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There are two methods of measuring on object of length $x$. The error of the first method is normally distributed with a mean of 0 and standard deviation of $0.0056x$. The error made by the second method is normally distributed with a mean of 0 and standard deviation of $0.0044x$. Furthermore, the two measurements are independent random variables.

What is the probability that their average value is within $0.005x$ of the object's length?

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Let $ e_1$ and $e_2 $ represent the measurement errors. The distribution of $e_1$ is N($0, 0.0056x$). The distribution of $e_2$ is N($0, 0.0044x$).

$F$ represents the average error, $ F= \frac{e_1 + e_2}{2} $

$F$ is distributed N$(0, \sqrt{\frac{0.0056^2x^2 + 0.0044^2x^2}{4}} )$

The above distribution of F simplifies to N$($0, 0.00356x$)$

The question asks for the probability $-0.005 \le F \le 0.005$

This is the same as $2P(F \le 0.005)$

$2P(F \le 0.005) - 1$ = $ 2P(Z\le \frac{0.005x}{0.00356x}) - 1$

Therefore, P($-0.005 \le F \le 0.005$) $ = 0.84 $ $ $

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  • $\begingroup$ Hummmm... $P\{-0.005 \leq F \leq +0.005\} \neq 2P\{F \leq 0.005\}$ $\endgroup$ Jan 18 '13 at 15:01
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Use the fact that the sum of two independent Normals is again Normal and that the variance of the sum is equal to the sum of their variances. That will allow you to find the variance of the average, and from there the probability.

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