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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 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. The transformation z = produces the distribution Z ~ N (0, 1). The value x comes from a normal distribution with mean μ and standard deviation σ .
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.
Notice that: 5 + (2)(6) = 17 (The pattern is μ + zσ = x )
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. Notice that: 5 + (–0.67)(6) is approximately equal to one (This has the pattern μ + (–0.67)σ = 1)
Summarizing, when z is positive, x is above or to the right of μ and when z is negative, x is to the left of or below μ . Or, when z is positive, x is greater than μ , and when z is negative x is less than μ .
What is the z -score of x , when x = 1 and X ~ N (12,3)?
Some doctors believe that a person can lose five pounds, on the 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). Fill in the blanks.
a. Suppose a person lost ten pounds in a month. The z -score when x = 10 pounds is z = 2.5 (verify). This z -score tells you that x = 10 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).
a. This
z -score tells you that
x = 10 is
2.5 standard deviations to the
right of the mean
five .
b. 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.
b.
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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