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.
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.