Standard Deviation

Standard Deviation exercise

• Calculate the standard deviation for Martin McTry.

  Hints

  The mean of the following list

  $\begin{array}{cccc} 2&4&6&8 \end{array}$

  is given by

  $\bar x=\frac{2+4+6+8}4=\frac{20}4=5$.

  The standard deviation of the list above is given by

  $\begin{array}{rcl} \sigma&=&~\sqrt{\frac{(5-2)^2+(5-4)^2+(5-6)^2+(5-8)^2}4}\\ &=&~\sqrt{\frac{9+1+1+9}4}\\ &=&~\sqrt 5\\ &\approx&~2.236 \end{array}$

  Keep PEMDAS in mind:

  • First, determine the values inside the parantheses.
  • Next, calculate the squares.
  • Then add the squares.
  • Divide the sum by the number of elements.
  • Take the square root of the quotient.

  Solution

  To calculate the standard deviation of Martin McTry's score, we first have to calculate the mean of those values.

  The sum of all elements divided by the number of the elements, which is $5$:

  $\bar x=\frac{21+18+16+24+21}{5}=\frac{100}5=20$.

  The standard deviation is the square root of the average of the squares of the differences between each element of the set and the mean of the set:

  $\begin{array}{rcl} \sigma&=&~\sqrt{\frac{(\bar x-21)^2+(\bar x-18)^2+(\bar x-16)^2+(\bar x-24)^2+(\bar x-21)^2}5}\\ &=&~\sqrt{\frac{1+4+16+16+1}5}\\ &=&~\sqrt{\frac{38}5}\\ &\approx&~2.757 \end{array}$.

• Explain the elements in the standard deviation formula.

  Hints

  The mean is the sum of all elements divided by the number of elements.

  The standard deviation gives information about the consistency of the elements of a list. It is dependent on the mean of the elements.

  The standard deviation is the square root of the average of the squares of the differences between each element of the set and the mean of the set.

  Solution

  The standard deviation is the square root of the average of the squares of the differences between each element of the set and the mean of the set:

  $\sigma=\sqrt{\frac{(\bar x- x_1)^2+...+(\bar x- x_n)^2}n}$

  • the Greek letter $\sigma$ is the common indicator for standard deviation
  • $\bar x$ is the mean of the elements
  • $x_i$, $i=1,~...,~n$ are the elements of the list
  • $n$ is the number of the elements of the list

• Calculate the standard deviation for the scoring.

  Hints

  To calculate the standard deviation, first you have to calculate the mean of the elements of the list.

  The mean is the sum of all elements divided by the number of elements.

  The standard deviation is the square root of the average of the squares of the differences between each element of the set and the mean of the set.

  Solution

  To calculate the standard deviation for the list, first we have to calculate the mean. The mean is the sum of all elements divided by the number of all elements:

  $\bar x=\frac{20+19+23+28+25+35}{6}=\frac{150}6=25$

  The standard deviation is the square root of the average of the squares of the differences between each element of the set and the mean of the set:

  $\begin{array}{rcl} \sigma&=&~\sqrt{\frac{(25-20)^2+(25-19)^2+(25-23)^2+(25-28)^2+(25-25)^2+(25-35)^2}6}\\ &=&~\sqrt{\frac{25+36+4+9+0+100}6}\\ &=&~\sqrt{\frac{174}6}\\ &\approx&~5.385 \end{array}$

• Compare the standard deviation of each player with each other.

  Hints

  First, calculate the mean per list. It is the sum of all elements divided by the number of all elements in one list.

  All players have the same mean.

  Here is the formula of the standard deviation:

  $\sigma=\sqrt{\frac{(22-x_1)^2+(22-x_2)^2+(22-x_3)^2+(22-x_4)^2}4}$

  The mean of each player is $22$.

  Solution

  The mean is the same for all four players:

  • Peter: $\bar x=\frac{22+23+24+19}4=22$
  • John: $\bar x=\frac{23+21+25+19}4=22$
  • Paul: $\bar x=\frac{25+21+25+17}4=22$
  • Luke: $\bar x=\frac{19+21+19+29}4=22$

  So, how can we decide who the most consistent player is? We determine the standard deviation for each player by using the standard deviation formula

  $\sigma=\sqrt{\frac{(22-x_1)^2+(22-x_2)^2+(22-x_3)^2+(22-x_4)^2}4}$

  $~$

  Peter

  $\begin{array}{rcl} \sigma&=&~\sqrt{\frac{(22-22)^2+(22-23)^2+(22-24)^2+(22-19)^2}4}\\ &=&~\sqrt{\frac{0+1+4+9}{4}}\\ &=&~\sqrt{\frac{14}4}\\ &\approx&~1.871 \end{array}$

  John

  $\begin{array}{rcl} \sigma&=&~\sqrt{\frac{(22-23)^2+(22-21)^2+(22-25)^2+(22-19)^2}4}\\ &=&~\sqrt{\frac{1+1+9+9}{4}}\\ &=&~\sqrt{\frac{20}4}\\ &\approx&~2.236 \end{array}$

