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maxonik [38]
3 years ago
9

Service calls arriving at an electric company follow a Poisson distribution with an average arrival rate of 5656 per hour. Find

the average and standard deviation of the number of service calls in a 1515-minute period. Round your answer to three decimal places, if necessary.
Mathematics
1 answer:
liberstina [14]3 years ago
4 0

Answer:

The average number of service calls in a 15-minute period is of 14, with a standard deviation of 3.74.

Step-by-step explanation:

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. The variance is the same as the mean.

Average rate of 56 calls per hour:

This means that \mu = 56n, in which n is the number of hours.

Find the average and standard deviation of the number of service calls in a 15-minute period.

15 minute is one fourth of a hour, which means that n = \frac{1}{4}. So

\mu = 56n = \frac{56}{4} = 14

The variance is also 14, which means that the standard deviation is \sqrt{14} = 3.74

The average number of service calls in a 15-minute period is of 14, with a standard deviation of 3.74.

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