Question

Let x be a continuous random variable with the probability distribution defined by
p(x)= {kx^2, 0<=x<=3
={0,else where

Evaluate k and find (i) P(1 ≤ x < 2), (ii) P(x ≤ 1),
(iii) P(x > 1). Also find the mean and variance of the distribution.

Answer 1

Answer on Question #45894 – Math – Statistics and Probability

Question.

Let $X$ be a continuous random variable with the probability distribution defined by

$$p(x) = \begin{cases} kx^2, & 0 \leq x \leq 3 \\ 0 \text{ elsewhere} \end{cases}. \text{ Evaluate } k \text{ and find } p(x) = \begin{cases} 0 & \text{EVAPOROUS} \\ 1 & \text{IF } 0 \text{ else} \end{cases}$$

(i) $P(1 \leq X < 2)$, (ii) $P(X \leq 1)$, (iii) $P(X > 1)$.

Also find the mean and variance of the distribution.

Solution.

According to the question, $p(x)$ is a probability distribution function $\Rightarrow \int_{-\infty}^{+\infty} p(x) dx = 1 \Rightarrow$

$$0 + \int_{0}^{3} kx^2 dx = 1 \Rightarrow$$ $$\Rightarrow \frac{k}{3}x^3 \Big|_{0}^{3} = 9k = 1 \Rightarrow k = \frac{1}{9}.$$

(i) $P(1 \leq X < 2) = \int_{1}^{2} \frac{1}{9}x^2 dx = \frac{x^3}{27}\Big|_{1}^{2} = \frac{7}{27};$

(ii) $P(X \leq 1) = \int_{0}^{1} \frac{1}{9}x^2 dx = \frac{x^3}{27}\Big|_{0}^{1} = \frac{1}{27};$

(iii) $P(X > 1) = \int_{1}^{3} \frac{1}{9}x^2 dx = \frac{x^3}{27}\Big|_{1}^{3} = \frac{26}{27}$

The mean: $EX = \int_{-\infty}^{+\infty} x p(x) dx = \frac{1}{9} \int_{0}^{3} x^3 dx = \frac{1}{36} x^4 \Big|_{0}^{3} = \frac{81}{36} = \frac{9}{4},$

$$EX^2 = \int_{-\infty}^{+\infty} x^2 p(x) dx = \frac{1}{9} \int_{0}^{3} x^4 dx = \frac{1}{45} x^5 \Big|_{0}^{3} = \frac{243}{45} = \frac{27}{5}.$$

The variance: $VarX = EX^2 - (EX)^2 = \frac{27}{5} - \frac{81}{16} = \frac{27}{80}$.

Answer. $k = \frac{1}{9}$

(i) $\frac{7}{27};$

(ii) $\frac{1}{27};$

(iii) $\frac{26}{27};$

$$EX = \frac{9}{4}; \quad VarX = \frac{27}{80}.$$

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