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5.3 The Standard Normal Distribution

LEARNING OBJECTIVES

  • Recognize the standard normal probability distribution and apply it appropriately.

The standard normal distribution is the normal distribution with [latex]\mu=0[/latex] and [latex]\sigma=1[/latex].  The normal random variable associated with the standard normal distribution is denoted [latex]Z[/latex].

For any normal distribution with mean [latex]\mu[/latex] and standard deviation [latex]\sigma[/latex], a [latex]z[/latex]-score is the number of the standard deviations a value [latex]x[/latex] is from the mean. For example, if a normal distribution has [latex]\mu=5[/latex] and [latex]\sigma=2[/latex], then for [latex]x=11[/latex]

[latex]\displaystyle{11=x=\mu+z\times\sigma=5+3\times 2}[/latex]

In this case, [latex]z=3[/latex]. We would say that [latex]11[/latex] is three standard deviations above (or to the right of) the mean.

The standard normal distribution is the normal distribution of these standardized [latex]z[/latex]-scores.   For any normal distribution with mean [latex]\mu[/latex] and standard deviation [latex]\sigma[/latex], we can transform the normal distribution to the standard normal distribution using the formula

[latex]\displaystyle{z=\frac{x-\mu}{\sigma}}[/latex]

where [latex]x[/latex] is a value from the normal distribution.  The [latex]z[/latex]-score is the number of standard deviations the value [latex]x[/latex] is above (to the right of) or below (to the left of) the mean [latex]\mu[/latex]. Values of [latex]x[/latex] that are larger than the mean have positive [latex]z[/latex]-scores, and values of [latex]x[/latex] that are smaller than the mean have negative [latex]z[/latex]-scores. If [latex]x[/latex] equals the mean, then [latex]x[/latex] has a [latex]z[/latex]-score of zero.

EXAMPLE

Suppose a normal distribution has mean [latex]\mu=5[/latex] and standard deviation [latex]\sigma=6[/latex].

For [latex]x=17[/latex], the [latex]z[/latex]-score is

[latex]\begin{eqnarray*}z&=&\frac{x-\mu}{\sigma}\\&=&\frac{17-5}{6}\\&=&2\end{eqnarray*}[/latex]

This says that [latex]x=17[/latex] is two standard deviations ([latex]2\times\sigma[/latex]) above or to the right of the mean [latex]\mu=5[/latex]. Notice that [latex]x=5+2\times 6=17[/latex]

For [latex]x=1[/latex], the [latex]z[/latex]-score is

[latex]\begin{eqnarray*}z&=&\frac{x-\mu}{\sigma}\\&=&\frac{1-5}{6}\\&=&-0.666...\end{eqnarray*}[/latex]

This says that [latex]x=1[/latex] is [latex]0.666...[/latex] standard deviations ([latex]-0.666...\times\sigma[/latex]) below or to the left of the mean [latex]\mu=5[/latex]. Notice that [latex]x=5+(-0.666...)\times 6=1[/latex]

NOTES

  • When [latex]z[/latex] is positive, [latex]x[/latex] is above or to the right of the mean [latex]\mu[/latex]. In other words, [latex]x[/latex] is greater than [latex]\mu[/latex].
  • When [latex]z[/latex] is negative, [latex]x[/latex] is below or to the left of the mean [latex]\mu[/latex]. In other words, [latex]x[/latex] is less than [latex]\mu[/latex].

TRY IT

What is the [latex]z[/latex]-score of [latex]x=1[/latex] for a normal distribution with [latex]\mu=12[/latex] and [latex]\sigma=3[/latex]?

 

Click to see Solution

 

[latex]\begin{eqnarray*}z&=&\frac{x-\mu}{\sigma}\\&=&\frac{1-12}{3}\\&=&-3.666...\end{eqnarray*}[/latex]

Video: “ck12.org normal distribution problems: z-score | Probability and Statistics | Khan Academy” by Khan Academy [7:48] is licensed under the Standard YouTube License.Transcript and closed captions available on YouTube.

