Question

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 29.2 and 30 hours studying? Write only a number as your answerRound to 4 decimal places (for example 0.0048 ). Do not write as a percentage .

Answer 1

Solution :

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

= P[(29.2 - 25) / 2.5 < ( - ) / < (30 - 25) / 2.5)]

= P(1.68 < Z < 2)

= P(Z < 2) - P(Z < 1.68)

= 0.9772 - 0.9535

= 0.0237

Probability = 0.0237

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