Question

(370831030)
properties of probability distributions
classifying data distributions

what type of distribution is the graph? negatively skewed
the mean is dropdown with options: greater than the median, less than the median, equal to the median
the mode is dropdown

Explanation:

For the mean and median relationship:

Step1: Recall skewness and mean-median

In a negatively skewed distribution, the tail is on the left (lower values). The mean is pulled towards the tail, so mean < median? Wait, no—wait, negatively skewed: mean < median? Wait, no, correction: In a negatively skewed distribution, the mean is less than the median? Wait, no, actually, let's think again. Wait, the graph here: the bars increase from left to right (except the last small one). Wait, the distribution has higher probabilities for higher x (1,2,3 have higher heights, 0 and 4 are small). Wait, no, the x-axis is 0,1,2,3,4. The height at 3 is the highest, then 2, then 1, then 0 and 4 are low. So the distribution is skewed to the left (negative skewness)? Wait, no, skewness direction: the tail is on the side of the lower values. Wait, if the peak is on the right (higher x), then the tail is on the left (lower x), so it's negatively skewed. In a negatively skewed distribution, mean < median? Wait, no, actually, let's recall: for a negatively skewed distribution, the mean is less than the median? Wait, no, maybe I got it reversed. Wait, positive skewness: tail on the right, mean > median. Negative skewness: tail on the left, mean < median? Wait, no, let's check the formula. Skewness: if mean < median < mode, it's negative skewness. So in negative skewness, mean is less than median. Wait, but the options here are "greater than the median", "less than the median", "equal to the median". Wait, the graph is negatively skewed (as selected). So in negative skewness, mean < median? Wait, no, wait the mode is the highest bar, which is at x=3. Median: let's find the median. Let's assume the probabilities: let's denote P(0)=a, P(1)=b, P(2)=c, P(3)=d, P(4)=e, with a,e small, b < c < d. The total probability is 1. The median is the middle value. Let's list the cumulative probabilities: P(0) = a, P(0)+P(1)=a+b, P(0)+P(1)+P(2)=a+b+c, P(0)+P(1)+P(2)+P(3)=a+b+c+d, P(0)+P(1)+P(2)+P(3)+P(4)=1. Since d is the largest, the median is likely at x=2 or x=3? Wait, maybe it's easier: in a negatively skewed distribution (tail left), mean < median. Wait, but the options: the question is "The mean is [option]". Wait, the selected skewness is negatively skewed. So in negative skewness, mean < median? Wait, no, wait I think I mixed up. Let's check a reference: In a negatively skewed distribution, the mean is less than the median, which is less than the mode (mean < median < mode). So the mean is less than the median. So the correct option is "less than the median".

Step2: For the mode:

The mode is the value with the highest probability (tallest bar). The tallest bar is at x=3, so the mode is 3.

Answer:

For the mean: less than the median
For the mode: 3