EXAMPLE

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 the amount of weight (in pounds) a person loses in a month has a normal distribution with [latex]\mu=5[/latex] and [latex]\sigma=2[/latex].  Fill in the blanks.

  1. Suppose a person lost ten pounds in a month. The [latex]z[/latex]-score when [latex]x=10[/latex] pounds is [latex]z=2.5[/latex] (verify). This [latex]z[/latex]-score tells us that [latex]x=10[/latex] 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 [latex]z[/latex]= __________. This [latex]z[/latex]-score tells us that [latex]x=–3[/latex] is ________ standard deviations to the __________ (right or left) of the mean.

Solution

  1. This [latex]z[/latex]-score tells us that [latex]x=10[/latex] is [latex]2.5[/latex] standard deviations to the right of the mean [latex]5[/latex].
  2. [latex]z=–4[/latex]. This [latex]z[/latex]-score tells us that [latex]x=–3[/latex] is [latex]4[/latex] standard deviations to the left of the mean.

EXAMPLE

Suppose [latex]X[/latex] is a normal random variable with [latex]\mu=5[/latex] and [latex]\sigma=6[/latex] and [latex]Y[/latex] is a normal random variable with [latex]\mu=2[/latex] and [latex]\sigma=1[/latex].

Suppose [latex]x=17[/latex]:

[latex]\begin{eqnarray*}z&=&\frac{x-\mu}{\sigma}\\&=&\frac{17-5}{6}\\&=&2\end{eqnarray*}[/latex]

The [latex]z[/latex]-score for [latex]x=17[/latex] is [latex]z=2[/latex], which means that [latex]17[/latex] is [latex]2[/latex] standard deviations to the right of the mean [latex]\mu=5[/latex].

Suppose [latex]y=4[/latex]:

[latex]\begin{eqnarray*}z&=&\frac{y-\mu}{\sigma}\\&=&\frac{4-2}{1}\\&=&2\end{eqnarray*}[/latex]

The [latex]z[/latex]-score for [latex]y=4[/latex] is [latex]z=2[/latex], which means that [latex]4[/latex] is [latex]2[/latex] standard deviations to the right of the mean [latex]\mu=2[/latex].

Therefore, [latex]x=17[/latex] and [latex]y=4[/latex] are both two (of their own) standard deviations to the right of their respective means. In other words, compared to the mean of their corresponding distributions, [latex]x=17[/latex] and [latex]y=4[/latex] have the same relative position.

NOTE

The [latex]z[/latex]-score allows us to compare data that are scaled differently by considering the data’s position relative to its mean. To understand the concept, suppose [latex]X[/latex] represents weight gains for one group of people who are trying to gain weight in a six-week period, and [latex]Y[/latex] measures the same weight gain for a second group of people. A negative weight gain would be a weight loss. Because [latex]x=17[/latex] and [latex]y=4[/latex] are each two standard deviations to the right of their means, they represent the same standardized weight gain relative to their means.

TRY IT

Fill in the blanks.

Jerome averages [latex]16[/latex] points a game with a standard deviation of [latex]4[/latex] points. Suppose Jerome scores [latex]10[/latex] points in a game. The [latex]z[/latex]–score when [latex]x=10[/latex] is [latex]–1.5[/latex].  This score tells us that [latex]x=10[/latex] is _____ standard deviations to the ______(right or left) of the mean______(What is the mean?).

 

Click to see Solution

 

1.5, left, 16

EXAMPLE

The heights of 15 to 18-year-old males from Chile from 2009 to 2010 follows a normal distribution with mean [latex]170[/latex] cm and standard deviation [latex]6.28[/latex] cm.

