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

MATH103 April 2010

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Question 07 (c)
The probability that a newly divided cell will divide again before t hours (where t ≥ 0) is given by $F(t)=1-e^{-t/12}$ (c) What is the median division time?

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
There is a direct relation between the median and the cumulative function.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The median $t_{\text{med}}$ is defined as $F(t_{\text{med}})={\frac {1}{2}}$ Since $F(t)$ is given we can easily calculate that Failed to parse (syntax error): {\displaystyle \frac12 = F(t_\text{med}) = 1-e^{-t_\text{med}/12}, \quad \text{i.e.} \quad e^{-t_\text{med}/12} = \frac12. } Taking the logarithm $\ln(x)$ on both sides yields the final answer $t_{\text{med}}=-12\ln(1/2).$ Note: From the class notes you may remember that the median waiting time is given by $\ln(2)/k$. When we rewrite $t_{\text{med}}=-12\ln(1/2)=12\ln(2)$ we observe that $k=1/12$, see also part (a). Hence our calculated answer agrees with the answer we expected.

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