容量分别是n1,n2.组成一个联合样本n1+n2,组合样本的方差是? n1与n2是相互独立的高斯随机变量,均值为0,方差为σ2,x...
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组合样本的方差是 S1^2 + S2^2This problem is not as simple as the answer suggests here. When combining m random variables: n1, n2, ..., nm each with an average value of X1, X2, ..., Xm and a standard deviation S1, S2, ..., Sm, the average of the new random variable N = n1 + n2 + ... + nm will have an average value of X = X1 + X2 + ... + Xm, this is expected. The standard deviation of the new random variable, though, is S = (S1^2 + S2^2 + ... + Sm^2)^(1/2). So the variance of the new random variable or Var(N) = S1^2 + S2^2 + ... + Sm^2
配对样本还是独立样本？
绛旓細缁勫悎鏍锋湰鐨勬柟宸槸 S1^2 + S2^2 This problem is not as simple as the answer suggests here. When combining m random variables: n1, n2, ..., nm each with an average value of X1, X2, ..., Xm and a standard deviation S1, S2, ..., Sm, the average of the new random...
绛旓細鐢卞崱鏂瑰垎甯冪殑鎬ц川鏈夛細(n1?1)S21蟽2锝瀀2锛坣1-1锛夛紝(n2?1)S22蟽2锝瀀2锛坣2-1锛夋晠锛欴锛(n1?1)S21蟽2锛=2锛坣1-1锛(n1?1)2蟽4DS21=2锛坣1-1锛塂S21=2蟽4n1?1鍚岀悊鍙緱锛欴S22=2蟽4n2?1鎵浠ユ湁锛欴锛圫12+S22锛=2蟽4n1?1+2蟽4n2?1=2(n1+n2?2)(n1?1)(n2?1)蟽4锛
绛旓細1绫7涓夛紝灏辨槸1绫7涓夈
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绛旓細鐢1鍙风鏉ヨ锛宎=0.3 b=0.24,褰撳彇n1=2,n2=2鏃秄(n1,n2)=1.1-1.08=0.02 鍚岀悊2鍙风涓 n1=1 n2=2 3鍙风涓 n1=1 n2=4 鎴杗1=3,n2=1 2.鍙皢f(n1,n2)-=min(1.1-n1*a-(n2-1)*b)>0 姣忚竟澶氭帓鍒椾袱涓崐涓墠涓嶄細鎺夈傚嵆鐪嬪仛1.1+b鐨勬鏂瑰舰 ...
绛旓細char *substr(const char *s,int n1,int n2)/*浠巗涓彁鍙栦笅鏍涓簄1~n2鐨勫瓧绗缁勬垚涓涓鏂板瓧绗︿覆锛岀劧鍚庤繑鍥炶繖涓柊涓茬殑棣栧湴鍧*/ { char *sp;sp=(char *)malloc(sizeof(char)*(n2-n1+2));int i,j=0;for (i=n1; i<=n2; i++) { sp[j++]=s[i];} sp[j]=0;return sp;} int ...
绛旓細N1:N2:N3=U1:U2:U3=4:2:1 鍥犱负姝ｅ父鍙戝厜锛屾墍浠2=6V,I2=U/R=6/3=2A R=U^2/P=6娆 鎵浠ュ浜巒3, R3=6*6/(6+6)=3娆 鎵浠3=4A I1*n1=I2*n2+I3*n3 甯﹀叆鏁版嵁寰桰1=2A 鍥犱负P=I^2R 寰桺2+P3=24W P1=12W 鏁匬鎬=36W ...
绛旓細鐞嗘兂姘斾綋鍐呰兘鍏紡锛歟=i/2 kT 瀵逛簬鍗曞師瀛愬垎瀛愶紝i=3 瀵逛簬鍙屽師瀛愬垎瀛愶紝i=5 鎵浠ュ鍣ㄥ唴瑁呮湁N1涓鍗曞師瀛愮悊鎯虫皵浣撳垎瀛愬拰N2涓鍒氭у弻鍘熷瓙鍒嗗瓙锛屽綋绯荤粺娓╁害T鏃跺唴鑳戒负3/2 N1kT+5/2 N2kT
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