Question

A certain kind of sheet metal has, on average, 9 defects per 11 square feet. Assuming a Poisson distribution, find the probability that a 14 square foot metal sheet has at least 10 defects. Round your answer to four decimals.

Answer 1

Answer:

The probability that there are at least 10 defects is 0,7067

Step-by-step explanation:

Using a rule of three, if there are 9 defects average per 11 square feet, then for 14 there should be 9/11 * 14 = 11.4545. Lets call X the total amount of defects per 14 square foots. The distribution of X is given by

$P_X(k) = \frac{e^{-11.4545} * {11.4545}^k }{k!}$

We can compute the probability for X being equal to 0,1,2,3,4,5,6,7,8 and 9 and substract it from 1

$P_X(0) = \frac{e^{-11.4545} * {11.4545}^0 }{0!} = 0.0000106012$

$P_X(1) = \frac{e^{-11.4545} * {11.4545}^1 }{1!} = 0.000121432$

$P_X(2) = \frac{e^{-11.4545} * {11.4545}^2 }{2!} = 0.000695472$

$P_X(3) = \frac{e^{-11.4545} * {11.4545}^3 }{3!} = 0.00265544$

$P_X(4) = \frac{e^{-11.4545} * {11.4545}^4 }{4!} = 0.007604218$

$P_X(5) = \frac{e^{-11.4545} * {11.4545}^5 }{5!} = 0.01742056$

$P_X(6) = \frac{e^{-11.4545} * {11.4545}^6 }{6!} = 0.033257437$

$P_X(7) = \frac{e^{-11.4545} * {11.4545}^7 }{7!} = 0.054421261$

$P_X(8) = \frac{e^{-11.4545} * {11.4545}^8 }{8!} = 0.077921351$

$P_X(9) = \frac{e^{-11.4545} * {11.4545}^9 }{9!} = 0.099172628$

Thus, P(X ≥10) = 1-(P(X <10)) = 1-0.2933 = 0.7067