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<document xmlns="http://cnx.rice.edu/cnxml" xmlns:md="http://cnx.rice.edu/mdml/0.4" xmlns:bib="http://bibtexml.sf.net/" xmlns:m="http://www.w3.org/1998/Math/MathML" id="new">
  <name>Normal Distribution: Z-scores</name>
  <metadata>
  <md:version>1.5</md:version>
  <md:created>2008/06/06 14:34:50 GMT-5</md:created>
  <md:revised>2008/07/21 03:07:26.246 GMT-5</md:revised>
  <md:authorlist>
      <md:author id="billowsky">
      <md:firstname>Barbara</md:firstname>
      
      <md:surname>Illowsky</md:surname>
      <md:email>illowskybarbara@deanza.edu</md:email>
    </md:author>
      <md:author id="sdean">
      <md:firstname>Susan</md:firstname>
      
      <md:surname>Dean</md:surname>
      <md:email>deansusan@deanza.edu</md:email>
    </md:author>
  </md:authorlist>

  <md:maintainerlist>
    <md:maintainer id="cnxorg">
      <md:firstname/>
      
      <md:surname>Connexions</md:surname>
      <md:email>cnx@cnx.org</md:email>
    </md:maintainer>
  </md:maintainerlist>
  
  <md:keywordlist>
    <md:keyword>elementary</md:keyword>
    <md:keyword>statistics</md:keyword>
  </md:keywordlist>

