The Probability Density Function(PDF) is the probability function which is represented for the density of a continuous random variable lying between a certain range of values. The probability Density function is defined by the formula,

P(a<p<b) = _a^b∫ f(p) dp

Questions on the probability distribution function

Question 1:

The pdf of a distribution is given as

\begin{array}{l} f (x) = {\begin{array}{c} x; f o r 0 < x < 1 \\ 2 - x; f o r 1 < x < 2 \\ 0; f o r x > 2 \end{array}} \end{array}

Calculate the density within the interval

\begin{array}{l} (0.5 < x < 1.5) \end{array}

Solution:

\begin{array}{l} P (0.5 < x < 1.5) = \int_{0.5}^{1.5} f (x) d x \end{array}

\begin{array}{l} = \int_{0.5}^{1} f (x) d x + \int_{1}^{1.5} f (x) d x \end{array}

\begin{array}{l} = \int_{0.5}^{1} x d x + \int_{1}^{1.5} (2 - x) d x \end{array}

\begin{array}{l} = {(\frac{x^{2}}{2})}_{0.5}^{1} + {((2 x - \frac{x^{2}}{2}))}_{1}^{1.5} \end{array}

= 0

Example 2: Let x be a random variable with PDF is given by

$\begin{array}{l} f (x) = {\begin{array}{c} k x^{2}; & | x | \leq 1 \\ 0; & o t h e r w i s e \end{array} \end{array}$

. Find the value of k and and P(x ≥ ½).

Solution:
Given,

\begin{array}{l} f (x) = {\begin{array}{c} k x^{2}; & | x | \leq 1 \\ 0; & o t h e r w i s e \end{array} \end{array}

To find the value of k, consider the below expression.

\begin{array}{l} \int_{- \infty}^{\infty} f (u) d u = 1 \\ \int_{- 1}^{1} c u^{2} d u = 1 \\ c [\frac{u^{3}}{3}]_{- 1}^{1} = 1 \\ c [\frac{1}{3} + \frac{1}{3}] = 1 \\ \frac{2}{3} c = 1 \\ c = \frac{3}{2} \end{array}

No finding P(x≥ ½),

\begin{array}{l} P (x \geq \frac{1}{2}) = \int_{\frac{1}{2}}^{1} c x^{2} d x \\ = \frac{3}{2} [\frac{x^{3}}{3}]_{\frac{1}{2}}^{1} \\ = \frac{3}{2} [\frac{1}{3} - \frac{1}{24}] \\ = \frac{3}{2} \times \frac{7}{24} \\ = \frac{7}{16} \end{array}

Therefore, k = 3/2 and P(x≥1/2) = 7/16.

Example 3: Suppose x be a random variable and PDF is give by

$\begin{array}{l} f (x) = {\begin{array}{c} x^{2} + 1; & x \geq 0 \\ 0; & x < 0 \end{array} \end{array}$

Find P(1 < x < 3).
Solution:
Given,

\begin{array}{l} f (x) = {\begin{array}{c} x^{2} + 1; & x \geq 0 \\ 0; & x < 0 \end{array} \end{array}

\begin{array}{l} P (1 < x < 3) = \int_{1}^{3} (x^{2} + 1) d x \\ = [\frac{x^{3}}{3} + x]_{1}^{3} \\ = [(\frac{27}{3} + 3) - (\frac{1}{3} + 1)] \\ = [(9 + 3) - \frac{4}{3}] \\ = \frac{36 - 4}{3} \\ = \frac{32}{3} \end{array}

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Questions on the probability distribution function

Question 1:

Solution:

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