Answer:
The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 1.1156 occur / millimeters
Step-by-step explanation:
Step 1:-
Given data A copper wire, it is known that, on the average, 1.5 flaws occur per millimeter.
by Poisson random variable given that λ = 1.5 flaws/millimeter
Poisson distribution [tex]P(X= r) = \frac{e^{-\alpha } \alpha ^{r} }{r!}[/tex]
Step 2:-
The probability that no flaws occur in a certain portion of wire
[tex]P(X= 0) = \frac{e^{-1.5 } \(1.5) ^{0} }{0!}[/tex]
On simplification we get
P(x=0) = 0.223 flaws occur / millimeters
Conclusion:-
The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 5 X P(X=0) = 5X 0.223 = 1.1156 occur / millimeters
The probability that no flaws occur in a certain portion of wire of length 5 millimeters is 0.055%
Poisson distribution is used when there are several occurrences in a distribution. It is given by:
[tex]f(x)=\frac{e^{-\lambda}\lambda^x}{x!} \\\\where\ \lambda=mean,x=number\ of \ occurences[/tex]
For a 5 mm wire:
E(X) = 5mm * 1.5 flaws per mm = 7.5 mm
P(X = 0) = [tex]\frac{e^{-7.5}7.5^0}{0!}=0.055\%[/tex]
The probability that no flaws occur in a certain portion of wire of length 5 millimeters is 0.055%
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