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Chapter 5: Problem 7

Approximately \(10 \%\) of all people are left-handed ("11 little-known facts," 2013). Consider a grouping of fifteen people. a. State the random variable. b. Argue that this is a binomial experiment Find the probability that c. None are left-handed. d. Seven are left-handed. e. At least two are left-handed. f. At most three are left-handed. g. At least seven are left-handed. h. Seven of the last 15 U.S. Presidents were left-handed. Is this unusual? What does that tell you?

Short Answer

Expert verified
X: number of left-handed in 15 people. This is a binomial experiment. P(X=0) 鈮� 0.2059, P(X=7) 鈮� 0.000003, P(X鈮�2) 鈮� 0.4502, P(X鈮�3) 鈮� 0.6206, P(X鈮�7) 鈮� 0.001. Having 7 left-handed U.S. Presidents is highly unusual.

Step by step solution

01

Define the Random Variable

The random variable in this scenario is defined as the number of left-handed people in a grouping of 15 people. Let this random variable be denoted by \( X \).
02

Determine if it is a Binomial Experiment

To argue that this is a binomial experiment, consider the following characteristics: (1) Each trial (selecting a person) has only two outcomes - left-handed or not left-handed. (2) The probability of being left-handed (\( p \)) is constant at 0.10. (3) We are concerned with a fixed number of trials, n = 15 people. (4) Each selection is independent. Therefore, this is a binomial experiment with parameters \( n = 15 \) and \( p = 0.10 \).
03

Calculate the Probability that None are Left-Handed

Use the binomial probability formula: \( P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \)For \( x = 0 \), \( P(X = 0) = \binom{15}{0} (0.10)^0 (0.90)^{15} = 1 \times 1 \times (0.90)^{15} \approx 0.2059 \)
04

Calculate the Probability that Seven are Left-Handed

Substitute \( x = 7 \) into the binomial formula\( P(X = 7) = \binom{15}{7} (0.10)^7 (0.90)^8 \)\( \binom{15}{7} = 6435 \)\( P(X = 7) = 6435 \times (0.10)^7 \times (0.90)^8 \approx 0.000003 \)
05

Calculate the Probability that At Least Two are Left-Handed

Use the complement rule: \( P(X \geq 2) = 1 - P(X < 2) \)\( P(X < 2) = P(X = 0) + P(X = 1) \)\( P(X = 1) = \binom{15}{1} (0.10)^1 (0.90)^{14} \approx 0.3439 \)\( P(X < 2) = 0.2059 + 0.3439 = 0.5498 \)\( P(X \geq 2) = 1 - 0.5498 = 0.4502 \)
06

Calculate the Probability that At Most Three are Left-Handed

Use the sum of binomial probabilities: \( P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \)For \( x = 2 \),\( P(X = 2) = \binom{15}{2} (0.10)^2 (0.90)^{13} \approx 0.0116 \)For \( x = 3 \),\( P(X = 3) = \binom{15}{3} (0.10)^3 (0.90)^{12} \approx 0.0592 \)\( P(X \leq 3) = 0.2059 + 0.3439 + 0.0116 + 0.0592 \approx 0.6206 \)
07

Calculate the Probability that At Least Seven are Left-Handed

Use the complement rule:\( P(X \geq 7) = 1 - P(X < 7) \)\( P(X < 7) = P(X \leq 6) \)Calculate cumulative probabilities for \(x = 0\) through \(x = 6\); given earlier results, sum the remaining probabilities:\( P(X \geq 7) \approx 1 - \sum_{x=0}^{6} P(X=x) \approx 1 - 0.999 \approx 0.001 \)
08

Determine if Seven Left-Handed U.S. Presidents is Unusual

Since \(P(X=7) \approx 0.000003\) for a population of 15, having seven out of 15 U.S. Presidents being left-handed is highly unusual and suggests that some other factors may influence this occurrence.

