Answer:
the probability of no defects in 10 feet of steel = 0.1353
Step-by-step 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
[tex]Y \sim P( \beta = 2)[/tex]
the probability mass function can be represented as follows:
[tex]\mathtt{P(y) = \dfrac{e^{- \beta} \ \beta^ \ y}{y!}}[/tex]
where;
y = 0,1,2,3 ...
Hence, the probability of no defects in 10 feet of steel
y = 0
[tex]\mathtt{P(y =0) = \dfrac{e^{- 2} \ 2^ \ 0}{0!}}[/tex]
[tex]\mathtt{P(y =0) = \dfrac{0.1353 \times 1}{1}}[/tex]
P(y =0) = 0.1353
