Calculus

Applications of Integrals

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Probability Density Function

Solved Problems

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Example 5

Let X be a normal random variable with the mean μ = 20 and the variance σ² = 25. Find the probability that the random variable X falls in the range [15, 30].

Example 6

IQ tests are designed to follow the distribution law N(100, 15). What is the percentage of individuals with an IQ-score higher than 120?

Example 7

A random variable \(X\) is defined by the linear \(PDF\) in the form \[f\left( x \right) = kx\] on the interval \(\left[ {0,10} \right].\)

  1. Find the value of \(k\);
  2. Determine the mean value \(\mu\) of \(X\);
  3. Calculate the probability \(P\left( {2 \le X \le 5} \right).\)

Example 8

A random variable \(X\) is defined by the quadratic \(PDF\) in the form \[f\left( x \right) = k{x^2}\] on the interval \(\left[ {0,3} \right].\)

  1. Find the value of \(k\);
  2. Determine the mean value \(\mu\) of \(X\);
  3. Calculate the probability \(P\left( {1 \le X \le 2} \right).\)

Example 9

The \(PDF\) of a random variable \(X\) is defined by \[f\left( x \right) = \begin{cases} {\frac{k}{{{x^2}}}}, & \text{if } {x \ge 1} \\ 0, & \text{if } x \lt 1 \end{cases}.\]

  1. Calculate the value of \(k\);
  2. Find the mean value \(\mu\) of \(X.\)

Example 10

Let \(X\) be a continuous random variable with \(PDF\) given by \[f\left( x \right) = \begin{cases} {\frac{k}{{1 + {x^2}}}}, & \text{if } {x \ge 0} \\ 0, & \text{if } x \lt 0 \end{cases}.\]

  1. Find the value of \(k\);
  2. Graph the \(PDF;\)
  3. Calculate the probability \(P\left( {0 \le X \le 1} \right);\)
  4. Determine the mean value \(\mu\) of \(X.\)

Example 5.

Let \(X\) be a normal random variable with the mean \(\mu = 20\) and the variance \({\sigma ^2} = 25.\) Find the probability that the random variable \(X\) falls in the range \(\left[ {15,30} \right].\)

Solution.

We compute the probability by the formula

\[P\left( {a \le X \le b} \right) = P\left( {\frac{{a - \mu }}{\sigma } \le Z \le \frac{{b - \mu }}{\sigma }} \right).\]

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

\[P\left( {15 \le X \le 30} \right) = P\left( {\frac{{15 - 20}}{5} \le Z \le \frac{{30 - 20}}{5}} \right) = P\left( { - 1 \le Z \le 2} \right) = P\left( {z = 2} \right) - P\left( {z = - 1} \right).\]

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

\[P\left( {z = 2} \right) = 0.9772,\;\;\; P\left( {z = - 1} \right) = 0.1587\]

Hence,

\[P\left( {15 \le X \le 30} \right) = 0.9772 - 0.1587 = 0.8185\]

Example 6.

\(IQ\) tests are designed to follow the distribution law \(N(100,15).\) What is the percentage of individuals with an \(IQ\)-score higher than \(120?\)

Solution.

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

\[P\left( {a \le X \le b} \right) = P\left( {\frac{{a - \mu }}{\sigma } \le Z \le \frac{{b - \mu }}{\sigma }} \right),\]

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

\[P\left( {120 \le X \le \infty } \right) = P\left( {\frac{{120 - 100}}{{15}} \le Z \le \frac{{\infty - 100}}{{15}}} \right) = P\left( {\frac{4}{3} \le Z \le \infty } \right) = P\left( {Z \ge \frac{4}{3}} \right).\]

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

\[P\left( {z = \frac{4}{3}} \right) \approx P\left( {z = 1.33} \right) = 0.9082 \approx 90.8\% \]

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

Example 7.

