Z-Scores

Example

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:

[latex]\displaystyle{z}=\frac{{x - \mu}}{{\sigma}}[/latex]= [latex]\displaystyle{z}=\frac{{17-5}}{{6}}={2}[/latex]

This means that x = 17 is two standard deviations (2σ) 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: [latex]\displaystyle{z}=\frac{{x - \mu}}{{\sigma}}[/latex] = [latex]\displaystyle {z}=\frac{{1-5}}{{6}} = -{0.67}[/latex]

(rounded to two decimal places)

This means that x = 1 is 0.67 standard deviations (–0.67σ) 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 zis 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.

Example

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.

1. 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 thatx = 10 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).
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.

Solution:

1. This z-score tells you that x = 10 is 2.5 standard deviations to the right of the mean five.
2. z= –4. This z-score tells you that x = –3 is 4 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? [latex]\displaystyle {z}=\frac{{y - \mu}}{{\sigma}} = \frac{{4-2}}{{1}}[/latex].

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

Example

The mean height of 15 to 18-year-old males from Chile from 2009 to 2010 was 170 cm with a standard deviation of 6.28 cm. Male heights are known to follow a normal distribution. Let X = the height of a 15 to 18-year-old male from Chile in 2009 to 2010. Then X ~ N(170, 6.28).

a. Suppose a 15 to 18-year-old male from Chile was 168 cm tall from 2009 to 2010. The z-score when x = 168 cm is z = _______. This z-score tells you that x = 168 is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).

b. Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a z-score of z = 1.27. What is the male’s height? The z-score (z = 1.27) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean.

Solution:

a. –0.32, 0.32, left, 170

b. 177.98, 1.27, right

try it

Use the information from the last example to answer the following questions.

1. Suppose a 15 to 18-year-old male from Chile was 176 cm tall from 2009 to 2010. The z-score when x = 176 cm is z = _______. This z-score tells you that x = 176 cm is ________ standard deviations to the ________ (right or left) of the mean _____ (What is the mean?).
2. Suppose that the height of a 15 to 18-year-old male from Chile from 2009 to 2010 has a z-score of z = –2. What is the male’s height? The z-score (z = –2) tells you that the male’s height is ________ standard deviations to the __________ (right or left) of the mean.

Solve the equation [latex]\displaystyle{z}=\frac{{x - \mu}}{{\sigma}}[/latex] for x. x = μ + (z)(σ) for x. x = μ + (z)(σ)

1. z=0.96. This z-score tells you that x = 176 cm is 0.96 standard deviations to the right of the mean 170 cm.
2. X = 157.44 cm, The z-score(z = –2) tells you that the male’s height is two standard deviations to the left of the mean.

Example

try it

In 2012, 1,664,479 students took the SAT exam. The distribution of scores in the verbal section of the SAT had a meanµ = 496 and a standard deviation σ = 114. Let X = a SAT exam verbal section score in 2012. Then X ~ N(496, 114).

Find the z-scores for x1 = 325 and x2 = 366.21. Interpret each z-score. What can you say about x1 = 325 and x2 = 366.21?

The z-score for x1 = 325 is z1 = –1.5.

The z-score for x2 = 366.21 is z2 = –1.14.

Student 2 scored closer to the mean than Student 1 and, since they both had negative z-scores, Student 2 had the better score.

Example

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

• About 68% of the x values lie between –1σ = (–1)(6) = –6 and 1σ = (1)(6) = 6 of the mean 50. The values 50 – 6 = 44 and 50 + 6 = 56 are within one standard deviation of the mean 50. The z-scores are –1 and +1 for 44 and 56, respectively.
• About 95% of the x values lie between –2σ = (–2)(6) = –12 and 2σ = (2)(6) = 12. The values 50 – 12 = 38 and 50 + 12 = 62 are within two standard deviations of the mean 50. The z-scores are –2 and +2 for 38 and 62, respectively.
• About 99.7% of the x values lie between –3σ = (–3)(6) = –18 and 3σ= (3)(6) = 18 of the mean 50. The values 50 – 18 = 32 and 50 + 18 = 68 are within three standard deviations of the mean 50. The z-scores are –3 and +3 for 32 and 68, respectively.

Example

From 1984 to 1985, the mean height of 15 to 18-year-old males from Chile was 172.36 cm, and the standard deviation was 6.34 cm. Let Y = the height of 15 to 18-year-old males in 1984 to 1985. Then Y ~ N(172.36, 6.34).

1. About 68% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
2. About 95% of the y values lie between what two values? These values are ________________. The z-scores are ________________, respectively.
3. About 99.7% of the y values lie between what two values? These values are ________________.  The z-scores are ________________, respectively.

SOLUTION: 1. 166.02 and 178.7, -1 and 1

2. 159.68 and 185.04, -2 and 2

3. 153.34 and 191.38, -3 and 3