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 that x = 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.

Now…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 cm, 1.27, right

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 from 1984 to 1985. Then Y ~ N(172.36, 6.34).

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).

Find the z-scores for x = 160.58 cm and y = 162.85 cm. Interpret each z-score. What can you say about x = 160.58 cm and y = 162.85 cm?

Solution:

The z-score for x = 160.58 is z = –1.5.

The z-score for y = 162.85 is z = –1.5.

Both x = 160.58 and y = 162.85 deviate the same number of standard deviations from their respective means and in the same direction.

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. About 68% of the values lie between the values 166.02 and 178.7 cm. The z-scores are –1 and 1, respectively.
2. About 95% of the values lie between the values 159.68 and 185.04 cm. The z-scores are –2 and 2, respectively.
3. About 99.7% of the values lie between the values 153.34 and 191.38 cm. The z-scores are –3 and 3, respectively.

Example

1. Find the area to the left of [latex]z=1.26[/latex].
2. Find the area to the right of [latex]z=-0.95[/latex].
3. Find the area between [latex]z=-1.74[/latex] and [latex]z=2.12[/latex]

Solution:

1. Start by drawing a sketch.

In Excel: Use =NORM.S.DIST(1.26,1)= 0.896.  89.6% of values fall to the left of [latex]z=1.26[/latex].

2. Start by drawing a sketch.

In Excel: Since the NORM.S.DIST function returns the area to the left of a z-score, we will need to use the fact that the total area under the standard normal curve is 1 to get to our answer.  Use =1-NORM.S.DIST(-0.95, 1) = 1- 0.171 = 0.829.  82.9% of values fall to the right of [latex]z=-0.95[/latex].

3. Start by drawing a sketch.

In Excel: Use =NORM.S.DIST(2.12,1) – NORM.S.DIST(-1.74,1)= 0.983 – 0.041 = 0.943.  94.3% of values fall between  [latex]z=-1.74[/latex] and [latex]z=2.12[/latex].  *Note, in this calculation we find the area to the left of the larger z-score then subtract the area to the left of the smaller z-score.