2. 职称英语
  3. [img]2014m9x/ct_egreqj_egreqbj_0031_20148[/img] [br] AFrom the histogram, you ca

[img]2014m9x/ct_egreqj_egreqbj_0031_20148[/img] [br] AFrom the histogram, you ca

游客2024-01-12  0
问题 [br]
选项
答案 A
解析 From the histogram, you can observe that
    all of the measurement intervals are the same size,
    the distribution has a peak at the measurement interval 6-10, and
    more of the measurement intervals are to the right of the peak than are to the left of the peak.
    Since in the histogram the 95 measurements have been grouped into intervals, you cannot calculate the exact value of either the average or the median; you must compare them without being able to determine the exact value of either one.
    The median of the 95 measurements is the middle measurement when the measurements are listed in increasing order. The middle measurement is the 48th measurement. From the histogram, you can see that the measurement interval 1-5 contains the first 15 measurements, and the measurement interval 6-10 contains the next 35 measurements(that is, measurements 16 through 50). Therefore, the median is in the measurement interval 6-10 and could be 6, 7, 8, 9, or 10.
    Estimating the average of the 95 measurements is more complicated.
    Since you are asked to compare the average and the median, not necessarily to calculate them, you may ask yourself if you can tell whether the average is greater than or less than the median. Note that visually the measurements in the first three measurement intervals are symmetric around the measurement interval 6-10, so you would expect the average of the measurements in just these three measurement intervals to lie in the 6-10 measurement interval. The 30 measurements in the remaining four measurement intervals are all greater than 10, some significantly greater than 10. Therefore, the average of the 95 measurements is greater than the average of the measurements in the first three measurementintervals, probably greater than 10. At this point it seems likely that the average of the 95 measurements is greater than the median of the 95 measurements. It turns out that this is true.
    To actually show that the average must be greater than 10, you can make the average as small as possible and see if the smallest possible average is greater than 10. To make the average as small as possible, assume that all of the measurements in each interval are as small as possible. That is to say, all 15 measurements in the measurement interval 1-5 are equal to 1, all 35 measurements in the measurement interval 6-10 are equal to 6, etc. Under this assumption, the average of the 95 measurements is

The value of the smallest possible average,1,015/95, is greater than 10.
    Therefore, since the average of the 95 measurements is greater than 10 and the median is in the measurement interval 6-10, it follows that the average is greater than the median, and the correct answer is Choice A
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