(B) The expected value $E(X)$ of a random variable $X$ is given by the formula $E(X) = \sum p_i x_i$.Using the provided table:$E(X) = (0 \times 0.35)

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(B) The expected value $E(X)$ of a random variable $X$ is given by the formula $E(X) = \sum p_i x_i$.Using the provided table:$E(X) = (0 \times 0.35) + (500 \times 0.25) + (1000 \times 0.15) + (1500 \times 0.10) + (2000 \times 0.08) + (2500 \times 0.05) + (3000 \times 0.02)$Calculating each term:$E(X) = 0 + 125 + 150 + 150 + 160 + 125 + 60$Summing the values:$E(X) = 770$Thus,the expected value of the maintenance cost is Rs. $770$.

Class 12 Mathematics Probability Probability distribution

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Rajesh has just bought a $VCR$ from Maharashtra Electronics and the shop offers an after-sales service contract for Rs. $1000$ for the next five years. Considering the experience of $VCR$ users,the following distribution of maintenance expenses for the next five years is formed:
Expenses $0$ $500$ $1000$ $1500$ $2000$ $2500$ $3000$
Probability $0.35$ $0.25$ $0.15$ $0.10$ $0.08$ $0.05$ $0.02$

The expected value of the maintenance cost is:

Expenses	$0$	$500$	$1000$	$1500$	$2000$	$2500$	$3000$
Probability	$0.35$	$0.25$	$0.15$	$0.10$	$0.08$	$0.05$	$0.02$

MHT CET 2021 All MHT CET

A
Rs. $800$
B
Rs. $770$
C
Rs. $700$
D
Rs. $900$

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$A$ random variate $X$ takes the values $0, 1, 2, 3$ and its mean is $1.3$. If $P(X=3) = 2 P(X=1)$ and $P(X=2) = 0.3$,then $P(X=0)$ is equal to:

DifficultTS EAMCET 2003

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$A$ random variable $X$ takes the values $0, 1, 2$. Its mean is $1.2$. If $P(X=0)=0.3$,then $P(X=1)=$

EasyMHT CET 2020

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For the probability distribution given by

$x = x_{i}$ $0$ $1$ $2$

$P_{i}$ $\frac{25}{36}$ $\frac{5}{18}$ $\frac{1}{36}$

the standard deviation $(\sigma)$ is

MediumKCET 2018

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The p.m.f of a random variable $X$ is $P(X=x)=\frac{1}{2^5}\binom{5}{x}$,where $x=0, 1, 2, 3, 4, 5$ and $P(X=x)=0$ otherwise. Then:

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In a book of $500$ pages,it is found that there are $250$ typing errors. Assume that Poisson law holds for the number of errors per page. Then,the probability that a random sample of $2$ pages will contain no error,is:

DifficultAP EAMCET 2006

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$x = x_{i}$	$0$	$1$	$2$
$P_{i}$	$\frac{25}{36}$	$\frac{5}{18}$	$\frac{1}{36}$

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$A$ random variate $X$ takes the values $0, 1, 2, 3$ and its mean is $1.3$. If $P(X=3) = 2 P(X=1)$ and $P(X=2) = 0.3$,then $P(X=0)$ is equal to:

$A$ random variable $X$ takes the values $0, 1, 2$. Its mean is $1.2$. If $P(X=0)=0.3$,then $P(X=1)=$

For the probability distribution given by $x = x_{i}$ $0$ $1$ $2$ $P_{i}$ $\frac{25}{36}$ $\frac{5}{18}$ $\frac{1}{36}$ the standard deviation $(\sigma)$ is

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In a book of $500$ pages,it is found that there are $250$ typing errors. Assume that Poisson law holds for the number of errors per page. Then,the probability that a random sample of $2$ pages will contain no error,is:

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For the probability distribution given by

$x = x_{i}$ $0$ $1$ $2$

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