  1. Suppose a 15 to 18-year-old male from Chile was [latex]168[/latex] cm tall from 2009 to 2010. The [latex]z[/latex]-score when [latex]x=168[/latex] cm is [latex]z[/latex] = _______.  This [latex]z[/latex]-score tells us that [latex]x=168[/latex] 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 [latex]z[/latex]-score of [latex]z=1.27[/latex]. What is the male’s height? The [latex]z[/latex]-score ([latex]z=1.27[/latex]) tells us that the male’s height is ________ standard deviations to the __________ (right or left) of the mean.

Solution

  1. [latex]–0.32[/latex], [latex]0.32[/latex], left, [latex]170[/latex]
  2. [latex]177.98[/latex], [latex]1.27[/latex], right

TRY IT

The heights of 15 to 18-year-old males from Chile from 2009 to 2010 follows a normal distribution with mean [latex]170[/latex] cm and standard deviation [latex]6.28[/latex] cm.

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

 

  1. [latex]\displaystyle{z=\frac{x-\mu}{\sigma}=\frac{176-170}{6.28}=0.96}[/latex]. This [latex]z[/latex]-score tells us that [latex]x=176[/latex] cm is [latex]0.96[/latex] standard deviations to the right of the mean [latex]170[/latex] cm.
  2. [latex]\displaystyle{x=\mu+z\times\sigma=170+(-2)\times 6.28=157.44}[/latex] cm. The [latex]z[/latex]-score ([latex]z=–2[/latex]) tells us that the male’s height is [latex]2[/latex] standard deviations to the left of the mean.

EXAMPLE

From 2009 to 2010, the heights of 15 to 18-year-old males from Chile from 2009 to 2010 follows a normal distribution with a mean of [latex]170[/latex] cm and a standard deviation of [latex]6.28[/latex] cm. Let [latex]X[/latex] be the height of a 15 to 18-year-old male from Chile in 2009 to 2010.

From 1984 to 1985, the heights of 15 to 18-year-old males from Chile follows a normal distribution with mean [latex]172.36[/latex] cm and standard deviation [latex]6.34[/latex] cm. Let [latex]Y[/latex] be the height of a 15 to 18-year-old male from Chile in 1984 to 1985.

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

Solution

The [latex]z[/latex]-score for [latex]x=160.58[/latex] is

[latex]\begin{eqnarray*}z&=&\frac{x-\mu}{\sigma}\\&=&\frac{160.58-170}{6.28}\\&=&-1.5\end{eqnarray*}[/latex]

The [latex]z[/latex]-score for [latex]y=162.58[/latex] is

[latex]\begin{eqnarray*}z&=&\frac{y-\mu}{\sigma}\\&=&\frac{162.85-172.36}{6.34}\\&=&-1.5\end{eqnarray*}[/latex]

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

TRY IT

In 2012, [latex]1,664,479[/latex] students took the SAT exam. The distribution of scores in the verbal section of the SAT followed a normal distribution with a mean of [latex]496[/latex] and a standard deviation of [latex]114[/latex].

Find the [latex]z[/latex]-scores for Student 1 with a score of [latex]325[/latex] and for Student 2 with a score of [latex]366.21[/latex]. Interpret each [latex]z[/latex]-score. What can we say about these two students’ scores?

 

Click to see Solution

 

For Student 1:

[latex]\begin{eqnarray*}z&=&\frac{x-\mu}{\sigma}\\&=&\frac{325-496}{114}\\&=&-1.5\end{eqnarray*}[/latex]

For Student 2:

[latex]\begin{eqnarray*}z&=&\frac{x-\mu}{\sigma}\\&=&\frac{366.21-496}{114}\\&=&-1.138...\end{eqnarray*}[/latex]

Student 2 scored closer to the mean than Student 1, and because they both had negative [latex]z[/latex]-scores, Student 2 had the better score.

Exercises

  1. What does a [latex]z[/latex]-score measure?
    Click to see Answer

    The number of standard deviations a value is from the mean of the distribution.