  <md:abstract/>
</metadata>
  <content>
    <para id="delete_me">If <m:math><m:mi>X</m:mi></m:math> is a normally distributed random variable and <m:math><m:mi>X</m:mi></m:math>~<m:math><m:mi>N(μ, σ)</m:mi></m:math>, then the z-score is:</para><equation id="element-521"><m:math>
<m:mi>z</m:mi>
<m:mo>=</m:mo>
<m:mfrac>
<m:mrow>
<m:mi>x</m:mi>
<m:mo>-</m:mo>
<m:mi>μ</m:mi>
</m:mrow>
<m:mrow>
<m:mi>σ</m:mi>
</m:mrow>
</m:mfrac>
</m:math>
</equation><para id="element-272"><emphasis>The z-score tells you how many standard deviations that the value <m:math><m:mi>x</m:mi></m:math> is above (to the
right of) or below (to the left of) the mean, <m:math><m:mi>μ</m:mi></m:math>.</emphasis> Values of <m:math><m:mi>x</m:mi></m:math> that are larger than the mean
have positive z-scores and values of <m:math><m:mi>x</m:mi></m:math> that are smaller than the mean have negative z-scores.</para><example id="element-49"><para id="element-161">Suppose <m:math><m:mi>X</m:mi></m:math> ~ <m:math><m:mi>N(5, 6)</m:mi></m:math>. This says that <m:math><m:mi>X</m:mi></m:math> is a normally distributed random
variable with mean <m:math><m:mi>μ = 5</m:mi></m:math> and standard deviation <m:math><m:mi>σ = 6</m:mi></m:math>. Suppose <m:math><m:mi>x = 17</m:mi></m:math>. Then:
</para><equation id="element-160"><m:math>
<m:mi>z</m:mi>
<m:mo>=</m:mo>
<m:mfrac>
<m:mrow>
<m:mi>x</m:mi>
<m:mo>-</m:mo>
<m:mi>μ</m:mi>
</m:mrow>
<m:mrow>
<m:mi>σ</m:mi>
</m:mrow>
</m:mfrac>
<m:mo>=</m:mo>
<m:mfrac>
<m:mrow>
<m:mn>17</m:mn>
<m:mo>-</m:mo>
<m:mn>5</m:mn>
</m:mrow>
<m:mrow>
<m:mn>6</m:mn>
</m:mrow>
</m:mfrac>
<m:mo>=</m:mo>
<m:mn>2</m:mn>
</m:math></equation><para id="element-935">This means that <m:math><m:mi>x = 17</m:mi></m:math> is <emphasis>2 standard deviations</emphasis> <m:math><m:mi>(2σ)</m:mi></m:math>
above or to the right of the mean
<m:math><m:mi>μ = 5</m:mi></m:math>. The standard deviation is <m:math><m:mi>σ = 6</m:mi></m:math>.</para><para id="element-314">Notice that:</para><equation id="element-668"><m:math>
		<m:mn>5</m:mn>
		<m:mo>+</m:mo>
		<m:mn>2</m:mn>
		<m:mo>⋅</m:mo>
		<m:mn>6</m:mn>
		<m:mo>=</m:mo>
		<m:mn>17</m:mn>
		<m:mspace width="20pt"/>
		<m:mtext>(The pattern is</m:mtext>
		<m:mspace width="5pt"/>
		<m:mi>μ</m:mi>
		<m:mo>+</m:mo>
		<m:mi>z</m:mi>
		<m:mi>σ</m:mi>
		<m:mo>=</m:mo>
		<m:mi>x</m:mi>
		<m:mo>.</m:mo>
		<m:mtext>)</m:mtext>
	</m:math>
</equation><para id="element-330">Now suppose 
<m:math><m:mi>x=1</m:mi></m:math>. Then:</para><equation id="element-870"><m:math>
<m:mi>z</m:mi>
<m:mo>=</m:mo>
<m:mfrac>
<m:mrow>
<m:mi>x</m:mi>
<m:mo>-</m:mo>
<m:mi>μ</m:mi>
</m:mrow>
<m:mrow>
<m:mi>σ</m:mi>
</m:mrow>
</m:mfrac>
<m:mo>=</m:mo>
<m:mfrac>
<m:mrow>
<m:mn>1</m:mn>
<m:mo>-</m:mo>
<m:mn>5</m:mn>
</m:mrow>
<m:mrow>
<m:mn>6</m:mn>
</m:mrow>
</m:mfrac>
<m:mo>=</m:mo>
<m:mo>-</m:mo>
<m:mn>0.67</m:mn>
<m:mspace width="20pt"/>
<m:mtext>  (rounded to two decimal places)</m:mtext>
</m:math>
</equation><para id="element-468"><emphasis> This means that <m:math><m:mi>x = 1</m:mi></m:math> is 0.67 standard deviations <m:math><m:mi>(- 0.67σ)</m:mi></m:math> below or to the left of
the mean <m:math><m:mi>μ = 5</m:mi></m:math>. Notice that:</emphasis></para><para id="element-469"><m:math>
<m:mn>5</m:mn>
<m:mo>+</m:mo>
<m:mo>(</m:mo>
<m:mn>-0.67</m:mn>
<m:mo>)</m:mo>
<m:mo>(</m:mo>
<m:mn>6</m:mn>
<m:mo>)</m:mo>
</m:math>
is approximately equal to 1
<m:math>
<m:mspace width="20pt"/>
</m:math>
(This has the pattern
<m:math>
<m:mi>μ</m:mi>
<m:mo>+</m:mo>
<m:mo>(</m:mo>
<m:mn>-0.67</m:mn>
<m:mo>)</m:mo>
<m:mi>σ</m:mi>
<m:mo>=</m:mo>
<m:mn>1</m:mn>
</m:math>
)
</para><para id="element-588">Summarizing, when <m:math><m:mi>z</m:mi></m:math> is positive, <m:math><m:mi>x</m:mi></m:math> is above or to the right of <m:math><m:mi>μ</m:mi></m:math> and when <m:math><m:mi>z</m:mi></m:math> is negative, <m:math><m:mi>x</m:mi></m:math> is to
the left of or below <m:math><m:mi>μ</m:mi></m:math>.</para>
</example><example id="element-305"><para id="element-979">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 <m:math><m:mi>X</m:mi></m:math> = the amount of weight lost (in pounds) by a person in a month.
Use a standard deviation of 2 pounds. <m:math><m:mi>X</m:mi></m:math>~<m:math><m:mi>N(5, 2)</m:mi></m:math>. Fill in the blanks.
</para><exercise id="element-984"><?solution_in_back?>
<problem>
  <para id="element-658">
Suppose a person <emphasis>lost</emphasis> 10 pounds in a month. The z-score when <m:math><m:mi>x = 10</m:mi></m:math> pounds is <m:math><m:mi>z = 2.5</m:mi></m:math>
(verify). This z-score tells you that <m:math><m:mi>x = 10</m:mi></m:math> is ________ standard deviations to the ________
(right or left) of the mean _____ (What is the mean?).
  </para>
</problem>

