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Chapter 6: Problem 46

Four of the 20 students \((20 \%)\) in a class are fraternity or sorority members. Five students are picked at random. Does \(X=\) the number of students in the sample who are fraternity or sorority members have the binomial distribution with \(n=5\) and \(p=0.20 ?\) Explain why or why not.

Short Answer

Expert verified
The distribution is not binomial due to lack of independence and non-constant probability in selections.

Step by step solution

01

Understanding Binomial Distribution

A random variable follows a binomial distribution if it represents the number of successes in a fixed number of independent Bernoulli trials. For a distribution to be binomial, each trial must only have two possible outcomes: 'success' or 'failure', trials must be independent, and the probability of success must remain constant across trials.
02

Defining Success and Failure

In this context, a 'success' is selecting a fraternity or sorority member, and a 'failure' is selecting a non-member. We want to examine if all the criteria for a binomial distribution apply to this scenario.
03

Identifying Number of Trials and Probability of Success

The number of trials, \(n\), is given as 5 since we are picking five students. The probability \(p\) of picking a fraternity or sorority member in a single draw should be consistent for a binomial distribution, here stated as 0.20.
04

Checking Independence of Trials

For the distribution to be binomial, each draw must be independent. However, drawing without replacement from a fixed group of students makes each trial dependent on the outcome of the previous ones, because the composition of the group changes as members are selected.
05

Assessing Constant Probability

The probability of picking a fraternity or sorority member changes after each draw, because each selection impacts the remaining population. Hence, the probability \(p=0.20\) is not constant throughout the selected trials without replacement.
06

Conclusion

Since the trials are not independent and the probability does not remain constant, the assumptions for the binomial distribution are not met. Therefore, the number of fraternity or sorority members in the sample does not follow a binomial distribution with \(n=5\) and \(p=0.20\).

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Key Concepts

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

Bernoulli Trials
To understand binomial distribution, it's crucial to first grasp the concept of Bernoulli trials. A Bernoulli trial is a random experiment where there are only two possible outcomes: "success" and "failure." Each trial is an independent event, following the same probability of success. For binomial distribution, you're counting the number of successes across several Bernoulli trials.
In our example, selecting a fraternity or sorority member is considered a success, while picking a non-member is a failure. Knowing this distinction helps set up the trial framework needed for the binomial model. However, certain rules must be maintained for the trials to be valid, which we'll explore next.
Independence Assumption
A key assumption for binomial distribution is that each Bernoulli trial must be independent. This means the result of one trial does not influence the result of another. For example, flipping a coin multiple times, each flip is unaffected by the previous one.
In our context of selecting students, independence is hindered because the selection is without replacement. Once a student is picked, it changes the composition of the remaining group, thus influencing subsequent selections. As each draw affects the next, this deviates from the independence required in binomial distribution.
Probability of Success
Another critical component is maintaining a constant probability of success across trials. In the framework of Bernoulli trials, the probability of success should remain the same for each attempted trial.
In our problem, the probability \( p = 0.20 \) was considered for the first draw. However, this probability does not stay constant since we are sampling without replacement. Each successful selection alters the remaining pool of students, thus affecting the likelihood of drawing another success. This variability challenges the assumptions of a binomial distribution where probabilities should be stable across trials.
Sampling Without Replacement
Sampling without replacement fundamentally alters how probabilities and independent events behave in a sample. Unlike replacing each chosen individual or item and maintaining the sample size, without replacement means each choice reduces the pool, thereby affecting subsequent events.
In our scenario, picking one fraternity/sorority member directly changes the probability of picking another in the very next trial. With a finite group of 20 students and only 4 members being of interest, ignoring replacement makes understanding and establishing a constant probability and independence for each student a challenge. This directly contradicts the needed criteria for establishing a traditional binomial distribution for the sample.

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Most popular questions from this chapter

From past experience, a wheat farmer living in Manitoba, Canada finds that his annual profit (in Canadian dollars) is \(\$ 80,000\) if the summer weather is typical, \(\$ 50,000\) if the weather is unusually dry, and \(\$ 20,000\) if there is a severe storm that destroys much of his crop. Weather bureau records indicate that the probability is 0.70 of typical weather, 0.20 of unusually dry weather, and 0.10 of a severe storm. In the next year, let \(X\) be the farmer's profit. a. Construct a table with the probability distribution of \(X\). b. What is the probability that the profit is \(\$ 50,000\) or less? c. Find the mean of the probability distribution of \(X\). Interpret. d. Suppose the farmer buys insurance for \(\$ 3000\) that pays him \(\$ 20,000\) in the event of a severe storm that destroys much of the crop and pays nothing otherwise. Find the probability distribution of his profit.

A World Health Organization study (the MONICA project) of health in various countries reported that in Canada, systolic blood pressure readings have a mean of 121 and a standard deviation of \(16 .\) A reading above 140 is considered to be high blood pressure. a. What is the \(z\) -score for a blood pressure reading of \(140 ?\) b. If systolic blood pressure in Canada has a normal distribution, what proportion of Canadians suffers from high blood pressure? c. What proportion of Canadians has systolic blood pressures in the range from 100 to \(140 ?\) d. Find the 90 th percentile of blood pressure readings.

SAT math scores follow a normal distribution with an approximate \(\mu=500\) and \(\sigma=100\). Also ACT math scores follow a normal distrubution with an approximate \(\mu=21\) and \(\sigma=4.7\). You are an admissions officer at a university and have room to admit one more student for the upcoming year. Joe scored 600 on the SAT math exam, and Kate scored 25 on the ACT math exam. If you were going to base your decision solely on their performances on the exams, which student should you admit? Explain.

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 the following random variables, explain why at least one condition needed to use the binomial distribution is unlikely to be satisfied. a. \(X=\) number of people in a family of size 4 who go to church on a given Sunday, when any one of them goes \(50 \%\) of the time in the long run (binomial, \(n=4, p=0.50\) ). (Hint: Is the independence assumption plausible?) b. \(X=\) number voting for the Democratic candidate out of the 100 votes in the first precinct that reports results, when \(60 \%\) of the population voted for the Democrat in that state (binomial, \(n=100, p=0.60\) ). (Hint: Is the probability of voting for the Democratic candidate the same in each precinct?) c. \(X=\) number of females in a random sample of four students from a class of size \(20,\) when half the class is female (binomial, \(n=4, p=0.50\) ).
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