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Chapter 7: Problem 55

You are to take a multiple-choice exam consisting of 100 questions with five possible responses to each question. Suppose that you have not studied and so must guess (select one of the five answers in a completely random fashion) on each question. Let \(x\) represent the number of correct responses on the test. a. What kind of probability distribution does \(x\) have? b. What is your expected score on the exam? (Hint: Your expected score is the mean value of the \(x\) distribution.) c. Compute the variance and standard deviation of \(x\). d. Based on your answers to Parts (b) and (c), is it likely that you would score over 50 on this exam? Explain the reasoning behind your answer.

Short Answer

Expert verified
a. The variable x follows a Binomial Distribution. b. The expected score on the exam is 20. c. The variance and standard deviation of x are 16 and 4, respectively. d. Based on the calculated mean and standard deviation, it appears to be statistically unlikely to score more than 50 on the exam by random guessing.

Step by step solution

01

Identify the Distribution

The situation described in the exercise is an example of a Binomial Distribution because there are 100 (n) independent trials (questions) with only two possible outcomes - correct (success) or incorrect (failure), determined by random guessing. This results in a Binom(100, 1/5) distribution for x.
02

Calculate Expected Score

The expected score or mean (碌) value for a binomial distribution is computed using the formula: 碌 = n * p. In this case, n is the number of trials (100 questions) and p is the probability of success (guessing the correct answer), which is 1/5. Therefore, 碌 = 100 * (1/5) = 20. The expected score on the exam is therefore 20.
03

Compute Variance and Standard Deviation

The variance (蟽虏) of a binomial distribution is computed using the formula: 蟽虏 = n * p * (1 - p). Here, n = 100, p = 1/5, so 蟽虏 = 100 * (1/5) * (1 - 1/5) = 16. The standard deviation (蟽) is simply the square root of the variance, which is 鈭�16 or 4.
04

Evaluate Probabilistic Outcome

Considering the expected score (20) and standard deviation (4), it seems unlikely to score over 50 by random selection, due to the nature of binomial distribution. Moreover, 50 is significantly more than one standard deviation away from the mean. But absolute certainty can't be provided without the precise computation of probability.

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Key Concepts

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

Probability Distribution
A probability distribution describes how the values of a random variable are distributed. In a more simplistic term, it tells us how likely each outcome is to occur. In the exercise provided, we deal with a **Binomial Distribution**. This is a specific type of probability distribution often used when there are two possible outcomes. For instance, every question on the exam has five choices, but only one is correct; hence it is treated as a success (correct) or failure (incorrect). When you guess on each question, there are 100 independent trials and therefore, this can be defined as a Binomial Distribution with 100 trials, where each trial has a success probability of 1/5. The probability distribution allows us to compute important metrics like the expected value and standard deviation, helping us understand more about what to expect from this random guessing scenario.
Expected Value
The **Expected Value** is a concept representing the average or mean of all possible outcomes. In probability, it gives us an idea of what the average result would be if we were to repeat the random process infinite times. In a binomial distribution, the expected value is calculated with the formula:
  • \( \mu = n \times p \)
where \( n \) is the number of trials (here, 100 questions) and \( p \) is the probability of success in each trial (guessing a question right, which is 1/5). Simplifying it gives:
  • \( \mu = 100 \times (1/5) = 20 \)
This means, if you guess randomly throughout the exam, on average, you should expect to get 20 questions correct. It's a way to prepare what you might reasonably achieve if you were to repeat the process many times.
Variance and Standard Deviation
**Variance** and **Standard Deviation** measure the spread of the probability distribution. Variance tells us how much the possible outcomes deviate from the expected value, and standard deviation is the square root of the variance. For a binomial distribution, the variance (\( \sigma^2 \)) is calculated using:
  • \( \sigma^2 = n \times p \times (1 - p) \)
Given the problem illustrates a guess for each of the 100 questions with a probability of success of 1/5, we have:
  • \( \sigma^2 = 100 \times (1/5) \times (4/5) = 16 \)
The standard deviation is simply:
  • \( \sigma = \sqrt{16} = 4 \)
This suggests that most of the time, your score will deviate by about 4 questions from the expected score of 20. Therefore, it's quite unlikely that you'd score much higher, for example, over 50, as it's significantly beyond just one standard deviation from 20.