<solution>
  <para id="element-603">This z-score tells you that <m:math><m:mi>x = 10</m:mi></m:math> is <emphasis>2.5</emphasis> standard deviations to the <emphasis>right</emphasis>
 of the mean <emphasis>5</emphasis>.
  </para>
</solution>
</exercise><exercise id="element-369"><?solution_in_back?>
<problem>
  <para id="element-195">
Suppose a person <emphasis>gained</emphasis> 3 pounds (a negative weight loss). Then <m:math><m:mi>z</m:mi></m:math> = __________. This
z-score tells you that <m:math><m:mi>x = -3</m:mi></m:math> is ________ standard deviations to the __________ (right or left)
of the mean.
  </para>
</problem>

<solution>
  <para id="element-639">
<m:math><m:mi>z</m:mi></m:math> = <emphasis>-4</emphasis>. This
z-score tells you that <m:math><m:mi>x = -3</m:mi></m:math> is <emphasis>4</emphasis> standard deviations to the <emphasis>left</emphasis>
of the mean.
  </para>
</solution>
</exercise><para id="element-137">Suppose the random variables <m:math><m:mi>X</m:mi></m:math> and <m:math><m:mi>Y</m:mi></m:math> have the following normal distributions:
<m:math><m:mi>X</m:mi></m:math> ~<m:math><m:mi> N(5, 6)</m:mi></m:math> and <m:math><m:mi>Y ~ N(2, 1)</m:mi></m:math>. If <m:math><m:mi>x = 17</m:mi></m:math>, then <m:math><m:mi>z</m:mi><m:mo> = </m:mo><m:mn>2</m:mn></m:math>. (This was previously shown.)
If <m:math><m:mi>y = 4</m:mi></m:math>, what is <m:math><m:mi>z</m:mi></m:math>?</para><equation id="element-958"><m:math>
<m:mi>z</m:mi>
<m:mo>=</m:mo>
<m:mfrac>
<m:mrow>
<m:mi>y</m:mi>
<m:mo>-</m:mo>
<m:mi>μ</m:mi>
</m:mrow>
<m:mrow>
<m:mi>σ</m:mi>
</m:mrow>
</m:mfrac>
<m:mo>=</m:mo>
<m:mfrac>
<m:mrow>
<m:mn>4</m:mn>
<m:mo>-</m:mo>
<m:mn>2</m:mn>
</m:mrow>
<m:mrow>
<m:mn>1</m:mn>
</m:mrow>
</m:mfrac>
<m:mo>=</m:mo>
<m:mn>2</m:mn>
<m:mspace width="20pt"/>
<m:mtext>where μ=2 and σ=1.</m:mtext>
</m:math></equation><para id="element-291">The z-score for <m:math><m:mi>y = 4</m:mi></m:math> is <m:math><m:mi>z = 2</m:mi></m:math>. This means that 4 is <m:math><m:mi>z = 2</m:mi></m:math> standard deviations to the right of the
mean. Therefore, <m:math><m:mi>x = 17</m:mi></m:math> and <m:math><m:mi>y = 4</m:mi></m:math> are both 2 (of <emphasis>their</emphasis>) standard deviations to the right of
<emphasis>their</emphasis> respective means.</para><para id="element-735"><emphasis>The z-score allows us to compare data that are scaled differently.</emphasis> To understand the
concept, suppose 

<m:math><m:mi>X</m:mi></m:math> ~<m:math><m:mi> N(5, 6)</m:mi></m:math> represents weight gains for one group of people who are trying
to gain weight in a 6 week period and <m:math><m:mi>Y</m:mi></m:math> ~<m:math><m:mi> N(2, 1)</m:mi></m:math> measures the same weight gain for a
second group of people. A negative weight gain would be a weight loss.
Since <m:math><m:mi>x = 17</m:mi></m:math> and <m:math><m:mi>y = 4</m:mi></m:math> are each 2 standard deviations to the right of their means, they
represent the same weight gain <emphasis>in relationship to their means</emphasis>.</para>
</example>   
  </content>
  
</document>
