5.4 The Standard Normal Distribution

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 tells us 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 tells us 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]

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 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 you 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 you that [latex]x = –3[/latex] is ________ standard deviations to the __________ (right or left) of the mean.

Solution:

1. This [latex]z[/latex]-score tells you 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 you that [latex]x = –3[/latex] is [latex]4[/latex] standard deviations to the left of the mean.

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 the the mean of their corresponding distributions, [latex]x=17[/latex] and [latex]y=4[/latex] have the same relative position.

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

a. 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 you that [latex]x = 168[/latex] 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 [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 you 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

From 2009 to 2010, the height 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.  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.