# AP Statistics Curriculum 2007 Normal Std

## Contents

## General Advance-Placement (AP) Statistics Curriculum - Standard Normal Variables and Experiments

### Standard Normal Distribution

The Standard Normal Distribution is a continuous distribution with the following density:

- Standard Normal
*density*function $ f(x)= {e^{-x^2 \over 2} \over \sqrt{2 \pi}}. $ - Standard Normal
*cumulative distribution*function \(\Phi(y)= \int_{-\infty}^{y}{{e^{-x^2 \over 2} \over \sqrt{2 \pi}} dx}.\) - Why are these two functions, \(f(x), \Phi(y)\) well-defined density and distribution functions, i.e., \(\int_{-\infty}^{\infty} {f(x)dx}=1\)? See the appendix below.

Note that the following exact *areas* are bound between the Standard Normal Density Function and the x-axis on these symmetric intervals around the origin:

- The area: -1.0 < x < 1.0 = 0.8413 - 0.1587 = 0.6826
- The area: -2.0 < x < 2.0 = 0.9772 - 0.0228 = 0.9544
- The area: -3.0 < x < 3.0 = 0.9987 - 0.0013 = 0.9974
- Note that the inflection points (\(f ''(x)=0\))of the Standard Normal density function are \(\pm\) 1.

- The Standard Normal distribution is also a special case of the more general normal distribution where the mean is set to zero and the variance is set to one. The Standard Normal distribution is often called the
*bell curve*because the graph of its probability density resembles a bell.

### Experiments

Suppose we decide to test the state of 100 used batteries. To do that, we connect each battery to a volt-meter by randomly attaching the positive (+) and negative (-) battery terminals to the corresponding volt-meter's connections. Electrical current always flows from + to -, i.e., the current goes in the direction of the voltage drop. Depending upon which way the battery is connected to the volt-meter we can observe positive or negative voltage recordings (voltage is just a difference, which forces current to flow from higher to the lower voltage.) Denote \(X_i\)={measured voltage for battery i} - this is random variable with mean of 0 and unitary variance. Assume the distribution of all \(X_i\) is Standard Normal, \(X_i \sim N(0,1)\). Use the Normal Distribution (with mean=0 and variance=1) in the SOCR Distribution applet to address the following questions. This Distributions help-page may be useful in understanding SOCR Distribution Applet. How many batteries, from the sample of 100, can we expect to have?

- Absolute Voltage > 1? $P(X>1) = 0.1586 $, thus we expect 15-16 batteries to have voltage exceeding 1.

- |Absolute Voltage| > 1? $P(|X|>1) = 1- 0.682689=0.3173 $, thus we expect 31-32 batteries to have absolute voltage exceeding 1.

- Voltage < -2? $P(X<-2) = 0.0227 $, thus we expect 2-3 batteries to have voltage less than -2.

- Voltage <= -2? $P(X<=-2) = 0.0227 $, thus we expect 2-3 batteries to have voltage less than or equal to -2.

- -1.7537 < Voltage < 0.8465? $P(-1.7537 < X < 0.8465) = 0.761622 $, thus we expect 76 batteries to have voltage in this range.

### Problems

### Appendix

The derivation below illustrates why the standard normal density function, \(f(x)= {e^{-x^2 \over 2} \over \sqrt{2 \pi}}\), represents a well-defined density function, i.e., \(f(x)\ge 0\) and \(\int_{-\infty}^{\infty} {f(x)dx}=1\).

- Clearly the exponential function \(e^{-x^2 \over 2}\) is always non-negative (in fact it's strictly positive for each real value argument).
- To show that \(\int_{-\infty}^{\infty} {f(x)dx}=1\), let \(A=\int_{-\infty}^{\infty} {f(x)dx}\). Then \(A^2=\int_{-\infty}^{\infty} {f(x)dx}\times \int_{-\infty}^{\infty} {f(w)dw}\). Thus,

\[A^2=\int_{-\infty}^{\infty} {\int_{-\infty}^{\infty} { {e^{-x^2 \over 2} \over \sqrt{2 \pi}} \times {e^{-w^2 \over 2} \over \sqrt{2 \pi}} dw}dx}\],

- Change variables from Cartesian to polar coordinates:
- \[x=r\cos(\theta)\]
- \[x=r\cos(\theta)\], \(0\le \theta\le 2\pi\)
- Hence,
- \[x^2+w^2=r^2\],
- \[e^{-x^2 \over 2} \times e^{-w^2 \over 2} =e^{-r^2 \over 2}\],
- \[dx=\cos(\theta)dr\], and
- \[dy=r\cos(\theta)dr\].

- Therefore, \(A^2=\int_{0}^{\infty} {\int_{0}^{2\pi} {{e^{-r^2 \over 2} \over 2 \pi}\cos^2(\theta)rdrd\theta}}\), and

\[A^2=\int_{0}^{\infty} {\int_{0}^{2\pi} {{e^{-r^2 \over 2} \over 2 \pi}d{\frac{r^2}{2}}d\theta}}=\int_{0}^{\infty} {{e^{-r^2 \over 2} \over \pi}d{\frac{r^2}{2}}} \times \int_{0}^{2\pi} {\cos^2(\theta)d\theta}=1\], since

- \[\int_{0}^{2\pi} {\cos^2(\theta)d\theta}=\int_{0}^{2\pi} {\frac{1+\cos(2\theta)}{2}d\theta}=\pi\], and
- \[\int_{0}^{\infty} {e^{-w}dw}=1\].

- SOCR Home page: http://www.socr.ucla.edu

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