Given the particular normal distribution specified below, what is the probability that a random observation falls within the specified range .004 greater and less than the average?
original Lower limit is .496
original upper limit is .504
$$\mu = .500$$ defective equipment cause the average to change to $$\mu = .449$$
$$\sigma = .002$$
new Lower limit is .445
new upper limit is .453
So how do you compensate for your average changing because of defective equipment?
OLD info When I use my ti 89 normal cdf function I get 2.24 which isn't even a probability. Is there something special I need to do because my average changes or is there something special about these numbers?
If $\mu=0.5$ and $\sigma=0.002$, $P(0.496<X<0.504)=P(\frac{0.496-0.5}{0.002}<Z<\frac{0.504-0.5}{0.002})=\Phi(2)-\Phi(-2)\approx 0.9545$, where $Z$ is the standard normal r.v. and $\Phi(\cdot)$ its cumulative density function.
If $\mu=0.499$ and $\sigma=0.002$, $P(0.496<X<0.504)=P(\frac{0.495-0.499}{0.002}<Z<\frac{0.503-0.499}{0.002})=\Phi(2)-\Phi(-2)\approx 0.9545$.