A normal distribution is a bell-shaped distribution. Theoretically, a normal distribution is continuous and may be depicted as a density curve, such as the one below. The distribution plot below is a standard normal distribution. A standard normal distribution has a mean of 0 and standard deviation of 1. This is also known as the z distribution. You may see the notation \(N(\mu, \sigma\)) where N signifies that the distribution is normal, \(\mu\) is the mean of the distribution, and \(\sigma\) is the standard deviation of the distribution. A z distribution may be described as \(N(0,1)\).
While we cannot determine the probability for any one given value because the distribution is continuous, we can determine the probability for a given interval of values. The probability for an interval is equal to the area under the density curve. The total area under the curve is 1.00, or 100%. In other words, 100% of observations fall under the curve.
For example, in Lesson 2 we learned about the Empirical Rule which stated that approximately 68% of observations on a normal distribution will fall within one standard deviation of the mean, approximately 95% will fall within two standard deviations of the mean, and approximately 99.7% will fall within three standard deviations of the mean.
Example: SAT-Math Scores Section
The distribution of SAT-Math scores can be described as \(N(500, 100)\). Let's apply the Empirical Rule to determine the SAT-Math scores that separate the middle 68% of scores, the middle 95% of scores, and the middle 99.7% of scores.
Middle 68%: \(500\pm1(100)=[400, 600]\)
Middle 95%: \(500\pm2(100)=[300, 700]\)
Middle 99.7%: \(500\pm 3(100)= [200, 800]\)
z scores Section
In Lesson 2 we wanted to describe one observation in relation to the distribution of all observations. We did this using a z score.
- z score
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Distance between an individual score and the mean in standard deviation units; also known as a standardized score.
- z score
- \(z=\dfrac{x - \overline{x}}{s}\)
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\(x\) = original data value
\(\overline{x}\) = mean of the original distribution
\(s\) = standard deviation of the original distribution
This equation could also be rewritten in terms of population values: \(z=\dfrac{x-\mu}{\sigma}\)
Example: IQ Scores Section
IQ scores are normally distributed with a mean of 100 and standard deviation of 15. Compute the z score for an individual with an IQ score of 120.
\(z=\dfrac{x- \mu}{\sigma}\)
Here, \(x=120\), \(\mu=100\), and \(\sigma=15\).
\(z=\dfrac{120-100}{15}=\dfrac{20}{15}=1.333\)
This individual's z score is 1.333. Their IQ is 1.333 standard deviations above the mean.