Question

Cloth is sold by the bolt where a bolt is 15 yards of material. If the expected number of blemishes per yard is 0.1 find the probability that:

(a) A yard of cloth contains 1 or more blemishes.
(b) A bolt contains at most 2 blemishes.
(c) A lot of 30 bolts contains at least 50 blemishes.

Answer 1

Answer:

(a) The probability that a yard of cloth contains 1 or more blemishes is 0.0952.

(b) The probability that a yard of cloth contains at most 1 blemishes is 0.9953.

(c) The probability that in a lot of 30 bolts contains at least 50 blemishes is 0.2468.

Step-by-step explanation:

Let X = number of blemishes per yard of material.

The expected value of X is:

E (X) = λ = 0.1

The random variable  X follows a Poisson distribution with parameter λ = 0.10

The probability function of a Poisson distribution is:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0, 1, 2, ...$

(a)

Compute the probability that a yard of cloth contains 1 or more blemishes as follows:

P (X ≥ 1) = 1 - P (X < 1)

             = 1 - P (X = 0)

             $=1-\frac{e^{-0.10}(0.10)^{0}}{0!}\\=1-0.90484\\=0.09516\\\approx0.0952$

Thus, the probability that a yard of cloth contains 1 or more blemishes is 0.0952.

(b)

Compute the probability that a yard of cloth contains at most 1 blemishes as follows:

P (X ≤ 1) = P (X = 0) + P (X = 1)

             $=\frac{e^{-0.10}(0.10)^{0}}{0!}+\frac{e^{-0.10}(0.10)^{1}}{1!}\\=0.90484+0.090484\\=0.995324\\\approx0.9953$

Thus, the probability that a yard of cloth contains at most 1 blemishes is 0.9953.

(c)

In a lot of 30 bolts the expected number of yards of material is:

$\lambda_{1}=0.10\times30\times15=45$

Compute the probability that in a lot of 30 bolts contains at least 50 blemishes as follows:

P (X ≥ 50) = 1 - P (X < 50)

                $=1-0.7532\\=0.2468$

**Use an online calculator or Excel to compute the probability.

Thus, the probability that in a lot of 30 bolts contains at least 50 blemishes is 0.2468.