     

  2. What is the [latex]z[/latex]-score of [latex]x=12[/latex] if it is two standard deviations to the right of the mean?
    Click to see Answer

    [latex]2[/latex]

     

  3. What is the [latex]z[/latex]-score of [latex]x=9[/latex] if it is [latex]1.5[/latex] standard deviations to the left of the mean?
    Click to see Answer

    [latex]-1.5[/latex]

     

  4. What is the [latex]z[/latex]-score of [latex]x=–2[/latex] if it is [latex]2.78[/latex] standard deviations to the right of the mean?
    Click to see Answer

    [latex]-2.78[/latex]

     

  5. What is the [latex]z[/latex]-score of [latex]x=7[/latex] if it is [latex]0.133[/latex] standard deviations to the left of the mean?
    Click to see Answer

    [latex]-0.133[/latex]

     

  6. Suppose [latex]X[/latex] is a normal random variable with a mean of [latex]2[/latex] and a standard deviation of [latex]6[/latex]. What value of [latex]x[/latex] has a [latex]z[/latex]-score of [latex]3[/latex]?
    Click to see Answer

    [latex]20[/latex]

     

  7. Suppose [latex]X[/latex] is a normal random variable with a mean of [latex]8[/latex] and a standard deviation of [latex]1[/latex]. What value of [latex]x[/latex] has a [latex]z[/latex]-score of [latex]-2.25[/latex]?
    Click to see Answer

    [latex]5.75[/latex]

     

  8. Suppose [latex]X[/latex] is a normal random variable with a mean of [latex]9[/latex] and a standard deviation of [latex]5[/latex]. What value of [latex]x[/latex] has a [latex]z[/latex]-score of [latex]-0.5[/latex]?
    Click to see Answer

    [latex]6.5[/latex]

     

  9. Suppose [latex]X[/latex] is a normal random variable with a mean of [latex]2[/latex] and a standard deviation of [latex]3[/latex]. What value of [latex]x[/latex] has a [latex]z[/latex]-score of [latex]-0.67[/latex]?
    Click to see Answer

    [latex]-0.01[/latex]

     

  10. Suppose [latex]X[/latex] is a normal random variable with a mean of [latex]4[/latex] and a standard deviation of [latex]2[/latex]. What value of [latex]x[/latex] is [latex]1.5[/latex] standard deviations to the left of the mean?
    Click to see Answer

    [latex]1[/latex]

     

  11. Suppose [latex]X[/latex] is a normal random variable with a mean of [latex]4[/latex] and a standard deviation of [latex]2[/latex]. What value of [latex]x[/latex] is [latex]2[/latex] standard deviations to the right of the mean?
    Click to see Answer

    [latex]8[/latex]

     

  12. Suppose [latex]X[/latex] is a normal random variable with a mean of [latex]8[/latex] and a standard deviation of [latex]9[/latex]. What value of [latex]x[/latex] is [latex]0.67[/latex] standard deviations to the left of the mean?
    Click to see Answer

    [latex]1.97[/latex]

     

  13. Suppose [latex]X[/latex] is a normal random variable with a mean of [latex]-1[/latex] and a standard deviation of [latex]2[/latex]. What is the [latex]z[/latex]-score of [latex]x=2[/latex]?
    Click to see Answer

    [latex]1.5[/latex]

     

  14. Suppose [latex]X[/latex] is a normal random variable with a mean of [latex]12[/latex] and a standard deviation of [latex]6[/latex]. What is the [latex]z[/latex]-score of [latex]x=3[/latex]?
    Click to see Answer

    [latex]-1.5[/latex]

     

  15. Suppose [latex]X[/latex] is a normal random variable with a mean of [latex]9[/latex] and a standard deviation of [latex]3[/latex]. What is the [latex]z[/latex]-score of [latex]x=9[/latex]?
    Click to see Answer

    [latex]0[/latex]

