The number of surface flaws in plastic panels used in the interior of automobile

Question

The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.03 flaws per square foot of plastic panel. Assume an automobile interior contains 10 square feet of plastic panel. (a) What is the probability that there are no surface flaws in an auto's interior? (b) If 10 cars are sold to a rental company, what is the probability that none of the 10 cars has any surface flaws? (c) If 10 cars are sold to a rental company, what is the probability that at most 1 car has any surface flaws? Round your answers to four decimal places (e.g. 98.7654). (a) The probability is (b) The probability is c) The probability is

Explanation / Answer

We know that the the Poisson probability is: P(x; ) = (e-) (x) / x!

where u is the avergae rate and x is the actual number of success

so

using this formula

a) P(X=0) , means x = 0 and u = 0.03 putting the values in the formula and solving for it we get

(e-) (x) / x! = (e^-0.03)(0.03^0) /0! =

0.970

b) This means we must calculate P(X>10)

again using the same formula and putting in the values we get

(e-) (x) / x! = (e^-0.03)(0.03^10) /10! = 0.0000 (close to zero )

c )

at most 1 car means P(x<=1)

P(0) + p(1)

putting the values in the formula

(e^-0.03)(0.03^0) /0! + (e^-0.03)(0.03^1) /1!

= 0.9995

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---