Standard Normal Distribution Question
If a set of measurements are normally distributed, what percentage of these differ from the mean by more than half of the standard deviation?
Attempt: P(z > 0.5) = 1-P(z < 0.5) = 1-0.6915 = 0.3085
The correct answer is 0.0617; I'm unsure of what I'm doing wrong.
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$\begingroup$Yes, I think you are supposed to be calculating $P[|z| > 0.5]$, since you want the observation to be more than half a standard deviation away from the mean (of the standard normal) in the positive and negative directions, not merely in one of these directions. This would be evaluated as $ 1 - \int_{-0.5}^{0.5} \frac{1}{\sqrt{2\pi}} e^{\frac{-x^2}{2}} dx = 0.617$, which is similar to your answer, except you seem to have an extra decimal point, but that would be a printing mistake, most likely.
(I did not need to calculate this value once you had done it : you may have noticed that $0.3085$ is half the correct answer, due to the symmetry of the standard normal).
$\endgroup$ $\begingroup$What you have to find are the regions more than "half away", in two directions, which includes both the regions $-\infty < x < -0.5$ and $0.5 < x < \infty$.
We can use an online calculator to get that the region in between $-0.5$ and $0.5$ is equal to around $0.3829$. Since the whole distribution sums to $1$, we can subtract $0.3829$ from $1$ to get $0.6171$.
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