12 4problems

# Problem Solving

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##### Example

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Glottochronology is the study of the age of languages and how long it takes for old words to be replaced by new words.

Prof. Graham studied the divergence of the English language from the German. He found that the life of an old Germanic word in English, (that is, the time taken for an old Germanic word to be replaced), is a random variable, T thousand years, with probability density function below, where A is a real constant.
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… … $f(t)=\Bigg\{ \begin{matrix} A(t-1)^2&\quad0 \leqslant {t} \leqslant 1 \\ 0 & \text{otherwise}\\ \end {matrix}$
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a) .. Show that $A = 3$.

b) .. Show that, according to this model the expected life (or mean life) of an old Germanic word in the English language is 250 years.

c) .. According to the model, what proportion of the old Germanic words have a life greater than 200 years?

Solution:

a) .. Show that $A = 3$.
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… … … Since $f(t)$ is a probability density function
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… … … $\displaystyle{ \int\limits_{0}^{1} \; f(t) \; dt} =1$
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… … … $A \displaystyle{ \int\limits_{0}^{1} \; (t-1)^2 \; dt} =1$
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… … … $A\left[\dfrac{(t-1)^3}{3}\right]_{0}^{1}=1$
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… … … $A \Bigg(\dfrac{(1-1)^3}{3}-\dfrac{(0-1)^3}{3} \Bigg)=1$
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… … … $A\times\dfrac{1}{3}=1$
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… … … $A=3$ … … {as required}
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b) .. Show that, according to this model the expected life (or mean life) of an old Germanic word in the English language is 250 years.
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… … … $\mu=E(X) = \displaystyle{ \int\limits_{-\infty}^{\infty} \; xf(x) \; dx}$
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… … … $\mu =E(T)= \displaystyle{ \int\limits_{0}^{1} \; t \times 3(t-1)^2 \; dt}$
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… … … … $= 3 \displaystyle{ \int\limits_{0}^{1} \; t(t-1)^2 \; dt}$
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… … … … $=3 \displaystyle{ \int\limits_{0}^{1} \; t^3-2t^2+t \; dt}$
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… … … … $=3\left[\dfrac{t^4}{4}-\dfrac{2t^3}{3}+\dfrac{t^2}{2}\right]_0^1$
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… … … … $=3\Bigg(\dfrac{1}{4}-\dfrac{2}{3}+\dfrac{1}{2}\Bigg) - \Big( 0 \Big)$
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… … … … $=3\Bigg(\dfrac{3}{12}-\dfrac{8}{12}+\dfrac{6}{12} \Bigg)$
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… … … … $=3\times\dfrac{1}{12}$
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… … … … $=\dfrac{1}{4}$
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… … … but T is measured in thousands of years, so
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… … … Expected life of the Germanic word is $\dfrac{1}{4}\times1000= 250$ years {as required}
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c) .. According to the model, what proportion of the old Germanic words have a life greater than 200 years?
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… … … T is measured in thousands of years so 200 years gives $T = 0.2$
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… … … $Pr (T>0.2)$
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… … … … $=3 \displaystyle{ \int\limits_{0.2}^{1} \;(t-1)^2\; dt}$
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… … … … $=3\left[\dfrac{(t-1)^3}{3}\right]_{0.2}^{1}$
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… … … … $=3\Bigg( \dfrac{(1-1)^3}{3}-\dfrac{(0.2-1)^3}{3}\Bigg)$
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… … … … $= 0.8^3$
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… … … … $=0.512$

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{The above example has been sourced from MAV Maths Methods Trial Exam 2, 2008.}

{Permission for use has been granted by the MAV.}

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