Suppose a continuous random variable X has the density function. Find; | Basic Statistics 2078

Basic Statistics20785 marksNumerical: continuous probability distributionContinuous probability distributions

Suppose a continuous random variable X has the density function. Find; (i) value of k, (ii) P (0 < x < 0.5), and, (iii) E(X) and (iv) E(2X + 4)

$f(x) = \begin{cases} k(1-x)^2; & \text{for } 0 < x < 1 \\ 0; & \text{elsewhere} \end{cases}$

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Suppose a continuous random variable X has the density function. Find; (i) value of k, (ii) P (0 < x < 0.5), and, (iii) E(X) and (iv) E(2X + 4)

$$f(x) = \begin{cases} k(1-x)^2; & \text{for } 0 < x < 1 \ 0; & \text{elsewhere} \end{cases}$$

[5]

Density: $f(x) = k(1-x)^2$ for $0 < x < 1$ ; $f(x) = 0$ elsewhere - Required: (i) $k$ , (ii) $P(0<X<0.5)$ , (iii) $E(X)$ , (iv) $E(2X+4)$ --- Total probability must equal 1: $\int{0}^{1} k(1-x)^2\,dx = 1$ $$k\left[-\frac{(1-x)^3}{3}\right]{0}^{1} = k\left[0-\left(-\frac{1}{3}\right)\right] = \fra...

Density: $f(x) = k(1-x)^2$ for $0 < x < 1$; $f(x) = 0$ elsewhere - Required: (i) $k$, (ii) $P(0<X<0.5)$, (iii) $E(X)$, (iv) $E(2X+4)$ --- Total probability must equal 1: $$\int{0}^{1} k(1-x)^2,dx = 1$$ $$k\left[-\frac{(1-x)^3}{3}\right]{0}^{1} = k\left[0-\left(-\frac{1}{3}\right)\right] = \fra...