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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 final exams was normally distributed with mean 35 and standard deviation 7. Find percentile corresponding to p = 36% of the number of hours studying.

Write only a number as your answer. Round to two decimal places (for example: 20.81).

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Answer #1

Solution :

Given that,  

mean = = 35

standard deviation = = 7

Using standard normal table ,

P(Z < z) = 36%

P(Z < -0.36) = 0.36

z = -0.36

Using z-score formula,

x = z * +

x = -0.36 * 7 + 35 = 32.48

Answer = 32.48

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