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?

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

step-by-step explanation:

ok one would be money, and the other would be refreshments or something similar. if each bag of popcorn is 5.00 and each drink is half of that that means drinks are 2.50 the proper way to do that would to look and see how much of each you can buy without exceeding a limit of 30.00

add a pic of the graph, then ill edit this "answer" and solve it for you : 3

step-by-step explanation: