Please answer both questions, thank you! Worksheet 5 Due 10-23-17 Nonmal Random
ID: 3335278 • Letter: P
Question
Please answer both questions, thank you!
Worksheet 5 Due 10-23-17 Nonmal Random Variables Name The lergths of a certain component are normally distributed with a mean of 250 mm and a stardard deviation of 0. mm Four of the components are joined end to end and fitted into a slot. 1. I the slots have length 1002.5 mm, what proportion of the time will four randomly selected components be able to fit into the slot? 2. Suppose that the lergths of the slots are also are normally distributed, with a mean of 1002.5 mm and a standard deviation of0.4 mm. What proportion of the time will four randomly selected components be able to fit into the slot?Explanation / Answer
1)here for four component mean length =4*250 =1000
and std deviation =0.6*(4)1/2 =1.2
therefore proprotion of time 4 component will be able to fit =P(X<1002.5)=P(Z<(1002.5-1000)/1.2)=P(Z<2.0833)
=0.9814
2)here let length of components is X and that of slot =Y
hence mean difference =(X-Y) =1000-1002.5 =-2.5
and std deviation of mean difference =(1.22+0.42)1/2 =1.2649
hence probability that components fit into slot =P(X-Y<0)=P(Z<(0-(-2.5))/1.2649)=P(Z<1.9764)=0.9759