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(C) The number of errors follows a Poisson distribution with parameter $\lambda = n 	imes p_{error}$.Given $n = 40$ pages and the rate of errors is $

Vedclass · Accepted Answer

$13 e^{-2}$

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(C) The number of errors follows a Poisson distribution with parameter $\lambda = n 	imes p_{error}$.Given $n = 40$ pages and the rate of errors is $1$ per $10$ pages, the average number of errors is $\lambda = 40 	imes \frac{1}{10} = 4$.The probability of having $X$ errors is given by $P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$.We need the probability that the errors are at most $2$, which is $p = P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$.$p = e^{-4} \left( \frac{4^0}{0!} + \frac{4^1}{1!} + \frac{4^2}{2!} ight) = e^{-4} (1 + 4 + 8) = 13 e^{-4}$.We are asked to find $e^2 p$.$e^2 p = e^2 	imes 13 e^{-4} = 13 e^{-2}$.

$X = x$	$-4$	$-2$	$0$	$2$	$4$	$6$	$8$	$10$
$F(X = x)$	$0.1$	$0.3$	$0.5$	$0.65$	$0.75$	$0.85$	$0.90$	$1$

Outcome	Probability
$\omega_{1}$	$0.1$
$\omega_{2}$	$0.2$
$\omega_{3}$	$0.3$
$\omega_{4}$	$0.4$
$\omega_{5}$	$0.5$
$\omega_{6}$	$0.6$
$\omega_{7}$	$0.7$

$A$ typist claims that he prepares a typed page with typo errors of $1$ per $10$ pages. In a typing assignment of $40$ pages, if the probability that the typo errors are at most $2$ is $p$, then $e^2 p=$

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