Question

Prepare a Pareto chart of the possible causes for a student to fail a final examination in a university course.

Answer 1

Answer:

Option 1

$\bar x_1 = 96.2$

$Mode = 99$

$\sigma_1 = 3.22$

Option 2

$\bar x_2 = 97.4$

$Mode = 97$

$\sigma_2 = 0.499$

Explanation:

Given

$Option\ 1: 99, 98, 99, 94, 92, 99, 98, 99, 94, 90$

$Option\ 2: 98, 97, 97, 97, 98, 98, 97, 97, 98$

Required

The mean, mode and standard deviation of both options

Option 1

Calculate mean

$\bar x = \frac{\sum x}{n}$

$\bar x_1 = \frac{99+ 98+ 99+ 94+ 92+ 99+ 98+ 99+ 94+ 90}{10}$

$\bar x_1 = \frac{962}{10}$

$\bar x_1 = 96.2$

Calculate mode

$Mode = 99$

Because it has a frequency of 4 (more than other element of the dataset)

Calculate standard deviation

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n}}$

$\sigma_1 = \sqrt{\frac{(99-96.2)^2 +.............+(90-96.2)^2}{10}}$

$\sigma_1 = \sqrt{\frac{103.6}{10}}$

$\sigma_1 = \sqrt{10.36}$

$\sigma_1 = 3.22$

Option 2

Calculate mean

$\bar x = \frac{\sum x}{n}$

$\bar x_2 = \frac{98+ 97+ 97+ 97+ 98+ 98+ 97+ 97+ 98}{9}$

$\bar x_2 = \frac{877}{9}$

$\bar x_2 = 97.4$

Calculate mode

$Mode = 97$

Because it has a frequency of 5 (more than other element of the dataset)

Calculate standard deviation

$\sigma_2 = \sqrt{\frac{(98-97.4)^2+..............+ (98-97.4)^2}{9}}$

$\sigma_2 = \sqrt{\frac{2.24}{9}}$

$\sigma_2 = \sqrt{0.2489}$

$\sigma_2 = 0.499$