2021年12月4日 星期六

109年初等考試-統計學大意詳解

109年公務人員初等考試試題

等 別: 初等考試
類 科: 統計
科 目: 統計學大意

解答0.10.1+0.4=15=0.2(C)
解答2X+Y=41(D)
解答(A)
解答
解答{X1N(μ,σ2)X2N(μ,σ2)μ(2X1X22)=μ(X1)12μ(X2)=μ12μ=12μμ(D)
解答(A):P(900<X<1500)=P(9001200100<z<15001200100)=P(3<z<3)1>89%(B)×:P(1100<X<1300)=P(1<z<1)=0.682675%(C)×:P(Z<z<Z)=0.95Z=1.96P(1004<X<1396)=0.95P(1000<X<1400)>95%(D)×:(A):P(900<X<1500)>99.7%(A)
解答[75,85]=80±55=z0.025×σ/n=1.96×20nn=(1.96×4)2=61.5n=62(C)
解答$0.440.520.04p1=0.440.44+0.52=0.4583,n1=700×0.44=308p2=10.4583=0.5417,n2=700×0.52=364=p2p1±zα/2×p1(1p1)n1+p2(1p2)n2=0.54170.4583±1.96×0.4583×0.5417308+0.5417×0.4583364=0.0834±1.96×0.0386=[0.00779,0.1590](A)
解答/(cluster)(C)
解答{E(ˉX2)=μ2ˉX+σ2ˉX=μ2+σ2nE(X2i/n2)=1n2E(X21++X2n)=1n2n(μ2+σ2)=1n(μ2+σ2)E(ˉX2)>E(X2i/n2)(D)
解答SK<0()>>(B)
解答n=z2α/2p(1p)E2{n1=1.9620.52/0.052=384n2=1.9620.52/0.0252=1536n2n1=1152(A)
解答H0:μ<18 v.s. Ha:μ18(A)
解答μ(0.6A+0.4B)=0.6μ(A)+0.4μ(B)=0.6×0.09+0.4×0.13=0.1060.206(A)
解答[(n1)s2χ2n1,α/2,(n1)s2χ2n1,1α/2]=[9×16χ29,0.025,9×16χ29,0.025]=[14419.0228,1442.7004]=[7.57,53.33](A)
解答F(D)
解答{b=3k=4{=b1=2=k1=3=(k1)(b1)=6=++=11SSE1=61=5=10(C)
解答pvalue0.05(C)
解答α(A)
解答D3pvalue=0.50010.05(C)
解答(C)
解答(D)
解答()(C)
解答186÷300=0.62(B)
解答P(μkσ<x<μ+kσ)11k2P(58.6<X<68.6)75%=1122k=2{μ2σ=58.6μ+2σ=68.6(μ,σ)=(63.6,2.5)(C)
解答{A=59z=59794.5=4.44B=55z=55693.7=3.78BCV{A:4.5/79=0.057B:3.7/69=0.054B(D)
解答{sX=x2n1(x)2n(n1)=1.024861123.62721312=0.033sY=y2n1(y)2n(n1)=1.27306124.05421312=0.027sX>sYB0.027(D)
解答P(BA)=P(BA)P(A)=0.4P(BA)=P(A)×0.4=0.8×0.4=0.32(B)
解答ABP(AB)=P(A)P(B)=0.32P(B)=0.32÷P(A)=0.32÷0.8=0.4(P(B),P(AB))=(0.4,0.32)(C)
解答1078(C)1070.2231
解答P(λ=1.5,X=0)=e1.5=0.223110351.5P(λ=1.5,X=3)=e1.51.533!=0.2231×1.536=0.1255(B)
解答{P(BA1)=P(BA1)/P(A1)=0.6P(BA1)=0.6×0.3=0.18P(BA2)=P(BA2)/P(A2)=0.3P(BA2)=0.3×0.7=0.21B=(BA1)(BA2)P(B)=0.18+0.21=0.39(C)
解答{P(BA1)=P(BA1)/P(A1)=0.6P(BA1)=0.6×0.3=0.18P(BA2)=P(BA2)/P(A2)=0.3P(BA2)=0.3×0.7=0.21B=(BA1)(BA2)P(B)=0.18+0.21=0.39P(A1B)=P(A1B)P(B)=0.180.39=613=0.46(D)
解答p=4350=1=1p41(D)
解答p=4350:(4350(10000+5000+1500+500)50×4)÷4=1.42857(A)
解答XB(p=0.1,n=100){μ=np=10σ=np(1p)=3P(X>11.5)=P(Z>11.5103)=P(Z>0.5)=10.6915=0.3085(B)()(B),(D)
解答100×0.308631(D)
解答P0.5(B)
解答AO1,1=50O1,2=24O1,3=18BO2,1=42O2,2=20O2,3=14Others O3,1=10O3,2=16O3,3=6{=50+42+10=102=24+20+16=60=18+14+6=38A=50+24+18=92B=42+20+14=76Others=10+16+6=32AE1,1=92102/200E1,2=9260/200E1,3=9238/200BE2,1=76102/200E2,2=7660/200E2,3=7638/200Others E3,1=32102/200E3,2=3260/200E3,3=3238/200χ2=3i,j=1(Oi,jEi,j)2Ei,j=8.03<χ24,0.05=9.488H0(C)
解答{b=n1=101=9c=1122.1804.062=318.038a=b1=91=8d=c/a=318.038/8=39.75475e=804.062/d=804.062/39.75475=20.226(a,b,c,d,e)=(8,9,318.038,39.75475,20.226)(C)

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