题目内容
360
360
.分析:设出公差,利用9个小长方形面积和为1,求出公差,然后求解中间一组的频数.
解答:解:设公差为d,那么9个小长方形的面积分别为0.02,0.02+d,0.02+2d,0.02+3d,0.02+4d,0.02+3d,0.02+2d,0.02+d,0.02,而9个小长方形的面积和为 1,可得
0.18+16d=1 解得d=
,
∴中间一组的频数为:1600×(0.02+4d)=360.
故答案为:360.
0.18+16d=1 解得d=
| 0.82 |
| 16 |
∴中间一组的频数为:1600×(0.02+4d)=360.
故答案为:360.
点评:本题考查频率分布直方图的应用,考查计算能力.
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