Question

A textile producer has established that a spinning machine stops randomly due to thread breakages at an average rate of 5 stoppages per hour.

a) What is a probability that in a given hour, 3 stoppages will occur on this spinning machine

b) What is the probability that no more than 1 stoppage will occur in a given half-hour interval?

Answer 1

a)

The following equation of Poisson Distribution is used to determine the probability that in a given hour, 3 stoppages occur in the spinning machine is:

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

$P(x) = \frac{e^{-5}*5^3}{3!}$

$P(x=3) = 0.1404$

So, the probability that in a given hour 3 stoppages occur on the spinning machine is 0.1404 or 14.04%.

b)

The probability of no more than 1 stoppage will occur in a given half-hour interval. This means the probability of at most 2 stoppages will occur in a given hour needs to be determined:

$P(x\eqslantless2)=P(x=0)+P(x=1)+P(x=2)$

$P(x\eqslantless2)=0.0067 + 0.0336 + 0.0842$

$P(x\eqslantless2)=0.1247$

So, the probability of no more than 1 stoppage will occur in a given half-hour interval is 0.1247 or 12.47%.