Question

When carpet is manufactured, small faults occur at random. The number of faults in Premium carpets can be modelled by a Poisson distribution with mean 0.5 faults per 20 m2. Mr Jones chooses Premium carpets to replace the carpets in his office building. The office building has 10 rooms, each with the area of 80 m2. Find the probability that the carpet laid in the first room has fewer than three faults.

Answer 1

Answer:

67.67% probability that the carpet laid in the first room has fewer than three faults.

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.

Mean 0.5 faults per 20 m2. The office building has 10 rooms, each with the area of 80 m2.

This means that for each room, $\mu = \frac{80*0.5}{20} = 2$

Find the probability that the carpet laid in the first room has fewer than three faults.

This is

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

In which

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

$P(X = 0) = \frac{e^{-2}*(2)^{0}}{(0)!} = 0.1353$

$P(X = 1) = \frac{e^{-2}*(2)^{1}}{(1)!} = 0.2707$

$P(X = 2) = \frac{e^{-2}*(2)^{2}}{(2)!} = 0.2707$

So

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.1353 + 0.2707 + 0.2707 = 0.6767$

67.67% probability that the carpet laid in the first room has fewer than three faults.