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Salvage value is an asset's anticipated book value after depreciation, based on what a firm anticipates to obtain in exchange for the equipment at the end of its lifecycle. As a result, the projected salvage value of an item is a crucial factor in determining a depreciation schedule.

Given:

\(f(x) = \frac{{32}}{{{{(x + 4)}^3}}}{\rm{ }}x > 0\)

Note: my book contained\(x < 0\), but then f is not a legitimate pdf. Thus, I assume that it should be \(x > 0\)instead.

f is a legitimate pdf if the integral over all its possible values is equal to 1:

\(\begin{aligned}\int_0^{ + \infty } f (x)dx &= \int_0^{ + \infty } {\frac{{32}}{{{{(x + 4)}^3}}}} dx =\left. {\left( {\frac{{32}}{{ - 2{{(x + 4)}^2}}}} \right)} \right|_0^{ + \infty } \\&= \left.{\left( {\frac{{ - 16}}{{{{(x + 4)}^2}}}} \right)}\right|_0^{ + \infty } \\ &= 0 + \frac{{16}}{{{4^2}}} \\&= 0 +1 =1\\\end{aligned}\)

Since the integral is equal to \(1,f(x)\)is a legitimate pdf.

Given:

\(f(x) = \frac{{32}}{{{{(x + 4)}^3}}}{\rm{ }}x > 0\)

The cdf at x is the integral of the pdf f(x) over all possible values up to x:

\(\begin{aligned}F(x)&= \int_{ - \infty }^x f (x)dx = \int_0^x {\frac{{32}}{{{{(x + 4)}^3}}}}dx = \left.{\left( {\frac{{32}}{{ - 2{{(x + 4)}^2}}}} \right)}\right|_0^x =\left.{\left( {\frac{{- 16}}{{{{(x + 4)}^2}}}}\right)} \right|_0^x \\&= \frac{{ - 16}}{{{{(x + 4)}^2}}} + \frac{{16}}{{{4^2}}}= -\frac{{16}}{{{{(x + 4)}^2}}} + 1 =1 - \frac{{16}}{{{{(x + 4)}^2}}} \\\end{aligned}\)

The cdf is zero for values smaller than or equal to zero:

\(F(x) = \left\{ {\begin{aligned}{}{1 - \frac{{16}}{{{{(x + 4)}^2}}}}&{x > 0}\\0&{x \le 0}\end{aligned}} \right.\)

Given:

\(f(x) = \frac{{32}}{{{{(x + 4)}^3}}}{\rm{ }}x > 0\)

The cdf at x is the integral of the pdf f(x) over all possible values up to x:

\(\begin{aligned}F(x)&= \int_{ - \infty }^x f (x)dx &= \int_0^x {\frac{{32}}{{{{(x + 4)}^3}}}}dx &= \left.{\left( {\frac{{32}}{{ - 2{{(x + 4)}^2}}}} \right)}\right|_0^x &=\left.{\left( {\frac{{- 16}}{{{{(x + 4)}^2}}}}\right)} \right|_0^x \\&= \frac{{ - 16}}{{{{(x + 4)}^2}}} + \frac{{16}}{{{4^2}}}&= -\frac{{16}}{{{{(x + 4)}^2}}} + 1 &=1 - \frac{{16}}{{{{(x + 4)}^2}}} \\\end{aligned}\)

The cdf is zero for values smaller than or equal to zero:

\(F(x) = \left\{ {\begin{aligned}{}{1 - \frac{{16}}{{{{(x + 4)}^2}}}}&{x > 0}\\0&{x \le 0}\end{aligned}} \right.\)

The probability in between two boundaries is the difference of the cdf evaluated at the boundaries:

\(\begin{aligned}P(2 < X < 5) &= F(5) - F(2) = 1 - \frac{{16}}{{{{(5 + 4)}^2}}} - \left( {1\frac{{16}}{{{{(2 + 4)}^2}}}} \right) \hfill \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, & = \frac{{16}}{{36}} - \frac{{16}}{{81}} = \frac{{36}}{{81}} - \frac{{16}}{{81}}= \frac{{20}}{{81}} \approx 0.2469 = 24.69\% \hfill \\\end{aligned}\)

\(f(x) = \frac{{32}}{{{{(x + 4)}^3}}}{\rm{ }}\;\;\;x > 0\)

Note: my book contained\(x < 0\), but then f is not a legitimate pdf. Thus, I assume that it should be \(x > 0\)instead.

The expected value of X is the integral of the product of each possible value x and its probability f(x) :

\(E(X) = \int_{ - \infty }^{ + \infty } x f(x)dx = \int_0^{ + \infty } {\frac{{32x}}{{{{(x + 4)}^3}}}} dx\)

Let \(u = x + 4,x = u - 4\)and \(du = dx\)

\(\begin{aligned}\int_{x = 0}^{x =+ \infty }{\frac{{32(u - 4)}}{{{u^3}}}}du &= \int_{x = 0}^{x = + \infty } {\frac{{32}}{{{u^2}}}} du - \int_{x = 0}^{x = + \infty } {\frac{{128}}{{{u^3}}}} du = \left. {\left( {\frac{{32}}{{ - 1{u^1}}}} \right)} \right|_{x = 0}^{x = + \infty } - \left. {\left( {\frac{{128}}{{ - 2{u^2}}}} \right)} \right|_{x = 0}^{x = + \infty } \\&= \left. {\left( {\frac{{ - 32}}{{x + 4}}} \right)} \right|_0^{ + \infty } - \left. {\left( {\frac{{ - 64}}{{{{(x + 4)}^2}}}} \right)} \right|_0^{ + \infty } = \frac{{32}}{4} - \frac{{64}}{{{4^2}}} = 8 - 4 = 4 \\\end{aligned}\)

Thus, the expected time of failure is 4 years.

Given:

\(f(x) = \frac{{32}}{{{{(x + 4)}^3}}}{\rm{ }}\;\;\;x > 0\)

Note: my book contained \(x < 0,\)but then f is not a legitimate pdf. Thus, I assume that it should be\(x > 0\)instead. The expected value of \(\frac{{100}}{{4 + X}}\)is the integral of the product of each possible value \(\frac{{100}}{{4 + x}}\) and its probability f(x) :

\(E(X) = \int_{ - \infty }^{ + \infty } {\frac{{100}}{{4 + x}}} f(x)dx = \int_0^{ + \infty } {\frac{{100}}{{4 + x}}} \frac{{32}}{{{{(x + 4)}^3}}}dx = \int_0^{ + \infty } {\frac{{3200}}{{{{(x + 4)}^4}}}} dx\)

Let \(u = x + 4\)and \(du = dx\)

\( = \int_{x = 0}^{x = + \infty } {\frac{{3200}}{{{u^4}}}} du = \left. {\left( {\frac{{3200}}{{ - 3{u^3}}}} \right)} \right|_{x = 0}^{x = + \infty } = \left. {\left( {\frac{{3200}}{{ - 3{{(x + 4)}^3}}}} \right)} \right|_{x = 0}^{x = + \infty } = \frac{{3200}}{{3{{(4)}^3}}} = \frac{{50}}{3} \approx 16.6667\)

Thus, the expected time salvage value is \(\frac{{50}}{3}\)or about \(16.6667\)years.