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 exam was normally distributed with mean 29 and standard deviation 5. Find percentile corresponding to P equals 59% of the number of hours studying. Round to 4 decimal places.
Homework Answers
µ = 29, σ = 5
P(x < a) = 0.59
Z score at p = 0.59 using excel = NORM.S.INV(0.59) = 0.2275
Value of X = µ + z*σ = 29 + (0.2275)*5 =
30.1377
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a , b , and c THANK YOU !!!!!
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