$A f u n c t i o n f (x) i s d e s i g n e d a s F o r f (x) t o b e a v a l i d p r o b a b i l i t y d e n s i t y f u n c t i o n t h e v a l u e o f A m u s t b e f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} 0 \frac{1}{A} (3 x + 5) 0 \end{matrix}; \begin{matrix} x < 3 3 < x < 6 x > 6 \end{matrix}$

63.5

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50

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55.5

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40.5

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Solution

The correct option is C $55.5$
$F o r f (x) t o b e p r o b a b i l i t y d e n s i t y f u n c t i o n = \int f (x) d x = 1 \frac{1}{A} \int_{3}^{6} (3 x + 5) d x = 1 \Rightarrow \frac{1}{A} {∣ ∣ ∣ \begin{matrix} \frac{3 x^{2}}{2} + 5 x \end{matrix} ∣ ∣ ∣}_{3}^{6} = 1 A = (\frac{3}{2} 6^{2} - \frac{3}{2} 3^{2}) + 5 (6 - 3) = \frac{3}{2} \times 27 + 15 = 55.5$

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A function f(x) is designed asFor f(x) to be a valid probability densityfunction the value of A must bef(x)=⎧⎪ ⎪⎨⎪ ⎪⎩01A(3x+5)0;x<33<x<6x>6

The correct option is C 55.5For f(x) to be probability density function=∫f(x)dx=11A∫63(3x+5)dx=1⇒1A∣∣∣3x22+5x∣∣∣63=1A=(3262−3232)+5(6−3)=32×27+15=55.5