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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 40 and standard deviation 7 . Find percentile corresponding to p = 62 % of the number of hours studying. Write only a number as your answer. Round to two decimal places (for example: 20.81).

Let X denotes the number of hours spent studying the week before final exams. Then X is a normally distributed random variable with mean 40 and standard deviation 7.

Let X0.62 denotes the percentile corresponding to p = 62% of the number of hours studying. It indicates that 62% of the number of hours spent studying the week before final exams will be lesser than X0.62. Representing this as probability,

P(X < X0.62) = 0.62

Using the standard normal transformation, Apply this to our case. From the standard normal distribution table,

P(Z < 0.305481) = 0.62

Now equate the similar terms of the equations and solve. X0.62 = 0.305481(7) + 40 = 42.14

The percentile corresponding to p = 62% of the number of hours studying is 42.14.

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