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问题: 数学问题

A,B两台测量仪器测量某一产品的直径多次,得分布列如下,是比较两种仪器的优劣。

解答:

A要比B好

(1)
利用数学期望来做:
A:
118*0.06+119*0.14+120*0.60+121*0.15+122*0.05
=7.08+16.66+72+18.15+6.1
=119.99
B:
118*0.09+119*0.15+120*0.52+121*0.16+122*0.08
=10.62+17.85+62.4+19.36+9.76
=119.99
A的数学期望=B的数学期望,比较不出来,再利用方差.

(2)
利用方差来做:
A:
(118-119.99)^2*0.06+(119-119.99)^2*0.14+(120-119.99)^2*0.60+(121-119.99)^2*0.15+(122-119.99)^2*0.05
=1.99^2*0.06+0.99^2*0.14+0.01^2*0.60+1.01^2*0.15+2.01^2*0.05
=0.237606+0.137214+0.00006+0.153015+0.202005
=0.7299
B:
(118-119.99)^2*0.09+(119-119.99)^2*0.15+(120-119.99)^2*0.52+(121-119.99)^2*0.16+(122-119.99)^2*0.08
=1.99^2*0.09+0.99^2*0.15+0.01^2*0.52+1.01^2*0.16+2.01^2*0.08
=0.356409+0.147015+0.000052+0.163216+0.323208
=0.9899

因为A的方差0.7299<B的方差0.9899

故A要比B好