# 6.1: Properties of a Normal Distribution

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Mathematically a normal distribution is defined by the equation

\[P(x) = \frac {1} {\sqrt{2 \pi \sigma^2}} e^{-(x - \mu)^2/(2 \sigma^2)} \nonumber\]

where \(P(x)\) is the probability of obtaining a result, \(x\), from a population with a known mean, \(\mu\), and a known standard deviation, \(\sigma\). Figure \(\PageIndex{1}\) shows the normal distribution curves for \(\mu = 0\) with standard deviations of 5, 10, and 20.

Because the equation for a normal distribution depends solely on the population’s mean, \(\mu\), and its standard deviation, \(\sigma\), the probability that a sample drawn from a population has a value between any two arbitrary limits is the same for all populations. For example, Figure \(\PageIndex{2}\) shows that 68.26% of all samples drawn from a normally distributed population have values within the range \(\mu \pm 1\sigma\), and only 0.14% have values greater than \(\mu + 3\sigma\).

This feature of a normal distribution—that the area under the curve is the same for all values of \(\sigma\)—allows us to create a probability table (see Appendix 1) based on the relative deviation, \(z\), between a limit, *x*, and the mean, \(\mu\).

\[z = \frac {x - \mu} {\sigma} \nonumber\]

The value of \(z\) gives the area under the curve between that limit and the distribution’s closest tail, as shown in Figure \(\PageIndex{3}\).

Suppose we know that \(\mu\) is 5.5833 ppb Pb and that \(\sigma\) is 0.0558 ppb Pb for a particular standard reference material (SRM). What is the probability that we will obtain a result that is greater than 5.650 ppb if we analyze a single, random sample drawn from the SRM?

**Solution**

Figure \(\PageIndex{4}\) shows the normal distribution curve given values of 5.5833 ppb Pb for \(\mu\) and of 0.0558 ppb Pb \(\sigma\). The shaded area in the figures is the probability of obtaining a sample with a concentration of Pb greater than 5.650 ppm. To determine the probability, we first calculate \(z\)

\[z = \frac {x - \mu} {\sigma} = \frac {5.650 - 5.5833} {0.0558} = 1.195 \nonumber\]

Next, we look up the probability in Appendix 1 for this value of \(z\), which is the average of 0.1170 (for \(z = 1.19\)) and 0.1151 (for \(z = 1.20\)), or a probability of 0.1160; thus, we expect that 11.60% of samples will provide a result greater than 5.650 ppb Pb.

Example \(\PageIndex{1}\) considers a single limit—the probability that a result exceeds a single value. But what if we want to determine the probability that a sample has between 5.580 g Pb and 5.625 g Pb?

**Solution**

In this case we are interested in the shaded area shown in Figure \(\PageIndex{5}\). First, we calculate \(z\) for the upper limit

\[z = \frac {5.625 - 5.5833} {0.0558} = 0.747 \nonumber\]

and then we calculate \(z\) for the lower limit

\[z = \frac {5.580 - 5.5833} {0.0558} = -0.059 \nonumber\]

Then, we look up the probability in Appendix 1 that a result will exceed our upper limit of 5.625, which is 0.2275, or 22.75%, and the probability that a result will be less than our lower limit of 5.580, which is 0.4765, or 47.65%. The total unshaded area is 71.4% of the total area, so the shaded area corresponds to a probability of

\[100.00 - 22.75 - 47.65 = 100.00 - 71.40 = 29.6 \% \nonumber\]