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The longest "run" of \(S\) 's in the 10 -trial sequence SSFSSSSFFS has length 4 , corresponding to the \(S\) 's on the fourth, fifth, sixth, and seventh trials. Consider a binomial experiment with \(n=4\), and let \(y\) be the length (number of trials) in the longest run of \(S\) 's. a. When \(p=0.5,\) the 16 possible outcomes are equally likely. Determine the probability distribution of \(y\) in this case (first list all outcomes and the \(y\) value for each one). Then calculate \(\mu_{y}\) b. Repeat Part (a) for the case \(p=0.6\).

Short Answer

Expert verified
The probability distribution of \(y\) and the mean \(\mu_{y}\) will depend on the given probability \(p\). For \(p = 0.5\), all outcomes are equally likely, but for \(p = 0.6\), sequences with more \(S\)'s have higher probability. Detailed calculations are needed to find exact values.

Step by step solution

01

List All Outcomes

First, list all outcomes of length 4 that can be made from the sequence S and F. There are 16 possible outcomes, because each trial has 2 outcomes and there are 4 trials, yielding \(2^4 = 16\) possible outcomes.
02

Calculate y value for Each Outcome and Create Probability Distribution

Determine the \(y\) value for each outcome, which is the length of the longest run of \(S\)'s. Then, count the frequency of each \(y\) value among all outcomes and divide by total number of outcomes to form a probability distribution.
03

Compute \(\mu_{y}\)

Compute \(\mu_{y}\) by multiplying each \(y\) value by its probability and summing up all these products.
04

Repeat Above Steps for \(p = 0.6\)

Repeat steps 1 to 3 for case \(p = 0.6\). In this case, sequences with more \(S\)'s are more likely to occur, thus the probability distribution of \(y\) and the mean \(\mu_{y}\) will change.

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Most popular questions from this chapter

A symptom validity test (SVT) is sometimes used to confirm diagnosis of psychiatric disorders. The paper "Developing a Symptom Validity Test for Posttraumatic Stress Disorder: Application of the Binomial Distribution" ( Journal of Anxiety Disorders [2008]: 1297-1302) investigated the use of SVTs in the diagnosis of post-traumatic stress disorder. One SVT proposed is a 60 -item test (called the MENT test), where each item has only a correct or incorrect response. The MENT test is designed so that responses to the individual questions can be considered independent of one another. For this reason, the authors of the paper believe that the score on the MENT test can be viewed as a binomial random variable with \(n=60 .\) The MENT test is designed to help in distinguishing fictitious claims of post-traumatic stress disorder. The items on the test are written so that the correct response to an item should be relatively obvious, even to people suffering from stress disorders. Researchers have found that a patient with a fictitious claim of stress disorder will try to "fake" the test, and that the probability of a correct response to an item for these patients is 0.7 (compared to 0.96 for other patients). The authors used a normal approximation to the binomial distribution with \(n=60\) and \(p=0.70\) to calculate various probabilities of interest, where \(x=\) number of correct responses on the MENT test for a patient who is trying to fake the test. a. Verify that it is appropriate to use a normal approximation to the binomial distribution in this situation. b. Approximate the following probabilities: i. \(\quad P(x=42)\) ii. \(P(x<42)\) iii. \(P(x \leq 42)\) c. Explain why the probabilities calculated in Part (b) are not all equal. d. The authors calculated the exact binomial probability of a score of 42 or less for someone who is not faking the test. Using \(p=0.96\), they found $$ P(x \leq 42)=.000000000013 $$ Explain why the authors calculated this probability using the binomial formula rather than using a normal approximation. e. The authors propose that someone who scores 42 or less on the MENT exam is faking the test. Explain why this is reasonable, using some of the probabilities from Parts (b) and (d) as justification.

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