Probability Density Function

Solved Problems

Solution.

We compute the probability by the formula

Substituting \(\mu = 20,\) \(\sigma = 5,\) \(a = 15,\) \(b = 30,\) we have

Using a \(Z-\)table, we find that

Hence,

Solution.

We need to calculate the \(Z-\)score. Using the formula

where \(X \sim N\left( {100,15} \right),\) we get

From a \(Z-\)table, we find that

Thus, about \(9.2\%\) of the population have an \(ID-\)score greater than \(120.\)

Solution.

1. To find the value of \(k,\) we integrate the \(PDF\) on the interval from \(0\) to \(10\) and equate it to \(1:\)
   \[\int\limits_0^{10} {f\left( x \right)dx} = 1,\;\; \Rightarrow \int\limits_0^{10} {kxdx} = 1,\;\; \Rightarrow \left. {k\frac{{{x^2}}}{2}} \right|_0^{10} = 1,\;\; \Rightarrow \frac{k}{2}\left( {100 - 0} \right) = 1,\;\; \Rightarrow 50k = 1,\;\; \Rightarrow k = \frac{1}{{50}}.\]

   So, the probability distribution is given by the function \(f\left( x \right) = \frac{x}{{50}}.\)

2. Calculate the mean value \(\mu\) of the distribution:
   \[\mu = \int\limits_0^{10} {xf\left( x \right)dx} = \int\limits_0^{10} {\frac{{{x^2}}}{{50}}dx} = \frac{1}{{50}}\int\limits_0^{10} {{x^2}dx} = \frac{1}{{50}}\left. {\frac{{{x^3}}}{3}} \right|_0^{10} = \frac{{1000}}{{150}} = \frac{{20}}{3}.\]
3. Find the probability \(P\left( {2 \le X \le 5} \right):\)
   \[P\left( {2 \le X \le 5} \right) = \int\limits_2^5 {f\left( x \right)dx} = \frac{1}{{50}}\int\limits_2^5 {xdx} = \left. {\frac{{{x^2}}}{{100}}} \right|_2^5 = \frac{1}{{100}}\left( {25 - 4} \right) = 0.21\]

Solution.

1. Since the integral of \(PDF\) over the domain must equal one, we have
   \[\int\limits_0^3 {f\left( x \right)dx} = 1,\;\; \Rightarrow \int\limits_0^3 {k{x^2}dx} = 1,\;\; \Rightarrow \left. {k\frac{{{x^3}}}{3}} \right|_0^3 = 1,\;\; \Rightarrow \frac{k}{3}\left( {27 - 0} \right) = 1,\;\; \Rightarrow k = \frac{1}{9}.\]

   Hence, the \(PDF\) is given by the function \(f\left( x \right) = \frac{{x^2}}{{9}}.\)

2. We can easily find the mean value \(\mu\) of the probability distribution:
   \[\mu = \int\limits_0^3 {xf\left( x \right)dx} = \int\limits_0^3 {\frac{{{x^3}}}{9}dx} = \frac{1}{9}\int\limits_0^3 {{x^3}dx} = \frac{1}{9}\left. {\frac{{{x^4}}}{4}} \right|_0^3 = \frac{{81}}{{36}} = \frac{9}{4}.\]
3. The probability \(P\left( {1 \le X \le 2} \right)\) is also determined through integration:
   \[P\left( {1 \le X \le 2} \right) = \int\limits_1^2 {f\left( x \right)dx} = \frac{1}{9}\int\limits_1^2 {{x^2}dx} = \left. {\frac{{{x^3}}}{{27}}} \right|_1^2 = \frac{1}{{27}}\left( {8 - 1} \right) = \frac{7}{{27}} \approx 0.26\]

Solution.

1. We determine the value of \(k\) from the condition
   \[\int\limits_1^\infty {f\left( x \right)dx} = 1.\]

   Then

   \[\int\limits_1^\infty {\frac{k}{{{x^2}}}dx} = 1,\;\; \Rightarrow \left. {\left( { - \frac{k}{x}} \right)} \right|_1^\infty = 1,\;\; \Rightarrow k\lim\limits_{b \to \infty } \left( { - \frac{1}{b} + 1} \right) = 1,\;\; \Rightarrow k = 1.\]
2. To find \(\mu,\) we take the integral
   \[\mu = \int\limits_1^\infty {xf\left( x \right)dx} = \int\limits_1^\infty {\frac{{\cancel{x}dx}}{{{x^{\cancel{2}}}}}} = \int\limits_1^\infty {\frac{{dx}}{x}} = \left. {\ln x} \right|_1^\infty = \infty .\]

   As you can see, the probability density function \(f\left( x \right) = \frac{1}{{{x^2}}}\) defined on the domain \(\left[ {1,\infty } \right)\) has no mean.

Solution.

This is a variation of the well-known Cauchy distribution.

1. We integrate the given \(PDF\) and equate it to \(1:\)
   \[\int\limits_{ - \infty }^\infty {f\left( x \right)dx} = 1,\;\; \Rightarrow \int\limits_0^\infty {\frac{{kdx}}{{1 + {x^2}}}} = 1,\;\; \Rightarrow k\int\limits_0^\infty {\frac{{dx}}{{1 + {x^2}}}} = 1,\;\; \Rightarrow k\left. {\arctan x} \right|_0^\infty = 1,\;\; \Rightarrow k\left[ {\frac{\pi }{2} - 0} \right] = 1,\;\; \Rightarrow k = \frac{2}{\pi }.\]

   Hence, the \(PDF\) has the form

   \[f\left( x \right) = \begin{cases} {\frac{2}{\pi\left({1 + {x^2}}\right)}}, & \text{if } {x \ge 0} \\ 0, & \text{if } x \lt 0 \end{cases}.\]
2. The \(PDF\) is sketched in Figure below:



3. Compute the probability \(P\left( {0 \le X \le 1} \right):\)
   \[P\left( {0 \le X \le 1} \right) = \int\limits_0^1 {f\left( x \right)dx} = \frac{2}{\pi }\int\limits_0^1 {\frac{{dx}}{{1 + {x^2}}}} = \frac{2}{\pi }\left. {\arctan x} \right|_0^1 = \frac{2}{\pi }\left( {\frac{\pi }{4} - 0} \right) = \frac{1}{2}.\]
4. Find the mean value \(\mu:\)
   \[\mu = \int\limits_{ - \infty }^\infty {xf\left( x \right)dx} = \frac{2}{\pi }\int\limits_0^\infty {\frac{{xdx}}{{1 + {x^2}}}} = \frac{1}{\pi }\int\limits_0^\infty {\frac{{d\left( {1 + {x^2}} \right)}}{{1 + {x^2}}}} = \frac{1}{\pi } \lim\limits_{b \to \infty } \int\limits_0^b {\frac{{d\left( {1 + {x^2}} \right)}}{{1 + {x^2}}}} = \frac{1}{\pi } \lim\limits_{b \to \infty } \left[ {\left. {\ln \left| {1 + {x^2}} \right|} \right|_0^\infty } \right] = \frac{1}{\pi } \lim\limits_{b \to \infty } \left[ {\ln \left( {1 + {b^2}} \right) - \ln 1} \right] = \infty .\]