X is a continuous random variable with probability density functi… - Vidyadip

MCQ

X is a continuous random variable with probability density function
$\begin{aligned}\mathrm{f}(x) & =\frac{x^2}{8}, 0 \leq x \leq 1 \\& =0, \quad \text { otherwise }\end{aligned}$
Then, the value of $\mathrm{P}(0.2 \leq \mathrm{X} \leq 0.5)$ is

✓
$\frac{0.117}{24}$
B
$\frac{0.112}{24}$
C
$\frac{0.117}{36}$
D
$\frac{0.112}{36}$

✓

Answer

Correct option: A.

$\frac{0.117}{24}$

(A)
$\mathrm{P}(0.2 \leq \mathrm{X} \leq 0.5)=\int_{0.2}^{0.5} \frac{x^2}{8} \mathrm{~d} x=\left[\frac{x^3}{24}\right]_{0.2}^{0.5}$
$\begin{aligned} & =\frac{1}{24}\left[(0.5)^3-(0.2)^3\right] \\ & =\frac{0.125-0.008}{24} \\ & =\frac{0.117}{24}\end{aligned}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from Probability Distributions→Probability Distributions — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

The angle between the lines represented by the equation $\left(x^2+y^2\right) \sin \theta+2 x y=0$ is

If the vectors $\hat{i}+2 \hat{k}, \hat{j}+\hat{k}$ and $\lambda \hat{i}+\mu \hat{j}$ are collinear, then

If $\alpha, \beta, \gamma$ are the angles made by a line with the co-ordinate axes, then $\sin ^2 \alpha+\sin ^2 \beta+\sin ^2 \gamma=$

$\int_0^1 x(1-x)^n d x=$

If n is any integer, then $\int_0^\pi e ^{\cos ^2 x} \cos ^3(2 n +1) x d x=$

If $0 \leq x<2 \pi$, then the number of real values of $x$, which satisfy the equation
$\cos x+\cos 2 x+\cos 3 x+\cos 4 x=0$, is

The probability distribution of a random variable X is given below.

X = k	0	1	2	3	4
P(X = k)	0.1	0.4	0.3	0.2	0

The variance of X is

Maximize $Z=7 x+11 y$, subject to $3 x+5 y \leq 26,5 x+3 y \leq 30, x \geq 0, y \geq 0$.

Solution of L.P.P. to minimize z = 2x + 3y such that x ≥ 0, y ≥ 0, 1 ≤ x + 2y ≤ 10 is _______.

Let $f(x)=5-|x-2|$ and $g(x)=|x+1|, x \in R$ If $f(x)$ attains maximum value at $\alpha$ and $g(x)$ attains minimum value at $\beta$, then $\lim _{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^2-5 x+6\right)}{x^2-6 x+8}$ is equal to