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).
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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