STEP II 2007 question 14 solution

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STEP II 2007 question 14 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP II 2007 question 14 solution

(i) The sketch should be straightforward: a curve (in the shape of $\ln{x}$ ) between x=1 and x=k, a straight horizontal line up until x=2k, and then a negative gradient straight line which reaches the x-axis at x=4k

(ii) Noticing the less than [B]or equal to[/B] signs, we can say that

$\ln{k}=a-2bk$ (by substituting x=2k into the pdf) and also

a-4bk=0

Solving these simultaneously yields:

$a=2\ln{k}, b=\frac{\ln{k}}{2k}$ as required for the first part.

Now the total area under the pdf must be equal to 1, and hence

$\displaystyle \int^{4k}_1{\ln{x} \text{ d} x+k\ln{k}+k\ln{k}=1$ (note the areas of two of the sections are simply rectangles and triangles and so integration is overkill)

Evaluating the integral and gathering like terms, we find

$3k\ln{k}-k=0 \Rightarrow k(3\ln{k}-1)=0$

Now as $k \neq 0$ we can say that $k=e^{\frac13}$ , and by the relations we derived earlier,

$a=\frac23\,,b=\frac{1}{6e^{1/3}}\,,k=e^{1/3}$

(iii) Using as estimate for $e^{1/3}$ close to 1, we can see that the median lies in the range $k \leq x \leq 2k$

Let us call the median $\gamma$ , then using the idea of rectangles and triangles under our pdf as before we have:

$\displaystyle \int^{e^{1/3}}_1{\ln{x}\text{ d} x+\frac13(\gamma-e^{1/3})=\frac13(2e^{1/3}-\gamma)+\frac13 e^{1/3}$ by the definition of the median that: