求算术平均值的标准差对某量独立测量8次,得802.40,802.50,802.38,802.48,802.42,802.

平均值:x=(802.40+802.50+802.38+802.48+802.42+802.46+802.45+802.43)/8 =802.44m
单次测量方差V1=(802.44-802.40)*(802.44-802.40)=16
V2=(802.44-802.50)*(802.44-802.50)=36
V3=(802.44-802.38)*(802.44-802.38)=36
V4=(802.44-802.48)*(802.44-802.48)=16
V5=(802.44-802.42)*(802.44-802.42)=4
V6=(802.44-802.46)*(802.44-802.46)=4
V7=(802.44-802.45)*(802.44-802.45)=1
V8=(802.44-802.43)*(802.44-802.43)=1
[vv]=16+36+36+16+4+4+1+1=114
标准差(中误差):m=±√[vv]/n-1
=±√114/7
=±4.04mm
极限误差:Δ=3m=3*4.04=12.12mm
算术平均值的标准差:M= ±m/√n
= ±4.04/2.83
= ± 1.43mm