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

Answer:

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

Step-by-step explanation:

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

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

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

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-1.6}*(1.6)^{0}}{(0)!} = 0.202$

$P(X = 1) = \frac{e^{-1.6}*(1.6)^{1}}{(1)!} = 0.323$

$P(X = 2) = \frac{e^{-1.6}*(1.6)^{2}}{(2)!} = 0.258$

$P(X = 3) = \frac{e^{-1.6}*(1.6)^{3}}{(3)!} = 0.138$

$P(X = 4) = \frac{e^{-1.6}*(1.6)^{4}}{(4)!} = 0.055$

$P(X = 5) = \frac{e^{-1.6}*(1.6)^{5}}{(5)!} = 0.018$

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

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