Production of watering hoses with an average of 4 defects per meter - probabilities

Question

A company produces watering hoses with an average of 4 defects per meter.

Let X be the random variable that expresses the probability that the hose we buy has a defect.

a) What is the type of distribution followed by the probability function of the random variable and what is the value of its parameter?

b) We buy half a meter of hose from this company. Express the probability (by determining the corresponding value of the parameter) that the hose we bought has less than 3 defects

c) We buy 2 meters of hose from this company. Express the probability (by determining the corresponding value of the parameter) that the hose we bought has more than 4 defects.

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I have done the following :

a) Do we have a binomial distribution with $p=\frac{4}{100}$ ? Or do we have Poisson?

b) Is the probability equal to $P(X<3)=\sum_{i=0}^2\binom{n}{i}p^i(1-p)^{n-i}$ ? But which is the $n$ ? Does it symbolize here the half meter?

c) Is the probability equal to $P(X>4)=1-P(X\leq 4)=1-\sum_{i=0}^4\binom{n}{i}p^i(1-p)^{n-i}$ ? But which is the $n$ ? Does it symbolize here the two meters?

Answer 1

a) The random variable's distribution is clearly a Poisson: $X\sim Po(4)$

b) If we buy 0.5 meter we have a rv $X\sim Po(2)$ thus

$$\mathbb{P}(X<3)=e^{-2}\left[1+2+\frac{2^2}{2!} \right]\approx 67.67\%$$

c) similarly, now the distribution is $X\sim Po(8)$ thus

$$\mathbb{P}(X>4)=1-e^{-8}\left[1+8+\frac{8^2}{2!}+ \frac{8^3}{3!} +\frac{8^4}{4!} \right]\approx 90.04\%$$

Answer 2

For $(a)$, you've been given the one parameter - the average - so we have a Poisson distribution with mean $4$.

For $(b)$, note that you bought half a meter. So, you have to account for this. So, let $Y$ be the defects in half a meter so that $Y$ ~ $Po(2)$.

For $(c)$, you bought two meters, so you have to also take this into consideration as in $(b)$.