Timber beams are widely used in home construction. When the load (measured in pounds) per unit length has a constant value over part of a beam, the load is said to be uniformly distributed over that part of the beam. Consider a cantilever beam with a uniformly distributed load between 90 and 105 pounds per linear foot. a. What is the probability that a beam load exceeds 99 pounds per linear foot? b. What is the probability that a beam load is less than 92 pounds per linear foot? c. Find a value L such that the probability that the beam load exceeds L is 0.4.
Accepted Solution
A:
Answer:a) 0.4b) 0.133c) [tex]L \geq 99[/tex] Step-by-step explanation:We are given the following information in the question:The load is said to be uniformly distributed over that part of the beam between 90 and 105 pounds per linear foot. a = 90 and b = 105Thus, the probability distribution function is given by [tex]f(x) = \displaystyle\frac{1}{b-a} = \frac{1}{105-90} = \frac{1}{15},\\\\90 \leq x \leq 105[/tex]a) P( beam load exceeds 99 pounds per linear foot)P( x > 99)[tex]=\displaystyle\int_{99}^{105} f(x) dx\\\\=\displaystyle\int_{99}^{105} \frac{1}{15} dx\\\\=\frac{1}{15}[x]_{99}^{105} = \frac{1}{15}(105-99) = 0.4[/tex]b) P( beam load less than 92 pounds per linear foot)P( x < 92)[tex]=\displaystyle\int_{90}^{92} f(x) dx\\\\=\displaystyle\int_{90}^{92} \frac{1}{15} dx\\\\=\frac{1}{15}[x]_{90}^{92} = \frac{1}{15}(92-90) = 0.133[/tex] c) We have to find L such that[tex]\displaystyle\int_{L}^{105} f(x) dx\\\\=\displaystyle\int_{L}^{105} \frac{1}{15} dx\\\\=\frac{1}{15}[x]_{L}^{105} = \frac{1}{15}(105-L) = 0.4\\\\\Rightarrow L = 99[/tex]The beam load should be greater than or equal to 99 such that the probability that the beam load exceeds L is 0.4.