We can estimate the probability of selecting a negative random
variable by evaluating the area under the curve in the region of
negative x. Approximating this area using a triangle will give us
an over-estimate of the actual probability.
Blue Curve: Pr(X< 0) $\approx$ 0.5 x 2 x 0.25 = 0.25
Red Curve: Pr(X< 0) $\approx$ 0.5 x 2 x 0.1 = 0.1
Black Curve: Pr(X< 0) $\approx$ 0.5 x 2 x 0.05 = 0.05
A normal distribution is symmetric about its mean. This allows us
to estimate the mean of each distribution by inspection:
$\mu_{Blue}$ = 1
$\mu_{Red}$ = 2
$\mu_{Black}$ = 3
We know that f(x) = $\frac{1}{ \sigma \sqrt{2 \pi}}
e^{\frac{-(x-\mu)^2}{2 \sigma ^2}}$
If we evaluate f(x) at x = $\mu$ the exponential will disappear
($e^0 = 1$)
We can then solve for $\sigma$
$f(\mu) = \frac{1}{ \sigma \sqrt{2 \pi}} $
$\sigma =\frac{1}{\sqrt{2 \pi} f(\mu)}$
Evaluating f($\mu$) from the curves and substituting $\mu$ into the
expression we find that: