2. The breaking strength of plastic bags used for packaging products is normally
distributed with a mean of 5 N/mm2 and a standard deviation of 1.5 N/mm2.
What proportion of the plastic bags has a breaking strength of:
(a) less than 3.17 N/mm2?
(b) at least 3.6 N/mm2?
(c) between 5 and 5.5 N/mm2?
(d) Between what two values symmetrically distributed around the mean will 95%
of the breaking strengths fall?
統計學...救命呀
2007-03-15 10:11 am
回答 (1)
2007-03-15 6:39 pm
✔ 最佳答案
(a) N(breaking strength<3.17)=N(z<((3.17-5)/1.5))=N(z<-1.22)=0.1112 (b) N(breaking strength>3.6)=N(z>((3.6-5)/1.5))=n(z>-0.9333)=1-N(z<-0.9333)=1-0.1753=0.8247
(c) N(5<breaking strength<5.5)=N((5-5)/1.5<z<((5.5-5)/1.5))
=N(0<z<0.3333)=N(z<0.3333)-N( z < 0 )=0.6306-0.5=0.1306
(d) N(z1)=0.025
z1= -1.96
(lower bound-5)/1.5=-1.96
lower bound=2.06
N(z2)=0.975
Z2= 1.96
(upper bound-5)/1.5=1.96
upper bound=7.94
The breaking strength is between 2.06N/mm2 and 7.94N/mm2.
參考: 自己
收錄日期: 2021-04-13 14:14:43
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070315000051KK00357