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If XX is a normally distributed random variable and XX~N(μ, σ)N(μ, σ), then the z-score is:

z = x - μ σ z= x - μ σ
(1)

The z-score tells you how many standard deviations that the value xx is above (to the right of) or below (to the left of) the mean, μμ. Values of xx that are larger than the mean have positive z-scores and values of xx that are smaller than the mean have negative z-scores. If xx equals the mean, then xx has a z-score of 00.

Example 1

Suppose XX ~ N(5, 6)N(5, 6). This says that XX is a normally distributed random variable with mean μ = 5μ = 5 and standard deviation σ = 6σ = 6. Suppose x = 17x = 17. Then:

z = x - μ σ = 17 - 5 6 = 2 z= x - μ σ = 17 - 5 6 =2
(2)

This means that x = 17x = 17 is 2 standard deviations (2σ)(2σ) above or to the right of the mean μ = 5μ = 5. The standard deviation is σ = 6σ = 6.

Notice that:

5 + 2 6 = 17 (The pattern is μ + z σ = x . ) 5+26=17(The pattern isμ+zσ=x.)
(3)

Now suppose x=1x=1. Then:

z = x - μ σ = 1 - 5 6 = - 0.67 (rounded to two decimal places) z= x - μ σ = 1 - 5 6 =-0.67 (rounded to two decimal places)
(4)

This means that x = 1x = 1 is 0.67 standard deviations (- 0.67σ)(- 0.67σ) below or to the left of the mean μ = 5μ = 5. Notice that:

5 + ( -0.67 ) ( 6 ) 5+(-0.67)(6) is approximately equal to 1 (This has the pattern μ + ( -0.67 ) σ = 1 μ+(-0.67)σ=1 )

Summarizing, when zz is positive, xx is above or to the right of μμ and when zz is negative, xx is to the left of or below μμ.

Example 2

Some doctors believe that a person can lose 5 pounds, on the average, in a month by reducing his/her fat intake and by exercising consistently. Suppose weight loss has a normal distribution. Let XX = the amount of weight lost (in pounds) by a person in a month. Use a standard deviation of 2 pounds. XX~N(5, 2)N(5, 2). Fill in the blanks.

Problem 1

Suppose a person lost 10 pounds in a month. The z-score when x = 10x = 10 pounds is z = 2.5z = 2.5 (verify). This z-score tells you that x = 10x = 10 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).

Solution

This z-score tells you that x = 10x = 10 is 2.5 standard deviations to the right of the mean 5.

Problem 2

Suppose a person gained 3 pounds (a negative weight loss). Then zz = __________. This z-score tells you that x = -3x = -3 is ________ standard deviations to the __________ (right or left) of the mean.

Solution

zz = -4. This z-score tells you that x = -3x = -3 is 4 standard deviations to the left of the mean.

Suppose the random variables XX and YY have the following normal distributions: XX ~ N(5, 6) N(5, 6) and Y ~ N(2, 1)Y ~ N(2, 1). If x = 17x = 17, then z = 2z = 2. (This was previously shown.) If y = 4y = 4, what is zz?

z = y - μ σ = 4 - 2 1 = 2 where μ=2 and σ=1. z= y - μ σ = 4 - 2 1 =2where μ=2 and σ=1.
(5)

The z-score for y = 4y = 4 is z = 2z = 2. This means that 4 is z = 2z = 2 standard deviations to the right of the mean. Therefore, x = 17x = 17 and y = 4y = 4 are both 2 (of their) 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 XX ~ N(5, 6) N(5, 6) represents weight gains for one group of people who are trying to gain weight in a 6 week period and YY ~ N(2, 1) 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 = 17x = 17 and y = 4y = 4 are each 2 standard deviations to the right of their means, they represent the same weight gain relative to their means.

The Empirical Rule

If XX is a random variable and has a normal distribution with mean µµ and standard deviation σσ then the Empirical Rule says (See the figure below)

  • About 68.27% of the xx values lie between -1σσ and +1σσ of the mean µµ (within 1 standard deviation of the mean).
  • About 95.45% of the xx values lie between -2σσ and +2σσ of the mean µµ (within 2 standard deviations of the mean).
  • About 99.73% of the xx values lie between -3σσ and +3σσ of the mean µµ (within 3 standard deviations of the mean). Notice that almost all the xx values lie within 3 standard deviations of the mean.
  • The z-scores for +1σσ and –1σσ are +1 and -1, respectively.
  • The z-scores for +2σσ and –2σσ are +2 and -2, respectively.
  • The z-scores for +3σσ and –3σσ are +3 and -3 respectively.
Empirical Rule
The Empirical Rule is also known as the 68-95-99.7 Rule.

Example 3

Suppose XX has a normal distribution with mean 50 and standard deviation 6.

  • About 68.27% of the xx values lie between -1σσ = (-1)(6) = -6 and 1σσ = (1)(6) = 6 of the mean 50. The values 50 - 6 = 44 and 50 + 6 = 56 are within 1 standard deviation of the mean 50. The z-scores are -1 and +1 for 44 and 56, respectively.
  • About 95.45% of the xx values lie between -2σσ = (-2)(6) = -12 and 2σσ = (2)(6) = 12 of the mean 50. The values 50 - 12 = 38 and 50 + 12 = 62 are within 2 standard deviations of the mean 50. The z-scores are -2 and 2 for 38 and 62, respectively.
  • About 99.73% of the xx values lie between -3σσ = (-3)(6) = -18 and 3σσ = (3)(6) = 18 of the mean 50. The values 50 - 18 = 32 and 50 + 18 = 68 are within 3 standard deviations of the mean 50. The z-scores are -3 and +3 for 32 and 68, respectively.

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