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Actual lengths of stay at a hospital’s emergency department in 2009 are shown in the following table (rounded to the nearest hour). Length of stay is the total of wait and ser- vice times. Some longer stays are also approximated as 15 hours in this table.

Respuesta :

The mean and the variance of the given discrete distribution are as follows:

  • Mean: 5.0563 minutes.
  • Variance: 7.9365 minutes².

What is the missing information?

The distribution of waiting times is missing, and is given by the image at the end of the answer.

The question is also incomplete, but it asks for the mean and the variance of the distribution.

What are the mean and the variance of a discrete distribution?

  • The mean of a discrete distribution is given by the sum of each outcome multiplied by it's respective probability.
  • The variance is given by the sum of the differences squared between each observation and the mean, multiplied by it's respective probability.

Hence, from the distribution off the image, considering the percentages as the probabilities, the mean is given by:

E(X) = 0.0299 x 1 + 0.1026 x 2 + 0.1496 x 3 + 0.2265 x 4 + 0.1581 x 5 + 0.1368 x 6 + 0.0769 x 7 + 0.0214 x 8 + 0.0299 x 9 + 0.0299 x 10 + 0.0385 x 15 =  5.0563 minutes.

Then, considering the mean found above, and it's formula, the variance is given as follows:

V(X) = 0.0299 x (1 - 5.0563)² + 0.1026 x (2 - 5.0563)² + 0.1496 x (3 - 5.0563)² + 0.2265 x (4 - 5.0563)² + 0.1581 x (5 - 5.0563)² + 0.1368 x (6 - 5.0563)² + 0.0769 x (7 - 5.0563)² + 0.0214 x (8 - 5.0563)² + 0.0299 x (9 - 5.0563)² + 0.0299 x (10 - 5.0563)² + 0.0385 x (15 - 5.0563)² = 7.9365 minutes².

More can be learned about the mean of a discrete distribution at https://brainly.com/question/27899440

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