A random variable \(X\) is defined by the linear \(PDF\) in the form \[f\left( x \right) = kx\] on the interval \(\left[ {0,10} \right].\)

  1. Find the value of \(k\);
  2. Determine the mean value \(\mu\) of \(X\);
  3. Calculate the probability \(P\left( {2 \le X \le 5} \right).\)

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 \(P\left( {2 \le X \le 5} \right):\)
    \[P\left( {2 \le X \le 5} \right) = \int\limits_2^5 {f\left( x \right)dx} = \frac{1}{{50}}\int\limits_2^5 {xdx} = \left. {\frac{{{x^2}}}{{100}}} \right|_2^5 = \frac{1}{{100}}\left( {25 - 4} \right) = 0.21\]

Example 8.

A random variable \(X\) is defined by the quadratic \(PDF\) in the form \[f\left( x \right) = k{x^2}\] on the interval \(\left[ {0,3} \right].\)

  1. Find the value of \(k\);
  2. Determine the mean value \(\mu\) of \(X\);
  3. Calculate the probability \(P\left( {1 \le X \le 2} \right).\)

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:
    \[P\left( {1 \le X \le 2} \right) = \int\limits_1^2 {f\left( x \right)dx} = \frac{1}{9}\int\limits_1^2 {{x^2}dx} = \left. {\frac{{{x^3}}}{{27}}} \right|_1^2 = \frac{1}{{27}}\left( {8 - 1} \right) = \frac{7}{{27}} \approx 0.26\]

Example 9.

The \(PDF\) of a random variable \(X\) is defined by \[f\left( x \right) = \begin{cases} {\frac{k}{{{x^2}}}}, & \text{if } {x \ge 1} \\ 0, & \text{if } x \lt 1 \end{cases}.\]

  1. Calculate the value of \(k\);
  2. Find the mean value \(\mu\) of \(X.\)

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.

Example 10.

Let \(X\) be a continuous random variable with \(PDF\) given by \[f\left( x \right) = \begin{cases} {\frac{k}{{1 + {x^2}}}}, & \text{if } {x \ge 0} \\ 0, & \text{if } x \lt 0 \end{cases}.\]

  1. Find the value of \(k\);
  2. Graph the \(PDF;\)
  3. Calculate the probability \(P\left( {0 \le X \le 1} \right);\)
  4. Determine the mean value \(\mu\) of \(X.\)

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. Probability density function f(x) = 2/(pi*(1+x^2)) for non-negative x
    Figure 6.
  4. Compute the probability \(P\left( {0 \le X \le 1} \right):\)
    \[P\left( {0 \le X \le 1} \right) = \int\limits_0^1 {f\left( x \right)dx} = \frac{2}{\pi }\int\limits_0^1 {\frac{{dx}}{{1 + {x^2}}}} = \frac{2}{\pi }\left. {\arctan x} \right|_0^1 = \frac{2}{\pi }\left( {\frac{\pi }{4} - 0} \right) = \frac{1}{2}.\]
  5. Find the mean value \(\mu:\)
    \[\mu = \int\limits_{ - \infty }^\infty {xf\left( x \right)dx} = \frac{2}{\pi }\int\limits_0^\infty {\frac{{xdx}}{{1 + {x^2}}}} = \frac{1}{\pi }\int\limits_0^\infty {\frac{{d\left( {1 + {x^2}} \right)}}{{1 + {x^2}}}} = \frac{1}{\pi } \lim\limits_{b \to \infty } \int\limits_0^b {\frac{{d\left( {1 + {x^2}} \right)}}{{1 + {x^2}}}} = \frac{1}{\pi } \lim\limits_{b \to \infty } \left[ {\left. {\ln \left| {1 + {x^2}} \right|} \right|_0^\infty } \right] = \frac{1}{\pi } \lim\limits_{b \to \infty } \left[ {\ln \left( {1 + {b^2}} \right) - \ln 1} \right] = \infty .\]
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