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Key Concepts

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

binomial experiment
A binomial experiment is a type of statistical experiment that involves a fixed number of trials. Each trial has only two possible outcomes: success or failure. In our example, the success is identifying a left-handed person, and the failure is identifying a right-handed person. The probability of success remains constant across trials. Here, the probability of a person being left-handed is 0.10 (or 10%). Another key point is that each trial is independent of the others, meaning that the outcome of one trial does not affect the outcome of another.
random variable
A random variable is a numerical outcome that results from a random process. In our scenario, the random variable is denoted by \( X \) and represents the number of left-handed people in a group of 15. The value of the random variable can be any number from 0 to 15. For example, if our group has 3 left-handed people, then the random variable \( X \) would be 3. The random variable helps us to set up the problem mathematically and apply binomial probability formulas to find various probabilities such as none being left-handed or any other specific number.
complement rule
The complement rule is an important concept in probability that helps us find the probability of the complement of an event. If we know the probability of an event happening, the probability of the event not happening (its complement) is simply 1 minus the probability of the event. For example, if the probability of having at least two left-handed people in a group of 15 is denoted as \( P(X \geq 2) \), we use the complement rule to calculate it as \( 1 - P(X < 2) \). This is useful because sometimes calculating the probability of a complement event is easier than the original event.
binomial probability formula
The binomial probability formula is used to find the probability of exactly \( x \) successes in \( n \) trials, given the probability of success \( p \). The formula is expressed as: \[ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \] Here, \( \binom{n}{x} \) is the binomial coefficient, calculating the number of ways to choose \( x \) successes from \( n \) trials. For example, to find the probability that exactly 7 out of 15 people are left-handed (i.e., \( X = 7 \)), we substitute the values into the formula: \[ P(X = 7) = \binom{15}{7} (0.10)^7 (0.90)^{8} \] This formula is the foundation of solving most binomial probability problems.
fixed number of trials
A fixed number of trials is a fundamental component of a binomial experiment. This means the experiment is conducted a specific number of times, and each trial is independent and identical in terms of the probability of success. In our example, we consider a group of 15 people, which implies we have 15 trials. Each trial assesses whether a person is left-handed (success) or not (failure). This fixed number is crucial to apply the binomial distribution since the formulas and probabilities rely on this consistency across trials. Without a fixed number of trials, the problem would not meet the criteria for a binomial experiment.

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Most popular questions from this chapter

According to an article in the American Heart Association's publication Circulation, \(24 \%\) of patients who had been hospitalized for an acute myocardial infarction did not fill their cardiac medication by the seventh day of being discharged (Ho, Bryson \& Rumsfeld, 2009). Suppose there are twelve people who have been hospitalized for an acute myocardial infarction. a. State the random variable. b. Write the probability distribution. c. Draw a histogram. d. Describe the shape of the histogram. e. Find the mean. f. Find the variance. g. Find the standard deviation.

The proportion of brown M\&M's in a milk chocolate packet is approximately \(14 \%\) (Madison, 2013). Suppose a package of M\&M's typically contains 52 M\&M's. a. State the random variable. b. Argue that this is a binomial experiment Find the probability that c. Six M\&M's are brown. d. Twenty-five M\&M's are brown. e. All of the M\&M's are brown. f. Would it be unusual for a package to have only brown M\&M's? If this were to happen, what would you think is the reason?

Suppose a random variable, \(x\), arises from a binomial experiment. Suppose \(n=6\), and \(p=0.13\). a. Write the probability distribution. b. Draw a histogram. c. Describe the shape of the histogram. d. Find the mean. e. Find the variance. f. Find the standard deviation.

According to an article in the American Heart Association's publication Circulation, \(24 \%\) of patients who had been hospitalized for an acute myocardial infarction did not fill their cardiac medication by the seventh day of being discharged (Ho, Bryson \& Rumsfeld, 2009). Suppose there are twelve people who have been hospitalized for an acute myocardial infarction. a. State the random variable. b. Argue that this is a binomial experiment Find the probability that c. All filled their cardiac medication. d. Seven did not fill their cardiac medication. e. None filled their cardiac medication. f. At most two did not fill their cardiac medication. g. At least three did not fill their cardiac medication. h. At least ten did not fill their cardiac medication. i. Suppose of the next twelve patients discharged, ten did not fill their cardiac medication, would this be unusual? What does this tell you?

Eyeglassomatic manufactures eyeglasses for different retailers. In March 2010 , they tested to see how many defective lenses they made, and there were \(16.9 \%\) defective lenses due to scratches. Suppose Eyeglassomatic examined twenty eyeglasses. a. State the random variable. b. Argue that this is a binomial experiment Find the probability that c. None are scratched. d. All are scratched. e. At least three are scratched. f. At most five are scratched. g. At least ten are scratched. h. Is it unusual for ten lenses to be scratched? If it turns out that ten lenses out of twenty are scratched, what might that tell you about the manufacturing process?
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