  Paul

  $\begin{array}{rcl} \sigma&=&~\sqrt{\frac{(22-25)^2+(22-21)^2+(22-25)^2+(22-17)^2}4}\\ &=&~\sqrt{\frac{9+1+9+25}{4}}\\ &=&~\sqrt{\frac{44}4}\\ &\approx&~3.317 \end{array}$

  Luke

  $\begin{array}{rcl} \sigma&=&~\sqrt{\frac{(22-19)^2+(22-21)^2+(22-19)^2+(22-29)^2}4}\\ &=&~\sqrt{\frac{9+1+9+49}{4}}\\ &=&~\sqrt{\frac{68}4}\\ &\approx&~4.123 \end{array}$

  So Peter is the most consistent player, and Luke shows the least consistency.

• Compare the results of the two players.

  Hints

  The bigger the standard deviation, the bigger the inconsistency.

  Have a look at the following example:

  Peter scores $40$ points in one game, but doesn't score at all in the next three games. He has a mean of $10$ points per game. Whereas Paul scores $10$ points per game in four games.

  Which player would you prefer?

  If Peter and Paul played for the same team, they would score $50$ points in one game and only $10$ points in each of the following three games.

  Solution

  You can calculate the mean and the standard deviation from the given data points.

  Let's have a look at the example:

  • Martin McTry's score has a mean of $20$ and a standard deviation of approximately $2.757$.
  • Lance Layton's score shows also a mean of $20$, but a standard deviation of $8.198$.

  Both have the same mean. So they've got the same number of average points. But the standard deviations are quite different from each other. The standard deviation of Lance is much bigger than Martin's.

  This shows that Martin is more consistent at scoring points than Lance. His actual number of scored points differ much less from the mean than those of Lance.

  If Cliff's decision of who to pick for the team is based on consistency, he should then prefer Martin McTry.

• Determine the standard deviation for each set of scores.

  Hints

  Keep PEMDAS in mind:

  • First, determine the values inside the parantheses.
  • Next, calculate the squares.
  • Then add the squares.
  • Divide the sum by the number of elements.
  • Take the square root of the quotient.

  Here, you can see the beginning of the calculation of the standard deviation for the first list:

  $\sigma=\sqrt{\frac{(\bar x-18)^2+(\bar x-22)^2+(\bar x-19)^2+(\bar x-24)^2+(\bar x-22)^2}5}$

  $\bar x$ is the mean of the list.

  The mean of the first list can be calculated this way:

  $\bar x=\frac{18+22+19+24+22}5$

  Solution

  Keep in mind:

  • The mean is the sum of all elements divided by the number of elements.
  • The standard deviation is the square root of the average of the squares of the differences between each element of the set and the mean of the set.

  Let's start with the first list

  $\begin{array}{l|c|c|c|c|c} \text{game}&\#1&\#2&\#3&\#4&\#5\\ \hline \text{score}&18&22&19&24&22 \end{array}$

  $\bar x=\frac{18+22+19+24+22}5=\frac{105}5=21$

  With the mean we can calculate the standard deviation:

  $\begin{array}{rcl} \sigma&=&~\sqrt{\frac{(21-18)^2+(21-22)^2+(21-19)^2+(21-24)^2+(21-22)^2}5}\\ &=&~\sqrt{\frac{9+1+4+9+1}5}\\ &=&~\sqrt{\frac{24}5}\\ &\approx&~2.19 \end{array}$

  Let's have a look at the next list

  $\begin{array}{l|c|c|c|c} \text{game}&\#1&\#2&\#3&\#4\\ \hline \text{score}&19&21&23&21 \end{array}$

  The mean is:

  $\bar x=\frac{19+21+23+21}4=\frac{84}4=21$

  Next, we can calculate the standard deviation:

  $\begin{array}{rcl} \sigma&=&~\sqrt{\frac{(21-19)^2+(21-21)^2+(21-23)^2+(21-21)^2}4}\\ &=&~\sqrt{\frac{4+0+4+0}4}\\ &=&~\sqrt2\\ &\approx&~1.41 \end{array}$

  There is just one list left

  $\begin{array}{l|c|c|c|c|c|c} \text{game}&\#1&\#2&\#3&\#4&\#5&\#6\\ \hline \text{score}&19&21&23&21&17&22 \end{array}$

  The mean is:

  $\bar x=\frac{19+21+23+21+17+22}6=\frac{123}6=20.5$

  Next, let's determine the standard deviation: $\begin{array}{rcl} \sigma&=&~\sqrt{\frac{(20.5-19)^2+(20.5-21)^2+(20.5-23)^2+(20.5-21)^2+(20.5-17)^2+(20.5-22)^2}6}\\ &=&~\sqrt{\frac{2.25+0.25+6.25+0.25+12.25+2.25}6}\\ &=&~\sqrt{\frac{23.5}6}\\ &\approx&~1.98 \end{array}$