Answer
The mean is 703.667, there is no mode, the median is 630.5, the midrange is 820.5. Not all seats meet the requirements.
Work Step by Step
The mean can be counted by summing all the data and dividing it by the number of data: $\frac{774+649+1210+546+431+612}{6}=703.667$, the mean is 703.667. The mode is the data occuring the most times but here all data occurs one time, therefore there is no mode. The median is the average of the 1 or 2 (here 2) middle data: (649+612)/2=630.5, the median is 630.5. The midrange is the average of the highest and the lowest data: (1210+431)/2=820.5, the midrange is 820.5. 1210 is more than 1000 that is the upper limit of the requirement, therefore not all seats meet the requirements.
