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...

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.

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thilina-g | College Teacher | (Level 1) Educator

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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.

Sources:

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