Question

question 13
several economics students are unprepared for a surprise true/false test with 20 questions, and all of their answers are guesses.
find the mean for the number of correct answers for such students. (round answer to one decimal place.)
$mu =$
find the standard deviation for the number of correct answers for such students. (round answer to two decimal places.)
$sigma =$
give the range for the usual number of correct answers. (enter answer as an interval using square - brackets only with whole numbers.)
usual values =
would it be unusual for a student to pass by guessing and getting at least 12 correct answers?
no, it would not be unusual
yes, it would be unusual

Explanation:

Step1: Identify the distribution parameters

This is a binomial - distribution problem. For a binomial distribution \(X\sim B(n,p)\), where \(n\) is the number of trials and \(p\) is the probability of success on a single trial. Here, \(n = 20\) (number of questions) and \(p=\frac{1}{2}=0.5\) (probability of getting a true - false question correct by guessing).

Step2: Calculate the mean

The mean of a binomial distribution is given by the formula \(\mu=np\). Substitute \(n = 20\) and \(p = 0.5\) into the formula: \(\mu=20\times0.5 = 10.0\).

Step3: Calculate the standard deviation

The standard deviation of a binomial distribution is \(\sigma=\sqrt{np(1 - p)}\). Substitute \(n = 20\) and \(p = 0.5\) into the formula: \(\sigma=\sqrt{20\times0.5\times(1 - 0.5)}=\sqrt{20\times0.5\times0.5}=\sqrt{5}\approx2.24\).

Step4: Find the range for the usual number of correct answers

Usual values are within \(2\) standard deviations of the mean. The lower limit is \(\mu - 2\sigma=10-2\times2.24 = 10 - 4.48 = 5.52\approx6\) (rounded up to the next whole number), and the upper limit is \(\mu + 2\sigma=10 + 2\times2.24=10 + 4.48 = 14.48\approx14\) (rounded down to the next whole number). So the interval is \([6,14]\).

Step5: Determine if getting at least 12 correct answers is unusual

Since \(12\) is within the interval \([6,14]\), it is not unusual.

Answer:

\(\mu = 10.0\)
\(\sigma\approx2.24\)
usual values = \([6,14]\)
no, it would not be unusual