# 2.3 Normal distribution

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This course is a short series of lectures on Introductory Statistics. Topics covered are listed in the Table of Contents. The notes were prepared by EwaPaszek and Marek Kimmel. The development of this course has been supported by NSF 0203396 grant.

## Normal distribution

The normal distribution is perhaps the most important distribution in statistical applications since many measurements have (approximate) normal distributions. One explanation of this fact is the role of the normal distribution in the Central Theorem.

Briefly, we say that X is $N\left(\mu ,{\sigma }^{2}\right)$

## Proof of the p.d.f. properties

Clearly, $f\left(x\right)>0$ . Let now evaluate the integral: $I=\underset{-\infty }{\overset{\infty }{\int }}\frac{1}{\sigma \sqrt{2\pi }}\mathrm{exp}\left[-\frac{{\left(x-\mu \right)}^{2}}{2{\sigma }^{2}}\right]dx,$ showing that it is equal to 1. In the integral, change the variables of integration by letting $z=\left(x-\mu \right)/\sigma$ . Then,

$I=\underset{-\infty }{\overset{\infty }{\int }}\frac{1}{\sqrt{2\pi }}{e}^{-{z}^{2}/2}dz,$ since $I>0$ , if ${I}^{2}=1$ , then $I=1$ .

Now ${I}^{2}=\frac{1}{2\pi }\left[\underset{-\infty }{\overset{\infty }{\int }}{e}^{-{x}^{2}/2}dx\right]\left[\underset{-\infty }{\overset{\infty }{\int }}{e}^{-{y}^{2}/2}dy\right],$ or equivalently,

${I}^{2}=\frac{1}{2\pi }\underset{-\infty }{\overset{\infty }{\int }}\underset{-\infty }{\overset{\infty }{\int }}\mathrm{exp}\left(-\frac{{x}^{2}+{y}^{2}}{2}\right)dxdy.$

Letting $x=r\mathrm{cos}\theta ,y=r\mathrm{sin}\theta$ (i.e., using polar coordinates), we have

${I}^{2}=\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{\infty }{\int }}{e}^{-{r}^{2}/2}rdrd\theta =\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}d\theta =\frac{1}{2\pi }2\pi =1.$

The mean and the variance of the normal distribution is as follows:

$E\left(X\right)=\mu$ and $Var\left(X\right)={\mu }^{2}+{\sigma }^{2}-{\mu }^{2}={\sigma }^{2}.$

That is, the parameters $\mu$ and ${\sigma }^{2}$ in the p.d.f. are the mean and the variance of X .

If the p.d.f. of X is

$f\left(x\right)=\frac{1}{\sqrt{32\pi }}\mathrm{exp}\left[-\frac{{\left(x+7\right)}^{2}}{32}\right],-\infty then X is $N\left(-7,16\right)$

That is, X has a normal distribution with a mean $\mu$ =-7, variance ${\sigma }^{2}$ =16, and the moment generating function

$M\left(t\right)=\mathrm{exp}\left(-7t+8{t}^{2}\right).$

show that the set of all natural number form semi group under the composition of addition
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The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
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2. (x) + (x + 2) = 60 2x + 2 = 60 2x = 58 x = 29 29, 30, & 31
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Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
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find the value of 2x=32
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X=16
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use the y -intercept and slope to sketch the graph of the equation y=6x
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4
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x-2y+3z=-3 2x-y+z=7 -x+3y-z=6
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Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
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(61/11,41/11,−4/11)
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