91Ó°ÊÓ

You should consider a cantilever beam with a uniformly distributed load between 100 and 115 pounds per linear foot.

Let x be the random variable uniformly distributed between 100 and 115 pounds.

The probability density function random variable x is given by

$f\left(x\right)=\frac{1}{d-c};c<x<d$

Here c=100 and d=115

So the pdf of x is:

$$\begin{gathered} f\left(x\right)=\frac{1}{115-100} \\ =\frac{1}{15} \\ =0.066667 \\ f\left(x\right)=\left\{\begin{array}{l}0.066667;100<x<115\\ 0;otherwise\end{array}\right. \end{gathered}$$

$$\begin{gathered} F\left(x\right)=P\left(X≤x\right) \\ ={∫}_{100}^{x}f\left(t\right)dt \\ ={∫}_{100}^{x}0.66667dt \\ =0.66667{∫}_{100}^{x}dt \\ =0.66667{\left[T\right]}_{100}^x \\ =0.66667\left(x-100\right) \\ F\left(x\right)=0.66667×\left(x-100\right) \end{gathered}$$

For continuous random variable x

$P\left(X<x\right)=P\left(X<x\right)$

The probability that a beam load exceeds110 pounds per linear foot is,

$$\begin{gathered} P\left(x>110\right)=1-P\left(x≤110\right) \\ =1-F\left(110\right) \\ =1-066667×\left(110-100\right) \\ =0.333333 \end{gathered}$$

Thus, the required probability is 0.333333.

The probability that a beam load is less than102 pounds per linear foot is,

$$\begin{gathered} P\left(x<102\right)=F\left(102\right) \\ =0.066667×\left(102×100\right) \\ =0.133333 \end{gathered}$$

Thus, the required probability is 0.133333.

To find a value Lsuch that the probability that the beamload exceeds Lis only .1, we require following calculation.

$$\begin{gathered} P\left(x>L\right)=0.10 \\ 1-P\left(x≤L\right)=0.10 \\ P\left(x≤L\right)=0.90 \\ F\left(L\right)=0.90 \\ 0.0666667×\left(L-100\right)=0.90 \\ L=100+\frac{0.90}{0.0666667} \\ L=113.5 \end{gathered}$$

Thus, the required value is 113.5 pounds.