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The standard normal distribution is a normal distribution of standardized values called z -scores . A z -score is measured in units of the standard deviation. For example, if the mean of a normal distribution is five and the standard deviation is two, the value x = 11 is three standard deviations above (or to the right of) the mean. The calculation is as follows:
x = μ + ( z )( σ ) = 5 + (3)(2) = 11
The z -score is three.
The mean for the standard normal distribution is zero, and the standard deviation is one. What this does is dramatically simplify the mathematical calculation of probabilities. Take a moment and substitute zero and one in the appropriate places in the above formula and you can see that the equation collapses into one that can be much more easily solved using integral calculus. The transformation z = produces the distribution Z ~ N (0, 1). The value x comes from a known normal distribution with known mean μ and known standard deviation σ . The z -score tells how many standard deviations a particular x is away from the mean.
If X is a normally distributed random variable and X ~ N(μ, σ) , then the z -score is:
The z -score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ . Values of x that are larger than the mean have positive z -scores, and values of x that are smaller than the mean have negative z -scores. If x equals the mean, then x has a z -score of zero.
Suppose X ~ N(5, 6) . This says that x is a normally distributed random variable with mean μ = 5 and standard deviation σ = 6. Suppose x = 17. Then:
This means that x = 17 is two standard deviations (2 σ ) above or to the right of the mean μ = 5. The standard deviation is σ = 6.
Now suppose x = 1. Then: z = = = –0.67 (rounded to two decimal places)
This means that x = 1 is 0.67 standard deviations (–0.67 σ ) below or to the left of the mean μ = 5.
Some doctors believe that a person can lose five pounds, on average, in a month by reducing his or her fat intake and by exercising consistently. Suppose weight loss has a normal distribution. Let X = the amount of weight lost(in pounds) by a person in a month. Use a standard deviation of two pounds. X ~ N (5, 2).
Suppose a person gained three pounds (a negative weight loss). Then z = __________. This z -score tells you that x = –3 is ________ standard deviations to the __________ (right or left) of the mean.
z =
–4 . This
z -score tells you that
x = –3 is
four standard deviations to the
left of the mean.
Suppose the random variables X and Y have the following normal distributions: X ~ N (5, 6) and Y ~ N (2, 1). If x = 17, then z = 2. (This was previously shown.) If y = 4, what is z ?
z = = = 2 where µ = 2 and σ = 1.
The z -score for y = 4 is z = 2. This means that four is z = 2 standard deviations to the right of the mean. Therefore, x = 17 and y = 4 are both two (of their own ) standard deviations to the right of their respective means.
The z -score allows us to compare data that are scaled differently. To understand the concept, suppose X ~ N (5, 6) represents weight gains for one group of people who are trying to gain weight in a six week period and Y ~ N (2, 1) measures the same weight gain for a second group of people. A negative weight gain would be a weight loss. Since x = 17 and y = 4 are each two standard deviations to the right of their means, they represent the same, standardized weight gain relative to their means .
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