Answer:
The distribution of the number of snacks grabbed is skewed right with a center around 18 and varies from 12 to 45. There are possible outliers at 38, 42, and 45.
Explanation:
First, we can see if the graph is symmetric. A symmetric graph is even on both sides of the center. As there are a lot more students that grabbed a small number of snacks, and the data is not even around the center (which is somewhere around 20 or 30 snacks). This means that the graph is not symmetric, making the second answer incorrect.
Next, we can check if the graph is skewed right or left. If the left of the graph represents a smaller amount of snacks and the right of it represents a higher number of snacks, we can see that most of the data is on the left of the graph. There are a few values to the right, but the overwhelming amount of data is on the left, making the distribution skewed to the right. This keeps the first and last answers possible
Moreover, we can find the center of the distribution. This is generally equal to the median, which is 18, so the center is around 18
After that, we can see what the values vary from. The lowest tens value is 1, and the lowest ones value in that is 2, making the lowest value 12. Similarly, the highest tens value is 4, and the highest ones value there is 5, making the range 12 to 45. This leaves the last answer, but we can check the outliers to make sure.
With the data, we can calculate the first quartile to be 15, the third quartile to be 21.5, and the interquartile range to be 21.5-15 = 6.75 . If a number is less than Q₁ - 1.5 * IQR or greater than Q₃ + 1.5 * IQR, it is a potential outlier. Applying that here, the lower bound for non-outliers is 15 - 6.5 * 1.5 = 5.25, and the upper bound if 21 + 6.5 * 1.5 = 30.75. No values are less than 5.25, but there are four values greater than 30.75 in 32, 38, 42, and 45. There are possible outliers at 38, 42, and 45, matching up with the last answer.