Solutions to Quiz 03

Notes:

  1. What I laid out here are recommended solutions. There could be other solutions.

For V26:

  1. You cannot substitute actual numbers here. You are supposed to prove this for any \(c\) and any list \(x_1,\ldots, x_n\). So, \[\begin{eqnarray*} \frac{1}{n}\sum_{i=1}^n c\cdot x_i &=& \frac{1}{n}\left(cx_1+cx_2+\ldots+cx_n\right) \\ &=& c\left(\frac{x_1+x_2+\ldots+x_n}{n}\right) \\ &=& c\cdot \overline{x}\end{eqnarray*}\]

  2. \(4*(-1.5)+50 = 44\) points – To see this, \(-1.5\) is the corresponding \(z\)-score for your midterm score. So, \[\frac{\mathrm{your\ score}-50}{4}=-1.5.\] Solve for your score.

  3. You are asked to find \(x_5\). From the given information, we have the first four deviations from the average (which is equal to 6): \(x_1-6=3\), \(x_2-6=2\), \(x_3-6=-4\), and \(x_4-6=0\). Therefore, you can obtain \(x_1=9\), \(x_2=8\), \(x_3=2\), and \(x_4=6\). To find \(x_5\), you only need to note that \[\frac{9+8+2+6+x_5}{5}=6.\] Thus, \(x_5=5\).

For V27:

  1. You cannot substitute actual numbers here. You are supposed to prove this for any \(c\) and any list \(x_1,\ldots, x_n\). So, \[\begin{eqnarray*} \frac{1}{n}\sum_{i=1}^n \left( x_i+a\right) &=& \frac{1}{n}\left((x_1+a)+(x_2+a)+\ldots+(x_n+a)\right) \\ &=& \frac{1}{n}\left[\left(x_1+x_2+\ldots+x_n\right) +na\right) \\ &=& \overline{x}+a\end{eqnarray*}\]

  2. \(5*(-2.5)+60 = 47.5\) points – To see this, \(-2.5\) is the corresponding \(z\)-score for your midterm score. So, \[\frac{\mathrm{your\ score}-60}{5}=-2.5.\] Solve for your score.

  3. You are asked to find \(x_5\). From the given information, we have the first four deviations from the average (which is equal to 5): \(x_1-5=5\), \(x_2-5=1\), \(x_3-5=-4\), and \(x_4-5=2\). Therefore, you can obtain \(x_1=10\), \(x_2=6\), \(x_3=1\), and \(x_4=7\). To find \(x_5\), you only need to note that \[\frac{10+6+1+7+x_5}{5}=5.\] Thus, \(x_5=1\).