     

  16. Suppose a normal distribution has a mean of [latex]6[/latex] and a standard deviation of [latex]1.5[/latex]. What is the [latex]z[/latex]-score of [latex]x=3.975[/latex]?
    Click to see Answer

    [latex]-1.35[/latex]

     

  17. In a normal distribution, [latex]x=5[/latex] and [latex]z=–1.25[/latex]. This tells you that [latex]x=5[/latex] is ____ standard deviations to the ____ (right or left) of the mean.
    Click to see Answer

    [latex]1.25[/latex], left

     

  18. In a normal distribution, [latex]x=3[/latex] and [latex]z=0.67[/latex]. This tells you that [latex]x=3[/latex] is ____ standard deviations to the ____ (right or left) of the mean.
    Click to see Answer

    [latex]0.67[/latex], right

     

  19. In a normal distribution, [latex]x=–2[/latex] and [latex]z=6[/latex]. This tells you that [latex]x=–2[/latex] is ____ standard deviations to the ____ (right or left) of the mean.
    Click to see Answer

    [latex]6[/latex], right

     

  20. In a normal distribution, [latex]x=–5[/latex] and [latex]z=-3.14[/latex]. This tells you that [latex]x=–5[/latex] is ____ standard deviations to the ____ (right or left) of the mean.
    Click to see Answer

    [latex]3.14[/latex], left

     

  21. In a normal distribution, [latex]x=6[/latex] and [latex]z=–1.7[/latex]. This tells you that [latex]x=6[/latex] is ____ standard deviations to the ____ (right or left) of the mean.
    Click to see Answer

    [latex]1.7[/latex], left

     

  22. The patient recovery time from a particular surgical procedure is normally distributed with a mean of [latex]5.3[/latex] days and a standard deviation of [latex]2.1[/latex] days.
    1. What is the median recovery time?
    2. What is the [latex]z[/latex]-score for a patient who takes ten days to recover?
    3. What is the [latex]z[/latex]-score for a patient who takes three days to recover?
    Click to see Answer
    1. [latex]5.3[/latex] days
    2. [latex]2.238[/latex]
    3. [latex]-1.095[/latex]

     

  23. The heights of the 430 National Basketball Association players were listed on team rosters at the start of the 2005–2006 season. The heights of basketball players have an approximate normal distribution with a mean of [latex]79[/latex] inches and a standard deviation of [latex]3.89[/latex] inches.
    1. Calculate the [latex]z[/latex]-score for a player with a height of [latex]77[/latex] inches. Interpret the [latex]z[/latex]-score.
    2. Calculate the [latex]z[/latex]-score for a player with a height of [latex]85[/latex] inches. Interpret the [latex]z[/latex]-score.
    3. If an NBA player reported his height had a [latex]z[/latex]-score of [latex]3.5[/latex], would you believe him? Explain your answer.
    Click to see Answer
    1. [latex]-0.514[/latex]; The height of [latex]77[/latex] inches is [latex]0.514[/latex] standard deviations to the left of the mean.
    2. [latex]1.542[/latex]; The height of [latex]85[/latex] inches is [latex]1.542[/latex] standard deviations to the right of the mean.
    3. No, because the player would need to be [latex]92.615[/latex] inches (over [latex]7[/latex] feet, [latex]7[/latex] inches) tall.

     

  24. The systolic blood pressure (given in millimetres) of males has an approximately normal distribution with a mean of [latex]125[/latex] and a standard deviation of [latex]14[/latex]. Systolic blood pressure for males follows a normal distribution.
    1. Calculate the [latex]z[/latex]-scores for the male systolic blood pressures [latex]100[/latex] and [latex]150[/latex] millimetres.
    2. If a male friend of yours said he thought his systolic blood pressure was [latex]2.5[/latex] standard deviations below the mean but that he believed his blood pressure was between [latex]100[/latex] and [latex]150[/latex] millimetres, what would you say to him?
    Click to see Answer
    1. [latex]-1.786[/latex], [latex]1.786[/latex]
    2. His systolic blood pressure cannot be between [latex]100[/latex] and [latex]150[/latex] millimetres because at [latex]2.5[/latex] standard deviations below the mean, his blood pressure is [latex]90[/latex] millimetres.

