For the probability density function $f ( x ) = \frac { 7 } { 6 ( x + 1 ) ^ { 2 } }$ on the interval $[ 0,6 ]$ find the probability that $x \leq 4$ . Round your answer to the nearest hundredth.
A)0.05
B)0.33
C)0.93
D)0.67
E)0.02

Question

Quizplus · Accepted Answer

To find the probability that $x \leq 4$, we need to integrate the given probability density function from the lower limit of the interval up to 4. The integral of $f(x) = \frac{7}{6(x+1)^2}$ from 0 to 4 is calculated as follows:$$\int_{0}^{4} \frac{7}{6(x+1)^2} dx = \left[ -\frac{7}{6(x+1)} \right]_{0}^{4} = -\frac{7}{6(4+1)} + \frac{7}{6(0+1)} = -\frac{7}{30} + \frac{7}{6} = \frac{7}{6} - \frac{7}{30} = \frac{35}{30} - \frac{7}{30} = \frac{28}{30} = 0.93$$Therefore, the probability that $x \leq 4$ is approximately 0.93.

For the Probability Density Function f(x)=76(x+1)2f ( x ) = \frac { 7 } { 6 ( x + 1 ) ^ { 2 } }f(x)=6(x+1)27​

For the Probability Density Function $f ( x ) = \frac { 7 } { 6 ( x + 1 ) ^ { 2 } }$