Calculate P(|Xn| > .08), n = 16

[Addez123]

Homework Statement

X1, X2.. are independend, equally distributed measuring errors with expected value .1 and standard deviation 8.
Xn = (X1 + X2 + ... + Xn) / n
Calculate P(|Xn| > .08) for n = 16

Relevant Equations

$$X + Y = N(u_x + u_y, \sqrt{\sigma_x^2 + \sigma_y^2}$$

Here's how I tried to solve it.
$$Xn = N(.1 * n, \sigma*\sqrt{n}) / n = N(.1, 8 / \sqrt{16}) = N(.1, 2)$$
$$P(|Xn| > .08) = 1 - P(-.08 < x < .08) = $$
$$1 - (1 - 2P(x < -.08)) = 2P(x < -.08) =$$
$$2*\phi(\frac {-.08 - .1}{2}) = 2\phi(-.09) =$$
$$2 * (1 - \phi(.09)) = 2 * (1 - .5358) = .9284$$

Answer in my book says .968, its too much to be a rounding error.

 

[mjc123]

Is P(x<-.08) + P(x>.08) = 2P(x<-.08) when μ≠0?

 

[Addez123]

Is P(x<-.08) + P(x>.08) = 2P(x<-.08) when μ≠0?

It's not. But when I calculate the probability I convert it to N(0, 1) distribution:
$$2*\phi(\frac {-.08 - .1}{2}) $$

 

[mjc123]

But that is not right. Do you see why? Your limits are not symmetrical about the mean.

 

[Addez123]

When working with P(x.. I assume the u = 0.
When I later calculate P I use standard normal distribution doing:
$$2*\phi(\frac {-.08 - .1}{2}) $$
instead of just
$$2*\phi(-.08) $$

I don't see how that's wrong?

 

[mjc123]

The limits are -.08 and +.08. The mean is 0.1. Translated into a standard normal distribution, you are looking at the range -.09 to -.01. The probability is not 2φ(-.09). It is φ(-.09) + 1-φ(-.01).

 

[Addez123]

Aha, that was the issue. Re did it and now I get the correct answer, thanks!