Calculate standard deviation given mean and percentage 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.
 A: So you can use R to get the answer:
target=function (sd){
  b=pnorm(0.0278, mean = 0.0276, sd = sd)
  a=pnorm(0.0275, mean = 0.0276, sd = sd)
  return(abs(b-a-0.98))
}
sd=optim(1,target)
sd$par

This gives: 
> sd$par
[1] 4.868167e-05

What we are doing is using numerical method to calculate $\sigma$ such that $$F(0.0278)-F(0.0275)=0.98$$ where $F()$ is cdf for $N(0.0276,\sigma)$
A: It has to be solved numerically. Here is a solution in R using a simple root finding algorithm. We simply solve the equation
$$
F_{\mu,\sigma}(b) - F_{\mu,\sigma}(a) - p = 0
$$
where $F_{\mu,\sigma}(\cdot)$ denotes the cumulative distribution function of the normal distribution with mean $\mu$ and standard deviation $\sigma$. $b$ and $a$ (with $b>a$) are the upper and lower bounds, respectively and $p$ ($0<p<1$) is the proportion of values that lies between $a$ and $b$.
The function find_sigma is very generic: It accepts fixed arguments for $a$, $b$, $\mu$ and $p$.
find_sigma <- function(sigma, a, b, mu, prop) {
  (pnorm(b, mean = mu, sd = sigma) - pnorm(a, mean = mu, sd = sigma)) - prop
}

uniroot(
  find_sigma
  , lower = .Machine$double.xmin
  , upper = 1
  , a = 0.0275  # lower bound
  , b = 0.0278  # upper bound
  , mu = 0.0276 # mean
  , prop = 0.98 # proportion between a and b
  , maxiter = 10000
  , tol = 1e-20
  # , extendInt = "yes"
)

$root
[1] 4.868168e-05

The standard deviation is $0.000048682$ as the other answers have found.
A: 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 $\sigma$).
A: I entered on WolframAlpha: 
integral_0.0275^0.0278 (1/sqrt(2 π))/a exp(-(((x - 0.0276)/sqrt(2))/a)^2) dx = 0.98 

Which got me
0.5 erf(0.0000707107/a) + 0.5 erf(0.000141421/a) = 0.98 

Solving which assuming a is real.
gives a = 0.0000486817
Multiple expansions of error function are listed on its Wikipedia page and on math.SE but they are not very useful accurate for hand calculations. 
