Question

1) [25 pts] The time it takes glyceraldehyde-3-phosphate dehydrogenase to convert aldehyde to carbolic acid follows a normal distribution with a mean of 19 milliseconds and standard deviation of 4.04 milliseconds. a) What is the probability that it will take glyceraldehyde-3-phosphate dehydrogenase between 15 milliseconds and 20 milliseconds to convert aldehyde to carbolic acid? b) What is the probability that it will take glyceraldehyde-3-phosphate dehydrogenase more than 20 milliseconds to convert aldehyde to carbolic acid? c) What is the probability that it will take glyceraldehyde-3-phosphate dehydrogenase less than 15 milliseconds to convert aldehyde to carbolic acid? 2) [ 15 pts] The average freshman GPA is 3.0 with a standard deviation of 0.4. If the GPA follows a normal distribution what is the 30th percentile GPA?| nmi and

Answer 1

1) Solution - Let X be the time takes glyceraldehyde-3 - phosphate dehydrogenase to convert aldehyde to carbolic acid. X follows normal distribution with mean M = 19 and standard deviation 6 = 4.04 milliseconds » X ( 3 , 4 4)) a) → The probability that it will take glyceraldehyde-3- phosphate dehydrogenase between 15 milliseconds and 20 milliseconds to convertanto carbolic acid is given by P (18 < X < 20) 15-1 X-M < 20-M 8 8 2 - P (15-19 <X-19 < 20-19 4.04 4.04 4.04 20 = P(-0.99 << < 0.25) where = X-19 = p<<<0.25) - P(Z <-0.99) = {1-(436)}-(230-94) 0.1641. (From statistical table) 1 -0.4013 P(15<x<20) = 0.4876 The probability that it will take 15 milliseconds and 20 milliseconds is 0.4376

b) -> The probability that it will take a more than 20 milliseconds to convert aldehyde into carbolic acid is given by. :P(x720) = P(x-4 > 20-1 2 2. -P X-19 4.04 20-19 4.04 P(Z >0.26 where Z=8-19 4.04. (From statistical table) .:|P(x > 20) = 0.4018 The probability that it will take more than 20. milliseconds to convert aldehyde into carbolic acid is 0.4013. c) The probability that it will take less than 15. milliseconds to convert aldehyde to carbolic acid, is given by P(x < 15) = P(x-4 < 15-M 3 2 P(X-19 < 15-19 4-04 P(Z <-0.99) where z = X-19 = P(2>0.99) (Due to symmetry ../P(x<15) From Statistical table) The probability is 0.1611 = 0 . 1611

percentile 2) solution: Let y be the GPA with mean 3.0 and and standard deviation 2=0.4.4 follows normal distribution. Assume the p is 30th of GPA such that. P(Y<p) = 0.80 ..P/4-3.0 P-3.0 = 0.30 0-4 "P(Z <Pi) = =0.30 where = Y-3.0 and Py = .. 4- P(72) = 0:30. =P-3.0 P(x > P1) 0.70 from statistical table. P(37 -0.524) = 0.70 Compairing equation (1) and (2) we get P1 = -0.524 P-3.0 0.4 .O. 524. P-3.0 = -0.2096 e 3-02096 P = 2.9904 The 30th percentile GPA ig 2.9904