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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

[Unparseable or potentially dangerous latex formula. Error 6 ] (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:

[Unparseable or potentially dangerous latex formula. Error 6 ] by the definition of the median that:

\displaystyle \int^{\gamma}_{-\infty}{f(x)} \text{ d} x=\int_{\gamma}^{\infty}{f(x)} \text{ d} x

Evaluating the integral and tidying up, we see

\frac23 \gamma=2e^{1/3}-1

\Rightarrow \gamma= \frac32(2e^{1/3}-1) as required

Solution by coffeym.

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