Question

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?

Answer 1

For each probability, transform the normal random variable X with mean 19 and s.d. 4.04 to a standard normal random variable Z with mean 0 and s.d. 1, using the rule

Z = (X - 19) / 4.04

(a) Pr[15 ≤ X ≤ 20] = Pr[(15 - 19)/4.04 ≤ (X - 19)/4.04 ≤ (20 - 19)/4.04]

… ≈ Pr[-0.9901 ≤ Z ≤ 0.2475]

… ≈ Pr[Z ≤ 0.2475] - Pr[Z ≤ -0.9901]

(since Z is a continuous random variable)

… ≈ 0.4367

(b) Pr[X > 20] = Pr[(X - 19)/4.04 > (20 - 19)/4.04]

… ≈ Pr[Z > 0.2475]

… ≈ 1 - P[Z ≤ 0.2475]

(taking the complement probability)

… ≈ 0.4023

(c) Pr[X < 15] = 1 - Pr[15 ≤ X ≤ 20] - Pr[X > 20]

(also taking the complement)

… ≈ 0.1611