A random variable X takes the values 1, 2, 3 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $2 P(X=1) = 3 P(X=2) = P(X=3) = 5 P(X=4) = k$.Then $P(X=1) = \frac{k}{2}, P(X=2) = \frac{k}{3}, P(X=3) = k, P(X=4) = \frac{k}{5}$.Since $\su

Vedclass · Accepted Answer

$\frac{421}{61}$

Vedclass · Answer

(A) Given $2 P(X=1) = 3 P(X=2) = P(X=3) = 5 P(X=4) = k$.Then $P(X=1) = \frac{k}{2}, P(X=2) = \frac{k}{3}, P(X=3) = k, P(X=4) = \frac{k}{5}$.Since $\sum P(X) = 1$, we have $\frac{k}{2} + \frac{k}{3} + k + \frac{k}{5} = 1$.$\Rightarrow k(\frac{15+10+30+6}{30}) = 1 \Rightarrow k(\frac{61}{30}) = 1 \Rightarrow k = \frac{30}{61}$.The probability distribution is:$x$$1$$2$$3$$4$$P(X=x)$$\frac{15}{61}$$\frac{10}{61}$$\frac{30}{61}$$\frac{6}{61}$Mean $\mu = E(X) = \sum x P(x) = 1(\frac{15}{61}) + 2(\frac{10}{61}) + 3(\frac{30}{61}) + 4(\frac{6}{61}) = \frac{15+20+90+24}{61} = \frac{149}{61}$.$E(X^2) = \sum x^2 P(x) = 1^2(\frac{15}{61}) + 2^2(\frac{10}{61}) + 3^2(\frac{30}{61}) + 4^2(\frac{6}{61}) = \frac{15+40+270+96}{61} = \frac{421}{61}$.Variance $\sigma^2 = E(X^2) - \mu^2$.We need $\sigma^2 + \mu^2 = E(X^2) - \mu^2 + \mu^2 = E(X^2)$.Therefore, $\sigma^2 + \mu^2 = \frac{421}{61}$.

$A$ random variable $X$ takes the values $1, 2, 3$ and $4$ such that $2 P(X=1) = 3 P(X=2) = P(X=3) = 5 P(X=4)$. If $\sigma^2$ is the variance and $\mu$ is the mean of $X$, then $\sigma^2 + \mu^2 =$

Similar Questions

If the function $f$ defined by $f(x) = \begin{cases} K(x-x^2) & \text{if } 0 < x < 1 \\ 0 & \text{otherwise} \end{cases}$ is the probability density function (p.d.f.) of a random variable $X$,then the value of $P(X < \frac{1}{2})$ is

$A$ random variable $X$ has the following probability distribution:
$X = 1, 2, 3, 4, 5$
$P(X) = 0.1, 0.2, 0.3, 0.2, 0.2$
For the events $E = \{ X \text{ is a prime number} \}$ and $F = \{ X < 4 \}$, find $P(E \cap F)$.

If the probability density function (p.d.f.) of a continuous random variable $X$ is given by $f(x) = \begin{cases} k(9 + 8x - x^2), & \text{for } -1 \leq x \leq 4 \\ 0, & \text{otherwise} \end{cases}$, then the value of $k$ is:

It is known that a box of $8$ batteries contains $3$ defective pieces and a person randomly selects two batteries from the box. If $X$ is the number of defective batteries selected,then $P(X \leq 1) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module