Respuesta :
Answer:
a) 0.4
b) 0.133
c) [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 = 105
Thus, 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.
