7.3: Probability from Gaussian distribution

6. Using standard probability density functions
   1. Some unique forms of pdfs (sec 16.3)
      1. Bernoulli -- only two f possible outcomes, but probability can be unequal
      2. Binomial -- multiple outcomes are possible, but distribution is uniform
      3. Poisson -- number of event opportunities becomes so large that likelihood is extremely small (i.e. earthquake \(\rightarrow\) tidal wave \(\rightarrow\) coastal nuclear power plant \(\rightarrow\) sea wall that is too short \ldots)
      4. Weibull -- time to failure of component (i.e. 99.999999% success of devices on spacecraft)
      5. Normal (Gaussian) -- standard bell curve
      6. Student t -- like Normal, but with finite number of data; sample instead of continuous population
   2. Normal distribution function (also referred to as Gaussian)
      1. If assuming an infinite number of possibilities (\(-\infty < x < \infty\) AND infinite number of decimal points), then function is \[ p(x) = \frac{1}{\sigma \sqrt{2\pi}}\mathrm{e}^{\frac{-(x-x')^{2}}{2\sigma^{2}}} \] where \(x'\) is the true mean, \(\sigma\) is the standard deviation, and \(\pi\) is an irrational but cool number NOT always related to circles
      2. Sketch is a bell curve with 1st central moment around \(x'\)
      3. Follows Central Limit Theorem: when a random process results from the summation of many elementary random processes, the result tends toward a Gaussian distribution. Examples include:
         • Noise of raindrops on a roof
         • Size distribution of grains on sandpaper
         • Static from analog radio/TV signals
   3. Using a normalized Gaussian distribution
      1. Clarify normalized relative to normal distribution: like dividing \(p(x)\) by the magnitude to get a unit vector
      2. Instead of using units of \(x\), define the variable as number of standard deviations \(x\) is from the mean \[ \beta = \frac{x-x'}{\sigma} \]
      3. Seeking a specific value through \(z\) substitution: \[ z_{1} = \frac{x_{1}-x'}{\sigma} \]
      4. Calculate the probability of \(x\) from the functional form of the probability density function (i.e. integrate \(p(x)\)) \[ P(x'-\delta x \leq x \leq x'+\delta x) = \int_{x'-\delta x}^{x'+\delta x} p(x) dx \]
      5. Instead of using range \(\delta x\), just define as specific values: \(+x_{1} = x'+\delta x\) and \(-x_{1}=x'-\delta x\) \[ P(- x_{1} \leq x \leq +x_{1}) = \int_{-x_{1}}^{+x_{1}} p(x) dx \]
      6. But can also use the \(\beta\) and \(z_{1}\) if solving for the \(x\) terms \[x = \beta \sigma + x' \] \[ x_{1} = z_{1}\sigma + x' \] \[ dx = \sigma d\beta \] \[ P(-z_{1}\sigma + x' \leq \beta\sigma+x' \leq z_{1}\sigma + x') = \int_{-z_{1}\sigma + x'}^{z_{1}\sigma + x'} \frac{1}{\sigma \sqrt{2\pi}}\mathrm{e}^{\frac{-\beta^{2}}{2}}\sigma d\beta \]
      7. Since \(x'\) everywhere as a shift, it can be removed
      8. Additionally, there is a symmetry to the function to make it one sided \[ P(0\leq \beta \leq z_{1}) = 2\int_{0}^{z_{1}} \frac{1}{\sqrt{2\pi}}\mathrm{e}^{-\beta^{2}/2} d\beta \] where integral is calculated as the symmetry of the cumulative distribution function (normcdf call in Matlab) 
         • Tables of values for integration to standard deviations below and above the mean
         • Conversion also possible through the similarity to error function \[ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{z_{1}} \mathrm{e}^{-\beta^{2}/2} d\beta = \frac{1}{2}\left[1+\mathrm{erf}\left(\frac{z_{1}}{\sqrt{2}}\right)\right]  \] where the integral output is an available function call in Matlab (erf)
         • To make single sided, half of the distribution is considered \[\frac{1}{\sqrt{2\pi}}\int_{0}^{z_{1}} \mathrm{e}^{-\beta^{2}/2} d\beta = \frac{1}{2}\left[\mathrm{erf}\left(\frac{z_{1}}{\sqrt{2}}\right)\right]   \]
         • Historic resource: table 17.2 on page 364 of Dunn & Davis textbook provides values of \(P(z_{1}) = normcdf(z_{1},0,1) - 0.5\)
         •  
      9. Example:

         Exercise \(\PageIndex{1}\)

         Electronic capacitors have Gaussian distribution with mean of 145 \(\mu\)F and standard deviation of 33 \(\mu\)F. The compute the probability that the capacitance is:

         • greater than 189 \(\mu\)F
         • less than 120 \(\mu\)F
         • between 121 and 167 \(\mu\)F

         Answer

         For the first scenario \(z_{1} = \frac{189-145}{33} = \frac{4}{3} \). Since seeking probability greater than value, the complementary cumulative is needed. From the table lookup at 1.33 decimal precision, probably is 9.17%. 

         Scenario #2: \(z_{1} = \frac{120-145}{33} = \frac{-25}{33} = -0.76 \). The cumulative integration less than -0.76 standard deviations is 22.36%.

         The third scenario has a top and bottom boundary for the integration

         • \(z_{low} = \frac{121-145}{33} = \frac{-8}{11} = -0.73 \) for the left boundary
         • \(z_{high} = \frac{167-145}{33} = \frac{2}{3} = 0.67 \) for the right boundary

         Multiple approaches could be considered:

         1. Using one-sided cumulative boundary function (\(\int_{0}^{z^{*}} p(x) dx\)) the absolute values of the standard deviations can be used: \(| z_{low}| \rightarrow 26.73\%\) probability and \(| z_{high}| \rightarrow 24.86\%\) for a total of 51.59% probability
         2. The difference of the full integration can be done: (\(\int_{-\infty}^{z_{high}} - \int_{-\infty}^{z_{low}}\)) where the lookup tables would have 74.86 - 23.27 = 51.59%
         3. Use the edges of the integration subtracted from unity (\(1 - \int_{-\infty}^{z_{low}} - \int_{z_{high}}^{\infty}\)) where the last term is the complementary cumulative: 100 - 23.27 - 25.14 = 51.59%