Calculate percentage of values in regions of Normal Distribution

Question

I found a question on The Data Science Manual -

What percentage of the standard normal distribution (Mean = 0, Std Dev = 1) is found in each region? (Z = Z-Score)
(a) Z > 1.13.
(b) Z < 0.18.
(c) Z > 8.
(d) |Z| < 0:5.

I know the 68-95-99.7 rule of normal distribution. But I don't know how it is derived. How do I solve this?

Answer 1

First, you should read Normal Distribution. The analytic formula of $N(0,1)$ is $$f(x)=\frac{1}{\sqrt{2\pi}} e^{-\frac{1}{2}x^2}$$ The $68-95-99.7$ rule is actually derived by calculating the area in between $1,2,3$ standard deviation. For $N(0,1)$, $\sigma=1$, $2\sigma=2$, and $3\sigma=3$.

Therefore, $$1\text{SD}=\frac{1}{\sqrt{2\pi}}\int_{-1}^1e^{-\frac{1}{2}x^2}dx=\left.\operatorname{erf}\left(\frac{x}{\sqrt{2}}\right)\right|^1_{-1}\approx0.682\bar{7}$$ $$2\text{SD}=\frac{1}{\sqrt{2\pi}}\int_{-2}^2e^{-\frac{1}{2}x^2}dx=\left.\operatorname{erf}\left(\frac{x}{\sqrt{2}}\right)\right|^2_{-2}\approx0.954\bar{5}$$ $$3\text{SD}=\frac{1}{\sqrt{2\pi}}\int_{-3}^3e^{-\frac{1}{2}x^2}dx=\left.\operatorname{erf}\left(\frac{x}{\sqrt{2}}\right)\right|^3_{-3}\approx0.997\bar{3}$$ where erf is the Error Function.

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Now consider the intervals of $z$-score you provided. Clearly, $z>1.13$ means integrating from $1.13$ to $+\infty$. Therefore, by simple calculations, you could get

1. $0.129$
2. $0.571$
3. $6.28\times10^{−16}$
4. $0.383$

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P.S. If you are studying data science, you should probably use a calculator and plug these intervals in.