21、39、16、23、17、28、31這七個(gè)數(shù)的平均數(shù)是________,中位數(shù)是________.

25    23
分析:先將這七個(gè)數(shù)從小到大排列,經(jīng)觀察即可得出中間的一個(gè)數(shù)就是這組數(shù)據(jù)的中位數(shù);
平均數(shù)就是用這幾個(gè)數(shù)據(jù)的總和除以數(shù)據(jù)的個(gè)數(shù)得到的商.
解答:這組數(shù)據(jù)的平均數(shù)是:
(21+39+16+23+17+28+31)÷7,
=175÷7,
=25.
這七個(gè)數(shù)從小到大排列是16,17,21,23,28,31,39,
經(jīng)觀察這組數(shù)據(jù)的中位數(shù)是23;
故答案為:25,23.
點(diǎn)評:本題考查了平均數(shù)和中位數(shù)的意義及求法,
一組數(shù)據(jù)的中位數(shù)與這組數(shù)據(jù)的排序及數(shù)據(jù)個(gè)數(shù)有關(guān),因此求一組數(shù)據(jù)的中位數(shù)時(shí),先將該組數(shù)據(jù)按從小到大(或按從大到。┑捻樞蚺帕校缓蟾鶕(jù)數(shù)據(jù)的個(gè)數(shù)確定中位數(shù):當(dāng)數(shù)據(jù)個(gè)數(shù)為奇數(shù)時(shí),則中間的一個(gè)數(shù)即為這組數(shù)據(jù)的中位數(shù);當(dāng)數(shù)據(jù)個(gè)數(shù)為偶數(shù)時(shí),則最中間的兩個(gè)數(shù)的平均數(shù)即為這組數(shù)據(jù)的中位數(shù).
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