z = (29.6-25)/(15/sqrt(36)
z = 1.84
z = (30-25)/(15/sqrt(36)
z = 2
A survey among freshmen at a certain university revealed that the number of hours spent studying the week before final exams was normally distributed with mean 25 and standard deviation 15. A sample of 36 students was selected. What is the probability that the average time spent studying for the sample was between 27.6 and 30 hours studying?
1 answer
1 answer