A random variable X has a density function f(x) = $\frac { 1 } { \ln 16 }$ ∙ $\frac { 1 } { x }$ , 1 ≤ x ≤ 16. Find a such that $\operatorname { Pr } ( 1 \leq X \leq a ) = \frac { 3 } { 4 }$ .
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The Answer of A random variable X has a density function f(x) = $\frac { 1 } { \ln 16 }$ ∙ $\frac { 1 } { x }$ , 1 ≤ x ≤ 16. Find a such that $\operatorname { Pr } ( 1 \leq X \leq a ) = \frac { 3 } { 4 }$ .
Enter just an integer, no labels.

A Random Variable X Has a Density Function F(x) = 1ln⁡16\frac { 1 } { \ln 16 }ln161​

A Random Variable X Has a Density Function F(x) = $\frac { 1 } { \ln 16 }$