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Some values have a normal distribution with mean .0276. What standard deviation is required so that 98% of values are between .0275 and .0278?

What I'm confused with is how to calculate the standard deviation when Z is between two intervals. I know that P(-.0001/σ < Z < .0002/σ) = .98, but I don't know where to go from here.

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$$ .0275=.0276-2*\sigma $$ $$ .0278=.0276+2*\sigma $$ The 2 comes from the fact the under a normal curve ~98% (97.5%) of the values are found within ~2(1.96) standard deviations of the mean. One just needs to solve for $\sigma$ and use the largest value, insuring that 98% is inclusive, as the answer.

So therefore a standard deviation of .0001 is required.

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  • $\begingroup$ This fails to take account of the fact that (a) the OP's interval is not symmetric about the mean (b) 1.96 cuts off 2.5% in each tail, not in total. $\endgroup$ – mdewey Apr 20 '17 at 12:34
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There is no simple way to calculate this, I believe. I'd suggest looking into numerical solutions for it.

Just to explain a bit, this is the normal distribution:

$f(x|\sigma, \mu)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$

The percentage of values in the interval [a, b] is then given by:

$F(a, b)=\int_a^b \! f(x|\sigma, \mu)dx = \int_{-\infty}^b \! f(x|\sigma, \mu)dx - \int_{-\infty}^a \! f(x|\sigma, \mu)dx$

You know F(a,b), you know a and b, and you know the mean $\mu$. What you need to do is to solve this equation for $\sigma$. However, the integral is the error function, you cannot solve it analytically, so you can't solve for $\sigma$.

Using a numerical approach, it gets easier - for example, you could calculate the normal distribution with a fixed $\sigma$ for 1000 x values, calculate the area between a and b, and then iteratively change $\sigma$ until you find the value that comes closest to 0.98.

There are also functions in most programming languages that calculate the cumulative normal distribution for given parameters, so if you want higher precision, you could use those (again with iterations of $\sigma$).

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  • $\begingroup$ Since I'm being downvoted, I assume that there's some mistake in my explanation. I'd appreciate it if someone could point it out to me! Maybe I've missed something obvious. $\endgroup$ – PoorYorick Oct 27 '16 at 12:55

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