How Do Typists Affect Error Variability in Typed Manuscripts?

[Chris L T521]

Thanks to those who participated in last week's POTW! Here's this week's problem!

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Problem: A manuscript is sent to a typing firm consisting of typists $A$, $B$ and $C$. If it is typed by $A$, then the number of errors made is a Poisson random variable with mean 2.6; if typed by $B$, then the number of errors is a Poisson random variable with mean 3; and if typed by $C$, then it is a Poisson random variable with mean 3.4. Let $X$ denote the number of errors in the typed manuscript. Assume that each typist is equally likely to do the work. Find $E[X]$ and $\mathrm{Var}\,[X]$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by magneto. You can find his solution below.

[sp]$\DeclareMathOperator{\E}{\mathsf{E}}
\DeclareMathOperator{\var}{\mathsf{Var}}$
Recall if a random variable $T$ is Poisson-distributed with parameter $\lambda$,
then the $\E[T] = \lambda = \var[T]$. Note that since $\var[T] = \E[T^2] - \E[T]^2 =
\E[T^2] - \lambda^2 = \lambda$, it implies, $\E[T^2] = \lambda^2 + \lambda$.

$(a)$ We have $E[X] = \Pr(A) \E[X|A] + \Pr(B) \E[X|B] + \Pr(C) \E[X|C] =
\frac 13 (2.6 + 3 + 3.4) = 3$.

$(b)$ $\var[X] = \E[X^2] - \E[X]^2$. We know $\E[X]^2 = 9$. $\E[X^2] = \Pr(A) \E[X^2|A] + \Pr(B) \E[X^2|B] + \Pr(C) \E[X^2|C] = \frac 13(2.6^2 + 2.6 + 3^2 + 3 + 3.4^2 + 3.4) = \frac{36.32}3 = 12.10\bar{6}.$

The variance is therefore $12.10\overline{6} - 9 = 3.10\overline{6}$.[/sp]