Chapter 2, Problem 1TP

To determine

The mean, median and mode for the set of 40 chemistry final scores.

Answer to Problem 1TP

The mean of the chemistry final scores is $87.55$, median of the chemistry final scores is 88, mode of the chemistry final scores is 94.

Explanation of Solution

The given data set of 40 scores that consists of chemistry final marks is given below:

93	85	99	77
94	99	86	76
95	99	97	84
91	89	77	87
97	83	80	98
75	94	81	85
78	92	89	94
76	94	96	94
90	79	80	94
77	86	83	81

Calculation:

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 the 40 final scores is given below.

$\begin{array}{c}{\displaystyle \sum _1^{40}X}=\left\{\begin{array}{c}93+94+95+91+97+75+78+76+90+77+85+99+99+89\\ +83+94+92+94+79+86+99+86+97+77+80+81+89+96\\ +80+83+77+76+84+87+98+85+94+94+92+81\end{array}\right\}\\ =3,502\end{array}$

Substitute 3502 for ${\displaystyle \sum _1^{40}X}$ and 40 for n in equation (1),

$\begin{array}{c}\text{Mean}=\frac{3,502}{40}\\ =87.55\end{array}$

Thus, the mean is 87.55.

Median:

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

99	99	99	98	97	97	96	95	94	94
94	94	94	93	92	92	91	90	89	89
87	86	86	85	85	84	83	83	81	81
80	80	79	78	77	77	77	76	76	75

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 40 so, substitute 40 for n in the above formula.

$\begin{array}{c}\text{Median}=\frac{{\left(\frac{40}{2}\right)}^{\text{th}}\text{term}+{\left(\frac{40}{2}+1\right)}^{\text{th}}\text{term}}{2}\\ =\frac{{\left(20\right)}^{\text{th}}\text{term}+{\left(21\right)}^{\text{st}}\text{term}}{2}\end{array}$ (2)

${20}^{\text{th}}\text{term}$ from the arranged data is 89 and the ${21}^{\text{st}}\text{term}$ is 87.

Substitute 89 for ${20}^{\text{th}}\text{term}$ and 87 for ${21}^{\text{st}}\text{term}$ in equation (2).

$\begin{array}{c}\frac{{20}^{\text{th}}\text{term}+{21}^{\text{st}}\text{term}}{2}=\frac{89+87}{2}\\ =\frac{176}{2}\\ =88\end{array}$

Thus, the median is 88.

Mode:

For calculation of mode, check the number that occurs most frequent. Since, 94 occur for 5 times in the table, thus 94 is the mode.