1 测定某矿样中CaO含量时,得如下数据:37.45%,37.20%,37.50%,37.25%,37.30%,计算此结果的平均值、平均偏差和标准差,写出正式报告。解=37.45+37.20+37.50+37.25+37.30/5=37.34%=(sum from i=1 to n |d<sub>i</sub>|...1 测定某矿样中CaO含量时,得如下数据:37.45%,37.20%,37.50%,37.25%,37.30%,计算此结果的平均值、平均偏差和标准差,写出正式报告。解=37.45+37.20+37.50+37.25+37.30/5=37.34%=(sum from i=1 to n |d<sub>i</sub>|)/n=(0.11+0.14+0.04+0.16+0.09)/5=0.11%S=(sum from i=1 to n d<sub>i</sub><sup>2</sup>/(n-1))<sup>1/2</sup> (((0.11)<sup>2</sup>+(0.14)<sup>2</sup>+(0.04)<sup>2</sup>+(0.16)<sup>2</sup>+(0.09)<sup>2</sup>)/(5-1))<sup>1/2</sup> =0.13%分析报告:n=5.=37.34%,S=0.13%对于误差和偏差、准确度和精密度的含义要清楚。2 下列各数的有效数字位数是几位?展开更多
文摘1 测定某矿样中CaO含量时,得如下数据:37.45%,37.20%,37.50%,37.25%,37.30%,计算此结果的平均值、平均偏差和标准差,写出正式报告。解=37.45+37.20+37.50+37.25+37.30/5=37.34%=(sum from i=1 to n |d<sub>i</sub>|)/n=(0.11+0.14+0.04+0.16+0.09)/5=0.11%S=(sum from i=1 to n d<sub>i</sub><sup>2</sup>/(n-1))<sup>1/2</sup> (((0.11)<sup>2</sup>+(0.14)<sup>2</sup>+(0.04)<sup>2</sup>+(0.16)<sup>2</sup>+(0.09)<sup>2</sup>)/(5-1))<sup>1/2</sup> =0.13%分析报告:n=5.=37.34%,S=0.13%对于误差和偏差、准确度和精密度的含义要清楚。2 下列各数的有效数字位数是几位?