Science:Math Exam Resources/Courses/MATH103/April 2010/Question 07 (b)

MATH103 April 2010

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[hide] Question 07 (b)
The probability that a newly divided cell will divide again before t hours (where t ≥ 0) is given by $F(t)=1-e^{-t/12}$ (b) What is the probability that the cell will divide between 3 and 6 hours? (Leave your answer in terms of $e$.)

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If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
If p(t) is the probability density function of a continuous random variable X, then the probability that an event $a_{1}\leq X\leq a_{2}$ happens is given by $\int _{a_{1}}^{a_{2}}p(t)\,dt.$

[show] Hint 2
How can you relate the given cumulative function F(t) to the probability density function p(t)? You don't need to calculate p(t) here if you use the Fundamental Theorem of Calculus.

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[show] Solution
Again, let X be the continuous random variable that denotes the waiting time. Since $F'(t)\ =p(t)$ we can use the Fundamental theorem of calculus to find the probability that the cell will divide between 3 and 6 hours: ${\begin{aligned}p(3\leq X\leq 6)&=\int _{3}^{6}p(t)\,dt\\&=F(6)-F(3)=(1-e^{-6/12})-(1-e^{-3/12})\\&=e^{-1/4}-e^{-1/2}\\&(\approx 17.2\%)\end{aligned}}$

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