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?

Answer 1

Let X = The time it takes glyceraldehyde-3-phosphate dehydrogenase to convert aldehyde to carbolic acid.

Its given, X~ N(19ms , 4.04²) $\Rightarrow$ (X - 19) / 4.04 ~ N(0,1). Hence we can compute the required probabilities using the Standard Normal Tables:
Let Z = (X - 19) / 4.04 $\Rightarrow$ Z~N(0,1)

a)
the probability that it will take glyceraldehyde-3-phosphate dehydrogenase between 15 milliseconds and 20 milliseconds to convert aldehyde to carbolic acid = P(15< X < 20) = P((15-19)/4.04 < (X - 19) / 4.04 < (20-19)/4.04) = P(-0.9900<Z<0.06126) = P(Z<0.06126) - P(Z<-0.9900) = 0.3633

b) the probability that it will take glyceraldehyde-3-phosphate dehydrogenase more than 20 milliseconds to convert aldehyde to carbolic acid = P(X>20) = P( (X - 19) / 4.04 > (20-19)/4.04 ) = P( Z > 0.06126 ) = 0.47557

c)
the probability that it will take glyceraldehyde-3-phosphate dehydrogenase less than 15 milliseconds to convert aldehyde to carbolic acid = P( X<15 ) = P( (X - 19) / 4.04 < (15-19)/4.04 ) = P( Z < 0.9900 ) = 0.1611

Thats all :)