is the mean greater than, less than, or equal to the median? greater than less than equal to

Explanation:

Step1: Count data - point frequency

Count the number of dots for each value. Assume the frequencies are (f_1,f_2,\cdots,f_n) for values (x_1,x_2,\cdots,x_n) respectively.

Step2: Calculate the mean

The mean (\bar{x}=\frac{\sum_{i = 1}^{n}x_if_i}{\sum_{i=1}^{n}f_i}).

Step3: Find the median

First, find the total number of data - points (N=\sum_{i = 1}^{n}f_i). If (N) is odd, the median is the (\left(\frac{N + 1}{2}\right))-th ordered data - point. If (N) is even, the median is the average of the (\frac{N}{2})-th and (\left(\frac{N}{2}+1\right))-th ordered data - points.

Step4: Compare mean and median

After calculating the mean and median, compare their values.

Let's assume the frequencies of the values (x = 10,15,20,25,30,35,40,45,50) are all (9) (since the number of dots in each column seems to be (9)). The total number of data - points (N=9\times9 = 81) (odd). The median is the (\left(\frac{81 + 1}{2}\right)=41)-th ordered data - point. The ordered data - set: Since the frequencies are equal for each value, the median value is (30). The mean (\bar{x}=\frac{(10\times9)+(15\times9)+(20\times9)+(25\times9)+(30\times9)+(35\times9)+(40\times9)+(45\times9)+(50\times9)}{81}) [ \begin{align*} \bar{x}&=\frac{9\times(10 + 15+20+25+30+35+40+45+50)}{81}\ &=\frac{9\times\frac{(10 + 50)\times9}{2}}{81}\ &=\frac{9\times270}{81}\ &=30 \end{align*} ]

Answer:

equal to