Question

Complete the mean and standard deviation of the following observations of marks of 5 students of a tutorial group: Marks out of $25:8,12,13,15,22$

• A. $14,4.604$
• B. $15,4.604$
• C. $14,5.604$
• D. None of these

Answer 1

The correct option is A

$14,4.604$

Explanation for the correct option:

Step 1: Find the mean of the following observations:

Divide the sum of all observations by the total number of observations to get the mean:

$\begin{array}{rcl}\mu & =& \frac{8+12+13+15+22}{5}\\ & =& \frac{70}{5}\\ & =& 14\end{array}$

Step 2: Find the value of the sum of the square of the difference of mean deducted from each term:

Subtract the mean from each term and square each term and get the sum of the obtained values.

$\begin{array}{rcl}\sum _i{\left(x_i-\mu \right)}^2& =& {\left(8-14\right)}^2+{\left(12-14\right)}^2+{\left(13-14\right)}^2+{\left(15-14\right)}^2+{\left(22-14\right)}^2\\ & =& {\left(-6\right)}^2+{\left(-2\right)}^2+{\left(-1\right)}^2+1^2+8^2\\ & =& 36+4+1+1+64\\ & =& 106\end{array}$

Step 3: Find the standard deviation:

Use the mean and the square difference of observation and mean to get the deviation.

$\begin{array}{rcl}\sigma & =& \sqrt{\frac{1}{5}\sum _i{\left(x_i-\mu \right)}^2}\\ & =& \sqrt{\frac{106}{5}}\\ & =& \sqrt{21.2}\\ & =& 4.604\end{array}$

Hence, option (A) is correct.