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JGalway
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Q The amount of time (in minutes) that an executive of a certain firm talks on the telephone is a random variable having the probability density:
$$f(x) = \begin{cases} \dfrac{x}{4}&\text{for $0 < x \le 2$}\\
\dfrac{4}{x^3}&\text{for $x > 2$}\\
0&\text{elsewhere}\end{cases}$$
with reference to part (b) of Exercise $4.59$, find the expected length of one of these telephone conversations that has lasted for 1 minute.
A: The formula from $4.59$(b) is $$E[u(x)|a<x \le b]= \frac{\int_a^b u(x)f(x)\, dx}{\int_a^b f(x)\, dx}$$
I tried $$E[x|x \ge 1]= \frac{\int_1^2 x(x/4)\, dx}{\int_1^2 x/4\, dx} + \frac{\int_2^\infty x(4/x^3)\, dx}{\int_2^\infty 4/x^3 \, dx}
= \frac{14/6}{9/6}+4=\text{5.55555 minutes}$$
but the back of the books says the answer is $2.95$ mins so i don't know where i went wrong.
$$f(x) = \begin{cases} \dfrac{x}{4}&\text{for $0 < x \le 2$}\\
\dfrac{4}{x^3}&\text{for $x > 2$}\\
0&\text{elsewhere}\end{cases}$$
with reference to part (b) of Exercise $4.59$, find the expected length of one of these telephone conversations that has lasted for 1 minute.
A: The formula from $4.59$(b) is $$E[u(x)|a<x \le b]= \frac{\int_a^b u(x)f(x)\, dx}{\int_a^b f(x)\, dx}$$
I tried $$E[x|x \ge 1]= \frac{\int_1^2 x(x/4)\, dx}{\int_1^2 x/4\, dx} + \frac{\int_2^\infty x(4/x^3)\, dx}{\int_2^\infty 4/x^3 \, dx}
= \frac{14/6}{9/6}+4=\text{5.55555 minutes}$$
but the back of the books says the answer is $2.95$ mins so i don't know where i went wrong.
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