Chapter 3 - Section 3.4 - Assess Your Understanding - Applying the Concepts - Page 173: 29

I) Let x be the pulse rate of the students. The population mean of the sample provided in Problem 21 of Section 3.1 is calculated as 72.22, and the population standard deviation of the sample provided in Problem 19 of Section 3.2 is calculated as 7.67. The z-scores of nine students are calculated as follows: The z-score of Perpectual Bempah when his pulse rate is 76 is calculated as \[\begin{align} & Z=\frac{x-\mu }{\sigma } \\ & =\frac{76-72.22}{7.67} \\ & =0.4928 \end{align}\] The z-score of Megan Brooks when her pulse rate is 60 is calculated as \[\begin{align} & Z=\frac{x-\mu }{\sigma } \\ & =\frac{60-72.22}{7.67} \\ & =-1.593 \end{align}\] The z-score of Jeff Honeycutt when his pulse rate is 60 is calculated as \[\begin{align} & Z=\frac{x-\mu }{\sigma } \\ & =\frac{60-72.22}{7.67} \\ & =-\ 1.593 \end{align}\] The z-score of Clarice Jefferson when her pulse rate is 81 is calculated as \[\begin{align} & Z=\frac{x-\mu }{\sigma } \\ & =\frac{81-72.22}{7.67} \\ & =1.144 \end{align}\] The z-score of Crystal Kurten Bach when her pulse rate is 72 is calculated as \[\begin{align} & Z=\frac{x-\mu }{\sigma } \\ & =\frac{72-72.22}{7.67} \\ & =-\ 0.028 \end{align}\] The z-score of Janette Lantka when her pulse rate is 80 is calculated as \[\begin{align} & Z=\frac{x-\mu }{\sigma } \\ & =\frac{80-72.22}{7.67} \\ & =1.014 \end{align}\] The z-score of Kevin McCarthy when his pulse rate is 80 is calculated as \[\begin{align} & Z=\frac{x-\mu }{\sigma } \\ & =\frac{80-72.22}{7.67} \\ & =1.014 \end{align}\] The z-score of Tammy Ohm when his pulse rate is 68 is calculated as \[\begin{align} & Z=\frac{x-\mu }{\sigma } \\ & =\frac{68-72.22}{7.67} \\ & =-\ 0.550 \end{align}\] The z-score of Kathy Wojdyla when her pulse rate is 73 is calculated as \[\begin{align} & Z=\frac{x-\mu }{\sigma } \\ & =\frac{73-72.22}{7.67} \\ & =0.1016 \end{align}\] (2) The mean of the z-scores is calculated as \[\begin{align} & \text{Mean}=\frac{0.4928+\left( -\ 1.593 \right)+\cdots +\left( -\ 0.55 \right)+\ 0.1016}{9} \\ & =\text{0}\text{.00029} \end{align}\] The standard deviation of the z-scores is calculated as \[\begin{align} & \text{S}\text{.D}\text{.}=\sqrt{\frac{{{\left( 0.4928-0.29 \right)}^{2}}+{{\left( -1.593-0.29 \right)}^{2}}+\cdots +{{\left( -0.55-0.29 \right)}^{2}}+{{\left( 0.1016-0.29 \right)}^{2}}}{9}} \\ & =\text{1}\text{.00009 } \end{align}\]

Given above.