Question

the graph shows the ages of different concertgoers who have backstage passes. which statement is true about the graph? a late arrival who is 21 years old with a back - stage pass will make the mean greater than the median. the two holders of back - stage passes whose ages are above 40 make the mean age higher than the median age. the ages of concert - goers with backstage passes are skewed left, so the mean age is less than the median age. a concert - goer who is 18 years old and wins a back - stage pass will pull the mean more than 2 years less than the median.

Explanation:

Step1: Recall mean - median relationship for skewed data

In a right - skewed distribution, the mean is greater than the median because the long tail on the right (higher values) pulls the mean in that direction. In a left - skewed distribution, the mean is less than the median as the long tail on the left (lower values) pulls the mean down.

Step2: Analyze the histogram

The histogram of ages of concert - goers with backstage passes has a few high - value outliers (ages above 40). These outliers pull the mean upwards. The median is the middle value when the data is ordered. Outliers have less of an effect on the median. So, the mean age will be higher than the median age due to these high - value outliers.

Step3: Analyze each option

• Option 1: A 21 - year - old is not an extreme value compared to the existing data, so it won't make the mean greater than the median in a significant way.
• Option 2: The two holders of back - stage passes whose ages are above 40 are outliers. They will pull the mean higher than the median, which is correct.
• Option 3: The data is skewed right (not left) because of the high - value outliers above 40.
• Option 4: An 18 - year - old is not an extreme outlier compared to the existing data range, so it won't pull the mean more than 2 years less than the median.

Answer:

The two holders of back - stage passes whose ages are above 40 make the mean age higher than the median age.