某校举行校园歌手大赛,5名参赛选手得分分别是9,8,7,9.3,x,y,已知这5名参赛选手得分平均数为9,方差为0.1,则|x-y|=多少
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亲亲,下午好|x-y|=2解析:已知样本数n=5,样本均值μ=9,样本方差σ^2=0.1则样本总和S=9+8+7+9.3+x+y=43.3由样本总和S=nμ得:43.3=5×9,即x+y=14.3样本方差σ^2=1/nΣ(x-μ)^2=1/5[(9-9)^2+(8-9)^2+(7-9)^2+(9.3-9)^2+(x-9)^2+(y-9)^2]由题意得:σ^2=0.1,即1/5[(x-9)^2+(y-9)^2]=0.1联立上式,得:(x-9)^2+(y-9)^2=2解得:|x-y|=2
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某校猛并盯举行校园歌手大赛枝和,5名参赛选手得分分别是9,8,7,9.3,x,y,已蔽没知这5名参赛选手得分平均数为9,方差为0.1,则|x-y|=多少
亲亲,下衡扮午好|x-y|=2解析握雀:已知样本数n=5,样本均值μ=9,样本方差σ^2=0.1则样本总和S=9+8+7+9.3+x+y=43.3由样本总和S=nμ得:43.3=5×9,即x+y=14.3样本方差σ^2=1/nΣ(x-μ)^2=1/5[(9-9)^2+(8-9)^2+(7-9)^2+(9.3-9)^2+(x-9)^2+(y-9)^2]由题意得:σ^2=0.1,段拦早即1/5[(x-9)^2+(y-9)^2]=0.1联立上式,得:(x-9)^2+(y-9)^2=2解得:|x-y|=2
已知正项数an的前n项和Sn满足4Sn=an的平方+2a,求an的通项公式
根据已知条燃拍宏件可得贺辩:4Sn=an2+2a即:4(a1+a2+a3+…+an)=an2+2a化简得:an2-2a-4a1-4a2-4a3-…-4an=0即:an2-2a-4Sn+4an=0化简得:an2-2a-4Sn=0解得:an=2±2√(4Sn+2a)即:an=2±2√(4Sn+2a)通项公式为:皮册an=2±2√(4Sn+2a)