All of the stats can be gleamed from the histogram. To obtain the mean or arithmetic average,
we need to find the the total number of video games played and divide it by the total number of students.
$$ \text{Mean}=\left(\frac{0+9+4+6+5+6}{20}\right) $$
$$ \text{Mean}=\frac{30}{20} $$
$$ \text{Mean}=1.5 $$
To find the median, we can count towards the middle number. Since there are 20 terms, we want to find the average of the 10th and 11th term.
In this case, both the 10th and 11th term is 1. Therefore,
$$ \text{Median}=1 $$
To find the mode, find the number with the highest frequency.
$$ \text{Mode}=1 $$
To find the range, simply take the largest number and subtract the smallest number.
$$ \text{Range} = 6-0 $$
$$ \text{Range} = 6 $$
It may be beneficial to convert the histogram into a simple list.
The list:
$$ 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 3, 5, 6$$
Calculating mean, median, mode, and range:
$$ \underbrace{0, 0, 0, 0, 0}_\text{$0\cdot5$}, \underbrace{1, 1, 1, 1, 1, 1, 1, 1, 1}_\text{$1\cdot 9$}, \underbrace{2, 2}_\text{$2\cdot 2$}, \underbrace{3, 3}_\text{$3\cdot 2$}, 5, 6$$
$$ \text{Mean}=\left(\frac{0+9+4+6+5+6}{20}\right) $$
$$ \text{Mean}=\frac{30}{20} $$
$$ \text{Mean}=1.5 $$
The median for 20 terms is the average of the 10th and 11th term. (Middle term for odd number of items)
$$ 0, 0, 0, 0, 0, 1, 1, 1, 1, \Large 1 \normalsize, \Large 1 \normalsize, 1, 1, 1, 2, 2, 3, 3, 5, 6$$
$$ \text{Median}=1 $$
The mode is the number that is present the most.
$$ 0, 0, 0, 0, 0, \colorbox{aqua}{1, 1, 1, 1, 1, 1, 1, 1, 1}, 2, 2, 3, 3, 5, 6$$
$$ \text{Mode}=1$$
The range is the difference between the largest number and smallest number on the list.
$$ \Large 0 \normalsize, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 3, 5, \Large 6$$
$$ \text{Range}= 6-0 $$
$$ \text{Range}=6 $$