If an average of five service calls are received per hour in a repair department. What is the probability that they receive fewer than three calls during a randomly chosen hour? choose the correct answer a) 0.4662 b) 0.6231 a) 0.1562 b) 0.6131 a) 0.1632 b) 0.6311 a) 0.1662 b) 0.6331

To solve this problem, we can use the Poisson distribution, which is commonly used to model the number of events occurring in a fixed interval of time or space.

The formula for the probability mass function of the Poisson distribution is:

P(X=k)=\frac{e^{-\lambda}\lambda^k}{k!}

where X is the random variable representing the number of events, k is the number of events we are interested in (in this case, fewer than three), and \lambda is the average number of events per interval (in this case, five).

To find the probability that they receive fewer than three calls during a randomly chosen hour, we can calculate the following:

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

Let's calculate each term:

P(X=0)=\frac{e^{-5}5^0}{0!}=e^{-5}\approx 0.0067

P(X=1)=\frac{e^{-5}5^1}{1!}=5e^{-5}\approx 0.0337

P(X=2)=\frac{e^{-5}5^2}{2!}=\frac{25e^{-5}}{2}\approx 0.0842

Now, we can add these probabilities together to get the final answer:

P(X<3)=0.0067+0.0337+0.0842=0.1246

Therefore, the probability that they receive fewer than three calls during a randomly chosen hour is approximately 0.1246.

Answer: a) 0.1562