Question

7. These values represent the expected number of paintings a person will produce over the next 10 days. 0, 0, 0, 1, 1, 1, 2, 2, 3, 5 a. What are the mean and standard deviation of the data? b. The artist is not pleased with these statistics. If the 5 is increased to a larger value, how does this impact the median, mean, and standard deviation?

Answer 1

The mean of a dataset is the average of the dataset.

• The mean is 1.5
• The standard deviation is 1.58
• The median remains unchanged, the mean increases, and the standard deviation decreases when 5 is increased to a larger value

The given parameters are:

$\mathbf{x = 0, 0, 0, 1, 1, 1, 2, 2, 3, 5}$

(a) Mean and Standard deviation

The mean of the dataset is calculated using:

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

So, we have:

$\mathbf{\bar x = \frac{0+ 0+ 0+ 1+ 1+ 1+ 2+ 2+ 3+ 5}{10}}$

$\mathbf{\bar x = \frac{15}{10}}$

$\mathbf{\bar x = 1.5}$

Hence, the mean is 1.5

The standard deviation is calculated using:

$\mathbf{\sigma_x = \sqrt{\frac{\sum(x - \bar x)^2}{n - 1}}}}$

So, we have:

$\mathbf{\sigma_x = \sqrt{\frac{(0 - 1.5)^2 + (0- 1.5)^2 + (0- 1.5)^2 + (1- 1.5)^2 + (1- 1.5)^2 + (1- 1.5)^2 + (2- 1.5)^2 + (2- 1.5)^2 + (3- 1.5)^2 + (5- 1.5)^2}{10 - 1}}}$

$\mathbf{\sigma_x = \sqrt{\frac{22.5}{9}}}$

$\mathbf{\sigma_x = \sqrt{2.5}}$

$\mathbf{\sigma_x = 1.58}$

Hence, the standard deviation is 1.58

(b) When 5 is increased to a larger value

When 5 is increased:

• The median remains unchanged
• The mean increases
• The standard deviation decreases

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