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

MATH103 April 2010

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Question 07 (a)
The probability that a newly divided cell will divide again before t hours (where t ≥ 0) is given by $F(t)=1-e^{-t/12}$ (a) If you start observing a cell immediately after a division, how long, on average, would you have to wait to see the next division?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

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.

Hint 1
Recall that the average value of a continuous random variable X with values between a and b and probability density function p(X=x) = p(x) is defined as $\int _{a}^{b}xp(x)\,dx$

Hint 2
How can you relate the given cumulative function F(t) to the probability density function p(t)?

Hint 3
For a quick solution: The given cumulative function is the typical cumulative function of a waiting time. The expected waiting time is already hidden in the exponential function.

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. To get from the cumulative function to the probability density function, you need to take the derivate, i.e. $p(t)=F'(t)={\frac {1}{12}}e^{-t/12}.$ Now we apply the definition of the expected value to the random variable X that denotes the waiting time: ${\overline {X}}=\int _{0}^{\infty }tp(t)\,dt.$ This is an improper integral, so we treat it with caution. To solve it, we apply integration by parts with u=t and dv = e-t/12dt and obtain ${\begin{aligned}{\overline {X}}&=\int _{0}^{\infty }tp(t)\,dt\\&=\lim _{T\rightarrow \infty }\left({\frac {1}{12}}\int _{0}^{T}te^{-t/12}\,dt\right)\\&=\lim _{T\rightarrow \infty }\left(\left.{\frac {1}{12}}t(-12e^{-t/12})\right|_{0}^{T}+{\frac {1}{12}}\int _{0}^{T}12e^{-t/12}\,dt\right)\\&=\lim _{T\rightarrow \infty }\left(-Te^{-T/12}+0-\left.12e^{-t/12}\right|_{0}^{T}\right)\\&=\lim _{T\rightarrow \infty }\left(-12e^{-T/12}+12e^{0}\right)\\&=12;\end{aligned}}$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. From class and the course notes you may remember that waiting time typically has a probability density function and cumulative function of the form $p(t)=ke^{-kt},\quad {\text{and}}\quad F(t)=1-e^{-kt},$ respectively. In this case, the expected waiting time is given by 1/k. Here k = 1/12 so that the expected value is 12.

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