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SAT官方每日一题附答案和解析[数学](2015年12月27日)

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答案:C

解析:

The histogram does not give the recorded heights for each cherry tree, but rather the intervals in which the heights lie. Thus, to find possible values for the average (arithmetic mean) height, one can find lower and upper bounds for the average height by calculating the averages of lower and upper interval bounds of the histogram with the frequencies given. A lower bound for the average, found by using the value at the left endpoint of the interval, is ((60*3) + (65*3) + (70*8) + (75*10) + (80*5) + (85*2)) over 31 is approximately equal to 72.74 feet, and so 72.74 + 5 = 77.74 feet would be an upper bound for the average height. Therefore, the average height of the 31 black cherry trees can be between 72.74 feet and 77.74 feet. Of the choices given, only 74 feet can be a possible value for the average height of the 31 black cherry trees.


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