90 degrees fahrenheit for select oklahoma cities\nwhich is an accurate comparison?\nthe mean is likely…

90 degrees fahrenheit for select oklahoma cities\nwhich is an accurate comparison?\nthe mean is likely greater than the median because the data is skewed to the right.\nthe mean is likely greater than the median because the data is skewed to the left.\nthe median is likely greater than the mean because the data is skewed to the right.\nthe median is likely greater than the mean because the data is skewed to the left.

90 degrees fahrenheit for select oklahoma cities\nwhich is an accurate comparison?\nthe mean is likely greater than the median because the data is skewed to the right.\nthe mean is likely greater than the median because the data is skewed to the left.\nthe median is likely greater than the mean because the data is skewed to the right.\nthe median is likely greater than the mean because the data is skewed to the left.

Answer

Explanation:

Step1: Analyze data skewness

The bar - chart shows that the tail of the distribution extends to the right (fewer cities with a higher number of days). So, the data is right - skewed.

Step2: Recall property of mean and median in skewed data

In right - skewed data, the mean is pulled in the direction of the tail. So, the mean is likely greater than the median.

Answer:

The mean is likely greater than the median because the data is skewed to the right.