Answer:
64.11% for 200 days.
t=67.74 days for R=95%.
t=97.2 days for R=90%.
Step-by-step explanation:
Given that
β=2
Characteristics life(scale parameter α)=300 days
We know that Reliability function for Weibull distribution is given as follows
![R(t)=e^{-\left((t)/(\alpha)\right)^\beta}](https://img.qammunity.org/2020/formulas/engineering/college/mqfjnhciks2luunrae9bityxti4q3by2cj.png)
Given that t= 200 days
![R(200)=e^{-\left((200)/(300)\right)^2}](https://img.qammunity.org/2020/formulas/engineering/college/pj0q9vbtndhtgf8zxsb45ek8n5u7181y1v.png)
R(200)=0.6411
So the reliability at 200 days 64.11%.
When R=95 %
![0.95=e^{-\left((t)/(300)\right)^2}](https://img.qammunity.org/2020/formulas/engineering/college/hxwbf53hecasz0y9gg66m0smmprlyfsv7g.png)
by solving above equation t=67.74 days
When R=90 %
![0.90=e^{-\left((t)/(300)\right)^2}](https://img.qammunity.org/2020/formulas/engineering/college/ol4njzc5oko38vd8m0mwuzg7uzeus7xx2i.png)
by solving above equation t=97.2 days