题目内容

有一批玉米杂交种子的标签标注品种纯度为98.0%,经某检验机构扦样并进行小区种植鉴定(监督抽查),共种植200株,鉴定结果为96.2%,请计算容许误差并对该批种子的纯度做出判定。(注:容许差距的计算公式为T=1.65;√1.92=1.39,√1.83=1.35,√0.98=0.99)

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