A function f(x) is designed as to be a valid probability density function the value of A must be

$f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} 0 & ; x < 3 \frac{1}{A} (3 x + 5) & ; 3 < x < 6. For f (x) 0 & ; x > 6 \end{matrix}$

55.5

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40.5

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63.5

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Solution

The correct option is A 55.5
For f(x) to be probability density function = $\infty \int - \infty f (x) d x = 1$

$\frac{1}{A} \infty \int - \infty (3 x + 5) d x = 1 \Rightarrow \frac{1}{A} {∣ ∣ \frac{3 x^{2}}{2} + 5 x ∣ ∣}_{3}^{6} = 1$

$A = (\frac{3}{2} 6^{2} - \frac{3}{2} 3^{2}) + 5 (6 - 3)$
$= \frac{3}{2} 27 + 15 = 55.5$

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A function f(x) is designed as to be a valid probability density function the value of A must be f(x)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩0;x<31A(3x+5) ;3<x<6. Forf(x)0;x>6

The correct option is A 55.5For f(x) to be probability density function = ∞∫−∞f(x)dx=1 1A∞∫−∞(3x+5)dx=1⇒1A∣∣3x22+5x∣∣63=1 A=(3262−3232)+5(6−3) =3227+15=55.5

A function f(x) is designed as to be a valid probability density function the value of A must be

$f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} 0 & ; x < 3 \frac{1}{A} (3 x + 5) & ; 3 < x < 6. For f (x) 0 & ; x > 6 \end{matrix}$