Question

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 (randomly select one of the five answers) 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. Calculate 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

a. The probability distribution of \(x\) is a binomial distribution. b. The expected score on the exam is 20. c. The variance of \(x\) is 16 and the standard deviation is 4. d. It is unlikely to score over 50 in the exam based on the calculated mean and standard deviation.

Step-by-step solution

Identify the distribution

This is a binomial distribution problem because we have a fixed number of trials (100 questions), with each one having two outcomes: success (correct answer) and failure (wrong answer). The trials are independent of each other since the outcome on one question doesn’t affect the outcome on others.

Compute the expected value

We calculate the expected value (\(E\)) or the mean score using the formula \(E = np\), where \(n\) represents the number of trials (questions), and \(p\) is the probability of success (randomly guessing the correct answer). Since there are 5 choices and only 1 is correct, \(p = 1/5 = 0.2\). So, \(E = 100 * 0.2 = 20\). Therefore, the expected score is 20.

Calculate the variance and standard deviation

Using the formula for binomial variance, \(Var = np(1-p)\), we calculate \(Var = 100 * 0.2 * 0.8 = 16\). And the standard deviation (\(\sigma\)) is the square root of the variance: \(\sigma = \sqrt{16} = 4\).

Determine the likelihood of scoring over 50

With a mean of 20 and a standard deviation of 4, scoring over 50 (which is more than 7 standard deviations away from the mean) is extremely unlikely in a normal distribution. Additionally, given the binomial distribution, the highest probabilities are around the mean, so scores far from the mean become drastically less likely.