Question

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.

Answer 1

Answer:

a) 0.4

b) 0.133

c) $L \geq 99$  

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

$f(x) = \displaystyle\frac{1}{b-a} = \frac{1}{105-90} = \frac{1}{15},\\\\90 \leq x \leq 105$

a) P( beam load exceeds 99 pounds per linear​ foot)

P( x > 99)

$=\displaystyle\int_{99}^{105} f(x) dx\\\\=\displaystyle\int_{99}^{105} \frac{1}{15} dx\\\\=\frac{1}{15}[x]_{99}^{105} = \frac{1}{15}(105-99) = 0.4$

b) P( beam load less than 92 pounds per linear​ foot)

P( x < 92)

$=\displaystyle\int_{90}^{92} f(x) dx\\\\=\displaystyle\int_{90}^{92} \frac{1}{15} dx\\\\=\frac{1}{15}[x]_{90}^{92} = \frac{1}{15}(92-90) = 0.133$

c) We have to find L such that

$\displaystyle\int_{L}^{105} f(x) dx\\\\=\displaystyle\int_{L}^{105} \frac{1}{15} dx\\\\=\frac{1}{15}[x]_{L}^{105} = \frac{1}{15}(105-L) = 0.4\\\\\Rightarrow L = 99$

The beam load should be greater than or equal to 99 such that the probability that the beam load exceeds L is 0.4.