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a survey among freshmen at a certain university revealed that the number of hours spent studying...

a survey among freshmen at a certain university revealed that the number of hours spent studying the week before finals 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 28.0 and 30 hours studying? round to four decimal places.

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Answer #1

Solution :

= / n = 15 / 36 = 15 / 6 = 2.5

= P[(28 - 25) / 2.5 < ( - ) / < (30 - 25) / 2.5)]

= P(1.2 < Z < 2)

= P(Z < 2) - P(Z < 1.2)

= 0.9772 - 0.8849

= 0.0923

Probability = 0.0923

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