Question

The number of knots in a particular type of wood has a Poisson distribution with an average of 1.6 knots in 10 cubic feet of the wood. Find the probability that a 10-cubic-foot block of the wood has at most 5 knots. (Round your answer to three decimal places.)

Answer 1

0.994 = 99.4% probability that a 10-cubic-foot block of the wood has at most 5 knots.

Explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

In which

x is the number of sucesses

e = 2.71828 is the Euler number

`\mu` is the mean in the given interval.

1.6 knots in 10 cubic feet of the wood.

This means that
`\mu = 1.6`

Find the probability that a 10-cubic-foot block of the wood has at most 5 knots.

`P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)`

In which

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 0) = (e^(-1.6)*(1.6)^(0))/((0)!) = 0.202`

`P(X = 1) = (e^(-1.6)*(1.6)^(1))/((1)!) = 0.323`

`P(X = 2) = (e^(-1.6)*(1.6)^(2))/((2)!) = 0.258`

`P(X = 3) = (e^(-1.6)*(1.6)^(3))/((3)!) = 0.138`

`P(X = 4) = (e^(-1.6)*(1.6)^(4))/((4)!) = 0.055`

`P(X = 5) = (e^(-1.6)*(1.6)^(5))/((5)!) = 0.018`

`P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0.202 + 0.323 + 0.258 + 0.138 + 0.055 + 0.018 = 0.994`

0.994 = 99.4% probability that a 10-cubic-foot block of the wood has at most 5 knots.