Answer:
the probability of no defects in 10 feet of steel = 0.1353
Explanation:
GIven that:
A roll of steel is manufactured on a processing line. The anticipated number of defects in a 10-foot segment of this roll is two.
Let consider β to be the average value for defecting
So;
β = 2
Assuming Y to be the random variable which signifies the anticipated number of defects in a 10-foot segment of this roll.
Thus, y follows a poisson distribution as number of defect is infinite with the average value of β = 2
i.e
![Y \sim P( \beta = 2)](https://img.qammunity.org/2021/formulas/mathematics/college/5rb4zfoev5291v7no6auxbice5r1lu7iie.png)
the probability mass function can be represented as follows:
![\mathtt{P(y) = (e^(- \beta) \ \beta^ \ y)/(y!)}](https://img.qammunity.org/2021/formulas/mathematics/college/g2q5rv68guwb0jjgr9vymuqqop8yr1se6t.png)
where;
y = 0,1,2,3 ...
Hence, the probability of no defects in 10 feet of steel
y = 0
![\mathtt{P(y =0) = (e^(- 2) \ 2^ \ 0)/(0!)}](https://img.qammunity.org/2021/formulas/mathematics/college/amn7whzyls17b1o1811o51a7sqk8p3utql.png)
![\mathtt{P(y =0) = (0.1353 * 1)/(1)}](https://img.qammunity.org/2021/formulas/mathematics/college/m6b0vbpw05yxk2juf5mtqfa8mtxum2ongy.png)
P(y =0) = 0.1353