Several psychology students are unprepared for a surprise true/false test with 18 questions, and all of their answers are guesses.
a. Find the mean and standard deviation for the number of correct answers for such students.
b. Would it be unusual for a student to pass by guessing (which requires getting at least 15 correct answers)? Why or why not?
a. mu equals
(Do not round.)
sigma equals
(Round to one decimal place as needed.)
b. Choose the correct answer below.
A.
Yes comma because 15 is greater than the maximum usual value.
B.
Yes, because 15 is within the range of usual values.
C.
No comma because 15 is within the range of usual values.
D.
Yes comma because 15 is below the minimum usual value.
To find the mean and standard deviation for the number of correct answers, we can use the binomial distribution formula. In this case, the probability of guessing a question correctly is 1/2 (since it is a true/false test), and there are 18 questions.
a. Mean (mu) = n * p
Mean (mu) = 18 * 1/2
Mean (mu) = 9
b. Standard Deviation (sigma) = sqrt(n * p * (1-p))
Standard Deviation (sigma) = sqrt(18 * 1/2 * (1 - 1/2))
Standard Deviation (sigma) = sqrt(18 * 1/2 * 1/2)
Standard Deviation (sigma) = sqrt(9)
Standard Deviation (sigma) = 3
So, the mean (mu) is 9 and the standard deviation (sigma) is 3.
To determine if it would be unusual for a student to pass by guessing (getting at least 15 correct answers), we can use z-scores to find the probability.
z-score = (x - mu) / sigma
z-score = (15 - 9) / 3
z-score = 2
Generally, z-scores below -2 or above 2 are considered unusual.
Since the z-score is 2, it indicates that getting at least 15 correct answers is two standard deviations above the mean. Thus, it would be unusual for a student to pass by guessing (getting at least 15 correct answers).
Therefore, the correct answer to part b is:
A. Yes, because 15 is greater than the maximum usual value.