Life expectancy expected value

Calculus Level 3

f ( x ) = π 14400 x sin ( π 120 x ) , 0 < x < 120 f(x)=\frac{\pi}{14400}x\sin\left(\frac{\pi}{120}x\right),\ \ 0<x<120

A life insurance actuary estimates the probabilities of X X , a person's life expectancy, with the probability density function as described above.

According to this model, what is E [ X ] ? \big\lfloor \text{E}[X]\big\rfloor?


The answer is 71.

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1 solution

Andy Hayes
Jun 19, 2016

Expected Value of the pdf:

E [ X ] = 0 120 x f ( x ) d x = 0 120 π 14400 x 2 sin ( π 120 x ) d x \text{E}[X]=\int_{0}^{120}{xf(x)\ dx}= \int_{0}^{120}{\frac{\pi}{14400}x^2\sin\left(\frac{\pi}{120}x\right)\ dx}

Using integration by parts , this is:

E [ X ] = 1 120 x 2 cos ( π 120 x ) + 2 π x sin ( π 120 x ) + 240 π 2 cos ( π 120 x ) 0 120 \text{E}[X]=\left. -\frac{1}{120}x^2\cos\left(\frac{\pi}{120}x\right)+\frac{2}{\pi}x\sin\left(\frac{\pi}{120}x\right)+\frac{240}{\pi^2}\cos\left(\frac{\pi}{120}x\right)\right |_{0}^{120}

This evaluates to:

E [ X ] = 120 480 π 2 71.366 \text{E}[X]=120-\frac{480}{\pi^2}\approx 71.366

Thus, E [ X ] = 71 \lfloor \text{E}[X]\rfloor =\boxed{71} .

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