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Most popular questions from this chapter

The states of Ohio, Iowa, and Idaho are often confused, probably because the names sound so similar. Each year, the State Tourism Directors of these three states drive to a meeting in one of the state capitals to discuss strategies for attracting tourists to their states so that the states will become better known. The location of the meeting is selected at random from the three state capitals. The shortest highway distance from Boise, Idaho to Columbus, Ohio passes through Des Moines, Iowa. The highway distance from Boise to Des Moines is 1350 miles, and the distance from Des Moines to Columbus is 650 miles. Let \(d_{1}\) represent the driving distance from Columbus to the meeting, with \(d_{2}\) and \(d_{3}\) representing the distances from Des Moines and Boise, respectively. a. Find the probability distribution of \(d_{1}\) and display it in a table. b. What is the expected value of \(d_{1} ?\) c. What is the value of the standard deviation of \(d_{1}\) ? d. Consider the probability distributions of \(d_{2}\) and \(d_{3}\). Is either probability distribution the same as the probability distribution of \(d_{1}\) ? Justify your answer. e. Define a new random variable \(t=d_{1}+d_{2}\). Find the probability distribution of \(t\). f. For each of the following statements, indicate if the statement is true or false and provide statistical evidence to support your answer. i. \(E(t)=E\left(d_{1}\right)+E\left(d_{2}\right)\) (Hint: \(E(t)\) is the expected value of \(t\) and another way of denoting the mean of \(t\).) ii. \(\sigma_{t}^{2}=\sigma_{d_{1}}^{2}+\sigma_{d_{3}}^{2}\)

Consider the following sample of 25 observations on the diameter \(x\) (in centimeters) of a disk used in a certain system: \(\begin{array}{lllllll}16.01 & 16.08 & 16.13 & 15.94 & 16.05 & 16.27 & 15.89 \\\ 15.84 & 15.95 & 16.10 & 15.92 & 16.04 & 15.82 & 16.15 \\ 16.06 & 15.66 & 15.78 & 15.99 & 16.29 & 16.15 & 16.19 \\ 16.22 & 16.07 & 16.13 & 16.11 & & & \end{array}\) The 13 largest normal scores for a sample of size 25 are \(1.965,1.524,1.263,1.067,0.905,0.764,0.637,0.519\) \(0.409,0.303,0.200,0.100\), and \(0 .\) The 12 smallest scores result from placing a negative sign in front of each of the given nonzero scores. Construct a normal probability plot. Does it appear plausible that disk diameter is normally distributed? Explain.

Determine the value \(z^{*}\) that a. Separates the largest \(3 \%\) of all \(z\) values from the others b. Separates the largest \(1 \%\) of all \(z\) values from the others c. Separates the smallest \(4 \%\) of all \(z\) values from the others d. Separates the smallest \(10 \%\) of all \(z\) values from the others

Symptom validity tests (SVTs) are 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 MENT 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 \(.7\) (compared to \(.96\) for other patients). The authors used a normal approximation to the binomial distribution with \(n=60\) and \(p=.70\) to compute 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. \(\quad P(x<42)\) iii. \(P(x \leq 42)\) c. Explain why the probabilities computed in Part (b) are not all equal. d. The authors computed the exact binomial probability of a score of 42 or less for someone who is not faking the test. Using \(p=.96\), they found \(p(x \leq 42)=.000000000013\) Explain why the authors computed 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.

A grocery store has an express line for customers purchasing at most five items. Let \(x\) be the number of items purchased by a randomly selected customer using this line. Give examples of two different assignments of probabilities such that the resulting distributions have the same mean but quite different standard deviations.
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