95% of the area under the standard normal distribution lies within 1.96 standard deviations away from the mean (0). This 1.96 number is used to construct 95% confidence intervals. I was just wondering... how is this 1.96 value derived? Looking at the standard normal distribution, I know the following statement is true:
$$\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^{-z}e^{-x^2/2}dx=\dfrac{1}{\sqrt{2\pi}}\int_{z}^{\infty}e^{-x^2/2}dx=0.025$$
Where z = 1.96. These areas are equal to 0.025 because they represent the leftover areas under the standard normal distribution below -1.96 and above 1.96, which adds up to 5% or 0.05. I want to know how I can mathematically show that z = 1.96.
I tried differentiating both sides with respect to z so I could make use of the fundamental theorem of calculus and solve for z to show that it had to equal 1.96, but that didn't work out for me... kept getting an undefined answer for z. Am I going about this the right way? Or is there a better way to explicitly show that z = 1.96? This is just something I've been curious about regarding z scores.