The health center of a small, private university requires students to have appointments for visits. Each appointment takes an average of minutes with a standard deviation of minutes. The health center makes appointments each day and is scheduled to be open for hours. What is the probability that the total time spent for students who have appointments for tomorrow will exceed hours? (Hint: hours minutes.) Carry your intermediate computations to at least four decimal places. Report your result to at least three decimal places.

Some data in the question are missing. The complete question is :

The health center of a small, private university requires students to have appointments for visits. Each appointment takes an average of 5.6 minutes with a standard deviation of 3.5 minutes. The health center makes 100 appointments each day and is scheduled to be open for 10 hours. What is the probability that the total time spent for 100 students who have appointments for tomorrow will exceed 10 hours? (Hint: 10 hours=600 minutes.) Carry your intermediate computations to at least four decimal places. Report your result to at least three decimal places.

Solution :

Let the time spent by students for the appointment be =

∴ E() = 5.6

   V() =

n = 100

The required probability

Now,

                            = 100 x 5.6

                            = 560

Standard deviation = 3.5 x 100

                                = 35

= 0.8735

= 0.874  (round to 3 decimal places)

answer: the answer is one of

step-by-step explanation:   we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following

step-by-step explanation:

compare