find the mean and standard deviation for the number of correct answers for such student.
The midterm exam in a nursing course consists of 75 t/f questions. Assume that an unprepared student makes random guesses for each of the answers.
1 Answer | Add Yours
An underprepared student will mark true or false randomly, therefore he has only two choices. Assuming he answers every question either true and false, this represents a binomial distribution of 75 instances that is n, and the probability of 0.5 of getting correct answer.
Therefore, p = 0.5 and n=75.
The expected value or mean of a binomial distribution is given by,
`E[X] = np`
`E[X] = 75 xx 0.5`
`E[X] = 37.5`
And standard deviation is given by,
`sigma[X] = sqrt(np(1-p))`
`sigma[X] = sqrt(75 xx 0.5 xx 0.5)`
`sigma[X] = 4.33`
Therefore mean is 37.5 and standard deviation is 4.33.
We’ve answered 319,189 questions. We can answer yours, too.Ask a question