Question

The number of knots in a particular type of wood has a Poisson distribution with an average of 1.7 knots in 10 cubic feet of the wood. Find the probability that a 10-cubic-foot block of the wood has at most 4 knots. (Round your answer to three decimal places.)

Answer 1

Answer:

0.972 = 97.2% probability that a 10-cubic-foot block of the wood has at most 4 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 time interval.

Average of 1.7 knots in 10 cubic feet of the wood.

This means that $\mu = 1.7$.

Find the probability that a 10-cubic-foot block of the wood has at most 4 knots.

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

In which

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

$P(X = 0) = \frac{e^{-1.7}*(1.7)^{0}}{(0)!} = 0.183$

$P(X = 1) = \frac{e^{-1.7}*(1.7)^{1}}{(1)!} = 0.311$

$P(X = 2) = \frac{e^{-1.7}*(1.7)^{2}}{(2)!} = 0.264$

$P(X = 3) = \frac{e^{-1.7}*(1.7)^{3}}{(3)!} = 0.15$

$P(X = 4) = \frac{e^{-1.7}*(1.7)^{4}}{(4)!} = 0.064$

$P(X \leq 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 0.183 + 0.311 + 0.264 + 0.15 + 0.064 = 0.972$

0.972 = 97.2% probability that a 10-cubic-foot block of the wood has at most 4 knots.