&\begin{array}{cc}\text { Number of Hours } & \text { Percentage } \\\\\hline 5 \text { or less } & 12 \% \\\6 & 29 \% \\\7 & 37 \% \\\8 \text { or more } & 24 \% \\\\\hline\end{array}\\\&\begin{array}{l} \text { Source: Strategy One survey for Tempur-Pedic of } \\\1004 \text { adults in April }\end{array} \end{aligned}$$ a. Are there any other values that the number of hours can attain? b. Explain why the total of the percentages is not \(100 \%\) c. Is this a discrete probability distribution? Is it a probability distribution? Explain.
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