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saw5 [17]
3 years ago
12

The time it takes glyceraldehyde-3-phosphate dehydrogenase to convert aldehyde to carbolic acid follows a normal distribution wi

th 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?
Mathematics
1 answer:
balandron [24]3 years ago
4 0

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

<em>Z</em> = (<em>X</em> - 19) / 4.04

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

… ≈ Pr[-0.9901 ≤ <em>Z</em> ≤ 0.2475]

… ≈ Pr[<em>Z</em> ≤ 0.2475] - Pr[<em>Z</em> ≤ -0.9901]

(since <em>Z</em> is a continuous random variable)

… ≈ 0.4367

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

… ≈ Pr[<em>Z</em> > 0.2475]

… ≈ 1 - P[<em>Z</em> ≤ 0.2475]

(taking the complement probability)

… ≈ 0.4023

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

(also taking the complement)

… ≈ 0.1611

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