     

  25. Kyle’s doctor told him that the [latex]z[/latex]-score for his systolic blood pressure is [latex]1.75[/latex]. The systolic blood pressure (given in millimetres) of males has an approximately normal distribution with a mean of [latex]125[/latex] and a standard deviation of [latex]14[/latex].
    1. Which of the following is the best interpretation of this standardized score?
      1. Kyle’s systolic blood pressure is [latex]175[/latex].
      2. Kyle’s systolic blood pressure is [latex]1.75[/latex] times the average blood pressure of men his age.
      3. Kyle’s systolic blood pressure is [latex]1.75[/latex] above the average systolic blood pressure of men his age.
      4. Kyles’s systolic blood pressure is [latex]1.75[/latex] standard deviations above the average systolic blood pressure for men.
    2. Calculate Kyle’s blood pressure.
    Click to see Answer
    1. iv
    2. [latex]149.5[/latex]

     

  26. Height and weight are two measurements used to track a child’s development. The World Health Organization measures child development by comparing the weights of children who are the same height and the same gender. In 2009, weights for all [latex]80[/latex] cm girls in the reference population had a mean of [latex]10.2[/latex] kg and a standard deviation of [latex]0.8[/latex] kg. Weights are normally distributed. Calculate the [latex]z[/latex]-scores that correspond to the following weights and interpret them.
    1. [latex]11[/latex] kg
    2. [latex]7.9[/latex] kg
    3. [latex]12.2[/latex] kg
    Click to see Answer
    1. [latex]1[/latex]; The weight of [latex]11[/latex] kg is [latex]1[/latex] standard deviations to the right of the mean.
    2. [latex]-2.875[/latex]; The weight of [latex]7.9[/latex] kg is [latex]2.875[/latex] standard deviations to the left of the mean.
    3. [latex]2.5[/latex]; The weight of [latex]12.2[/latex] kg is [latex]2.5[/latex] standard deviations to the right of the mean.

     

  27. In 2005, 1,475,623 students heading to college took the SAT. The distribution of scores in the math section of the SAT follows a normal distribution with a mean of [latex]520[/latex] and a standard deviation of [latex]115[/latex].
    1. Calculate the [latex]z[/latex]-score for an SAT score of [latex]720[/latex]. Interpret it using a complete sentence.
    2. What math SAT score is [latex]1.5[/latex] standard deviations above the mean? What can you say about this SAT score?
    3. For 2012, the SAT math test had a mean of [latex]514[/latex] and a standard deviation of [latex]117[/latex]. The ACT math test is an alternative to the SAT and is approximately normally distributed with a mean of [latex]21[/latex] and a standard deviation of [latex]5.3[/latex]. If one person took the SAT math test and scored [latex]700[/latex] and a second person took the ACT math test and scored [latex]30[/latex], who did better with respect to the test they took?
    Click to see Answer
    1. [latex]1.739[/latex]; The score of [latex]720[/latex] is [latex]1.739[/latex] standard deviations to the right of the mean.
    2. [latex]692.5[/latex]; The score of [latex]692.5[/latex] kg is [latex]1.5[/latex] standard deviations to the right of the mean.
    3. SAT [latex]z[/latex]-score is [latex]1.565[/latex]; ACT [latex]z[/latex]-score is [latex]1.698[/latex]; The person who wrote the ACT test did better because the [latex]z[/latex]-score for their score is higher.

     

“5.4 The Standard Normal Distribution” and “5.6 Exercises” from Introduction to Statistics by Valerie Watts is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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