Question
A monthly demand for commodity is a continuous random distribution with probability distribution function given as:
where N is normalization constant. Obtain the value of N so that the function is normalized and hence, find the mean and the variance.
Answer :
Word Count : 426
The probability distribution function is given as: [ p(x) = \begin{cases} 3N(x^2 - 1) & 1 < x < 2 \ 0 & \text{elsewhere} \end{cases} ] Step 1: Find the normalization constant (N) The normalization condition is: [ \int_{-\infty}^{\infty} p(x) , dx = 1 ] Since (p(x) = 0) outside (1 < x < 2), [ \int_1^2 3N(x^2 - 1) , dx = 1 ] Compute the integral: [ \int_1^2 (x^2 - 1) , dx = \int_1^2 x^2 , dx - \int_1^2 1 , dx ] [ \int_1^2 x^2 , dx = \left[\frac{x^3}{3}\right]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3} ] [ \int_1^2 1 , dx = _____ _____ ____ ______ __________ ____ ___ _________ ___ _____ ______ ________.
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The probability distribution function is given as: [ p(x) = \begin{cases} 3N(x^2 - 1) & 1 < x < 2 \ 0 & \text{elsewhere} \end{cases} ] Step 1: Find the normalization constant (N) The normalization condition is: [ \int_{-\infty}^{\infty} p(x) , dx = 1 ] Since (p(x) = 0) outside (1 < x < 2), [ \int_1^2 3N(x^2 - 1) , dx = 1 ] Compute the integral: [ \int_1^2 (x^2 - 1) , dx = \int_1^2 x^2 , dx - \int_1^2 1 , dx ] [ \int_1^2 x^2 , dx = \left[\frac{x^3}{3}\right]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3} ] [ \int_1^2 1 , dx = _____ _____ ____ ______ __________ ____ ___ _________ ___ _____ ______ ________.
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