Chapter 2, Problem 2TP

To determine

The mean, median and mode for all the three sets of scores.

Answer to Problem 2TP

The mean of Score 1 is $5.6$, median of Score1 is $5.5$, mode of Score1 is 5.

The mean of Score 2 is $27.6$, median of Score2 is 25, mode of Score2 is 25. 34.

The mean of Score 3 is $144.3$, median of Score3 is $149.5$, mode of Score3 is 154.

Explanation of Solution

Given info:

The data of three sets of scores is given below:

Score 1	Score 2	Score 3
3	34	154
7	54	167
5	17	132
4	26	145
5	34	154
6	15	145
7	14	113
8	24	156
6	25	154
5	23	123

Calculation:

Mean for score 1:

The formula to calculate the mean is,

$\begin{array}{c}\overline{\text{X}}=\frac{{\displaystyle \sum _1^{n}X}}{n}\\ =\frac{X_1+X_2+X_3+\mathrm{...}+X_n}{n}\end{array}$ (1)

Where,

• $X$ is the number of final scores.
• $n$ is the total number of final scores.

The sum of Score1 is given below.

$\begin{array}{c}{\displaystyle \sum _1^{10}X}=3+7+5+4+5+6+7+8+6+5\\ =56\end{array}$

Substitute 56 for ${\displaystyle \sum _1^{10}X}$ and 10 for n in equation (1),

$\begin{array}{c}\text{Mean}=\frac{56}{10}\\ =5.6\end{array}$

Thus, the mean is 5.6.

Median for score 1:

For calculation of median, arrange the data in ascending or descending order. Now data is arranged in descending order then,

8

7

7

6

6

5

5

5

4

3

The formula to calculate the median for even data is shown below.

$\text{Median}=\frac{{\left(\frac{n}{2}\right)}^{\text{th}}\text{term}+{\left(\frac{n}{2}+1\right)}^{\text{th}}\text{term}}{2}$

The total number of terms in the data is 10 so, substitute 10 for n in the above formula.

$\begin{array}{c}\text{Median}=\frac{{\left(\frac{10}{2}\right)}^{\text{th}}\text{term}+{\left(\frac{10}{2}+1\right)}^{\text{th}}\text{term}}{2}\\ =\frac{{\left(5\right)}^{\text{th}}\text{term}+{\left(6\right)}^{\text{th}}\text{term}}{2}\end{array}$ (2)

$5^{\text{th}}\text{term}$ from the arranged data is 6 and the $6^{\text{th}}\text{term}$ is 5.

Substitute 6 for $5^{\text{th}}\text{term}$ and 5 for $6^{\text{th}}\text{term}$ in equation (2).

$\begin{array}{c}\frac{5^{\text{th}}\text{term}+6^{\text{th}}\text{term}}{2}=\frac{6+5}{2}\\ =\frac{11}{2}\\ =5.5\end{array}$

Thus, the mean is 5.5.

Mode for score 1:

For calculation of mode, check the number that occurs most frequent. Since 5 occur 3 times in the table, thus 5 is the mode of Score 1.

Mean for score 2:

The sum of Score2 is given below.

$\begin{array}{c}{\displaystyle \sum _1^{10}X}=34+54+17+26+34+25+14+24+25+23\\ =276\end{array}$

Substitute 276 for ${\displaystyle \sum _1^{10}X}$ and 10 for n in equation (1),

$\begin{array}{c}\text{Mean}=\frac{276}{10}\\ =27.6\end{array}$

Thus, the mean is 27.6.

Median for score 2:

For calculation of median, arrange the data in ascending or descending order. Now data is arranged descending order.

54

34

34

26

25

25

24

23

17

14

$5^{\text{th}}\text{term}$ from the arranged data is 25 and the $6^{\text{th}}\text{term}$ is 25.

Substitute 25 for $5^{\text{th}}\text{term}$ and 25 for $6^{\text{th}}\text{term}$ in equation (2).

$\begin{array}{c}\frac{5^{th}\text{term}+6^{\text{th}}\text{term}}{2}=\frac{25+25}{2}\\ =\frac{50}{2}\\ =25\end{array}$

Thus, the median is 25.

Mode for score 2:

For calculation of mode, check the number that occurs most frequent. Since 25 and 34 both occurs twice in the table, thus 25 and 34 are the bimodal of Score 2.

Mean for score 3:

The sum of Score3 is given below.

$\begin{array}{c}{\displaystyle \sum _1^{10}X}=154+167+132+145+154+145+113+156+154+123\\ =1,443\end{array}$

Substitute 1443 for ${\displaystyle \sum _1^{10}X}$ and 10 for n in equation (1),

$\begin{array}{c}\text{Mean}=\frac{1,443}{10}\\ =144.3\end{array}$

Thus, the mean is 144.3.

Median for score 3:

For calculation of median, arrange the data in ascending or descending order. Now data is in descending order then,

167

156

154

154

154

145

145

132

123

113

$5^{\text{th}}\text{term}$ from the arranged data is 154 and the $6^{\text{th}}\text{term}$ is 145.

Substitute 154 for $5^{\text{th}}\text{term}$ and 145 for $6^{\text{th}}\text{term}$ in equation (2).

$\begin{array}{c}\frac{5^{\text{th}}\text{term}+6^{\text{th}}\text{term}}{2}=\frac{154+145}{2}\\ =\frac{299}{2}\\ =149.5\end{array}$

Thus, the median is 149.5.

Mode for score 3:

For calculation of mode, check the number that occurs most frequent. Since 154 occur 3 times in the table, thus 154 is the mode of Score 3.