Mathematics, 05.03.2021 04:30 dbn4everloved

A certain type of aluminum screen has, an average, one flaw in a 100-foot roll. Assume the flaw distribution approximately follows the Poisson distribution. (a) Find the probability that a 100-foor roll has at most 2 flaws. (b) Suppose that I bought 10 200-foot rolls, find the probability that there are exactly 3 rolls that have no flaws in them. (c) Suppose that a store had many 200-foot rolls in storage, a clerk is doing inspections. What is the probability that the 15th roll he inspected was the 3rd one that have no flaws in it

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Answer from: klairecarregal6016

a) 0.9197 = 91.97% probability that a 100-foor roll has at most 2 flaws.

b) 0.1074 = 10.74% probability that there are exactly 3 rolls that have no flaws in them.

c) 0.0394 = 3.94% probability that the 15th roll he inspected was the 3rd one that have no flaws in it.

Step-by-step explanation:

To solve this question, we need to understand the Poisson and the Binomial distribution.

Poisson distribution:

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.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

(a) Find the probability that a 100-foor roll has at most 2 flaws.

A certain type of aluminum screen has, an average, one flaw in a 100-foot roll, which means that $\mu = 1$ . Only one roll means that the Poisson distribution is used.

This is:

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

In which

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

$P(X = 0) = \frac{e^{-1}*1^{0}}{(0)!} = 0.3679$

$P(X = 1) = \frac{e^{-1}*1^{1}}{(1)!} = 0.3679$

$P(X = 2) = \frac{e^{-1}*1^{2}}{(2)!} = 0.1839$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.3679 + 0.3679 + 0.1839 = 0.9197$

0.9197 = 91.97% probability that a 100-foor roll has at most 2 flaws.

(b) Suppose that I bought 10 200-foot rolls, find the probability that there are exactly 3 rolls that have no flaws in them.

Two parts.

Probability that a single 200-foot roll has no flaw.

This is P(X = 0) , Poisson(single 200-foot roll) when $\mu = \frac{200*1}{100} = 2$ . So

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

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

0.1353 probability that a single 200-foot roll has no flaw.

Probability that on 10 200-foot rolls, 3 have no flaws.

Multiple 200-foot rolls means that we use the binomial distribution.

0.1353 probability that a single 200-foot roll has no flaw means that p = 0.1353

10 200-foot rolls means that n = 10

We want P(X = 3) . So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 3) = C_{10,3}.(0.1353)^{3}.(0.8647)^{7} = 0.1074$

0.1074 = 10.74% probability that there are exactly 3 rolls that have no flaws in them.

0.1353 probability that a single 200-foot roll has no flaw means that p = 0.1353

2 with flaws in the first 14, which is P(X = 2) when n = 14

The 15th has no flaw, with probability of 0.1353. So

$P = 0.1353*P(X = 2) = 0.1353*(C_{14,2}.(0.1353)^{2}.(0.8647)^{12}) = 0.1353*0.2911 = 0.0394$

0.0394 = 3.94% probability that the 15th roll he inspected was the 3rd one that have no flaws in it

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