Question

If marks scored by five students in statistics test of 100 marks, are given in following table.

Student	1	2	3	4	5
Marks(/100)x	46	34	52	78	65

Find standard deviation and arithmetic mean $\overline{x}$.

Answer 1

Arithmetic mean $\overline{x} = 55$.

Standard deviation $\sigma = \sqrt{232} \approx 15.23$.

Answer 2

Given the marks of five students in a statistics test: $x = \{46, 34, 52, 78, 65\}$. The number of students is $n=5$.

First, calculate the arithmetic mean ($\overline{x}$). The formula for the arithmetic mean is $\overline{x} = \frac{\sum x_i}{n}$. Sum of marks $\sum x_i = 46 + 34 + 52 + 78 + 65 = 275$. $\overline{x} = \frac{275}{5} = 55$.

Next, calculate the standard deviation ($\sigma$). We will use the formula for the population standard deviation, as the given data represents the scores of the specified five students. The formula for the population standard deviation is $\sigma = \sqrt{\frac{\sum (x_i - \overline{x})^2}{n}}$. First, calculate the deviations from the mean $(x_i - \overline{x})$:

$46 - 55 = -9$

$34 - 55 = -21$

$52 - 55 = -3$

$78 - 55 = 23$

$65 - 55 = 10$

Next, square the deviations $(x_i - \overline{x})^2$:

$(-9)^2 = 81$

$(-21)^2 = 441$

$(-3)^2 = 9$

$(23)^2 = 529$

$(10)^2 = 100$

Now, sum the squared deviations $\sum (x_i - \overline{x})^2$: $\sum (x_i - \overline{x})^2 = 81 + 441 + 9 + 529 + 100 = 1160$.

Finally, calculate the standard deviation: $\sigma = \sqrt{\frac{1160}{5}} = \sqrt{232}$. The value of $\sqrt{232}$ is approximately $15.23$.

Explanation of the solution:

1. The arithmetic mean $\overline{x}$ is calculated as the sum of all marks divided by the number of students: $\overline{x} = \frac{\sum x_i}{n}$.
2. The standard deviation $\sigma$ is calculated as the square root of the variance, where the variance is the average of the squared deviations from the mean: $\sigma = \sqrt{\frac{\sum (x_i - \overline{x})^2}{n}}$.