## Related questions with answers

The total sustained load on the concrete footing of a planned building is the sum of the dead load plus the occupancy load. Suppose that the dead load

$X _ { 1 }$

has a gamma distribution with

$\alpha _ { 1 } = 50 \text { and } \beta _ { 1 } = 2$

, whereas the occupancy load

$X _ { 2 }$

has a gamma distribution with

$\alpha _ { 2 } = 20 \operatorname { and } \beta _ { 2 } = 2$

. (Units are in kips.) Assume that

$X _ { 1 } \text { and } X _ { 2 }$

are independent. Find a value for the sustained load that will be exceeded with probability less than 1/16.

Solution

Verified$\textbf{Given:}$ The dead load $X_{1}$ has a gamma distribution with $\alpha_{1} = 50$ and $\beta_{1} = 2$ wheareas occupancy load $X_{2}$ has a gamma distribution with $\alpha_{1} = 20$ and $\beta_{1} = 2$.

We have to calculate maximum value of sustained load which will be exceeded with probability less than 1/16.

Let $X_{1} + X_{2} = Y$

$\textbf{Results obtained in part (a):}$

$E(Y) = \mu = 140$

And

$V(Y) = \sigma^{2} = 280 \Rightarrow \sigma = 16.733$

According to Tchebysheff’s Theorem,

$P(|Y - \mu| \geq k\sigma) \leq \dfrac{1}{k^{2}} = \dfrac{1}{16} = \dfrac{1}{4^{2}}$

Hence, $k = 4$. So,

$|Y - \mu| \geq 4\sigma$

$Y \leq \mu - 4 \sigma \ \text{And} \ Y \geq \mu + 4 \sigma$

$Y \leq 140 - 4 \times 16.733 \ \text{And} \ Y \geq 140 + 4 \times 16.733$

Hence

$Y \leq 73.067 \ \text{And} \ Y \geq 206.933$

Hence, $\boxed{\text{Required sustained load} = 206.933 \ \text{kips}}$

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