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?

Answer 1

The detailed solution is given in the pictures below.

Please go through them carefully specially the notations.

Hope the solution helps. Thank you.

(Please do comment if further help is required)

het x be the time (in milliseconds) taken by glycer aldehyde -3-phosphate dehydrogenase to convert aldehyde to carobolic acid. XA NC 19, 4.04 It ig is given that x-19 ~ N(01) (a) p ( 15 < XC 2o = P(xc 20) - P(x P(x <15) = P(x-19 15-19 D 6.04 F 20-19 4.04 :) - P(x-19 T hrou Gron 4:04 20-19 4.04 15-19 f 4.04 ) -0149 2 (n) = P(N(0, 1) (2) to & (0.2475) - I(-0.9961) = 0.4367, using nopmal 2 Ans: 0.4367 table values

P(x > 20) = 1- P(x = 20) = 1-2 (2 29.04 ) = 1-21-2475) Ans Eo.40226 Capprox). 190) = $10.9901) Ane ~NCorl). C) PLX <15) = ( 15-19 ) = = 0.16106-Corone) (2) cet y be the capt of any specified student. Y ~ N(3, 0142) => (7-3)%0.4 Let the required 30th percentile CPA be y Thus, P(124) = 0.3=2(4-3) or, y = 3 + 0.4 x 81 (0.3) or, y = 3 + 0.48 (-0.524) , using normal table 2.7904 is the required percentile. 30 or, 0.3= P(Y-3 & 4 22 or ie. 217904 Ang.