Chapter 2, Question 3

Let \(f(x)=\frac{C}{x}\), where \(C\) is a constant, be a density function on the interval \([2,10]\).

A.) Find \(C\)

We know that the integral of a density function over its entire interval must be 1. So

\[ \int_2^{10}f(x)dx=1\\ \int_2^{10}\frac{C}{x}dx=1\\ C\int_2^{10}\frac{1}{x}dx=1\\ C\big(\ln(x)\big|_2^{10}\big)=1\\ C(\ln(10)-\ln(2))=1\\ C(\ln(\frac{10}{2}))=1\\ C=\frac{1}{\ln(5)}\\ \approx 0.6213349 \]

1/log(5)
## [1] 0.6213349

B.) Find \(P(E)\) where \(E=[a,b]\) is a subinterval of \([2,10]\)

Since \(P(E)=\int_E f(x)dx\), we can write this as \[ \int_a^bf(x)dx =\\ \int_a^b\frac{C}{x}dx=\\ C(\ln(b)-\ln(a))=\\ C(\ln(\frac{b}{a}))=\frac{1}{\ln(5)}\ln(\frac{b}{a}) \]

C.) Find \(P(X>5)\), \(P(X<7)\), and \(P(X^2-12X+35>0)\)

\[ P(X>5)=P([5,10]) \]

Using B.), we can write this as \[ P([5,10])=C\ln(\frac{10}{5})\\ =\frac{\ln(2)}{\ln(5)}\\\approx 0.4306766 \]

log(2)/log(5)
## [1] 0.4306766

Similarly, for \(P(X<7)\), \[ P(X<7)=P([2,7]) \] So, again we have \[ P([2,7])=C\ln(\frac{7}{2})\\ =C\ln\frac{7}{2}\\\approx 0.7783854 \]

log(7/2)/log(5)
## [1] 0.7783854

Finally, for \(P(X^2-12x+35>0)\), we need to factor the polynomial first. So it becomes \(P((X-7)(X-5))>0\). If both terms are positive, the result is true, if both terms are negative, the result is also true. So this gives us an interval of \(E=[2,5]\cup[7,10]\).

\[ P(E)=P([2,5])+P([7,10])\\ =C\ln(\frac{5}{2})+C\ln(\frac{10}{7})\\ =C(\ln(\frac{5\times10}{2\times7}))\\ =\frac{\ln(\frac{50}{14})}{\ln(5)}\\ \approx .790938 \]

log(50/14)/log(5)
## [1] 0.790938