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Suppose that a particular large hotel has 790 rooms. Furthermore, suppose that the hotel's marketing group's forecast is normally distributed with a mean demand of 730 rooms and a standard deviation of 35 rooms for this coming weekend. What is the probability that the hotel will sell out all 790 rooms this weekend? (please round your answer to 4 decimal places)

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nareenasad80

Answer:

The probability that the hotel will sell out all 790 rooms this weekend is approximately 0.9564 or 95.64% (rounded to 4 decimal places).

Step-by-step explanation:

To find the probability that the hotel will sell out all 790 rooms this weekend, you can use the standard normal distribution and the Z-score formula.

The Z-score is calculated using the formula:

Z = X - μ​ / σ

Where:

  •    X is the value (in this case, the number of rooms sold),
  •    μ is the mean,
  •    σ is the standard deviation.

In this case, X = 790, μ = 730, and σ = 35.

[tex]Z = \frac{790-730}{35} = \frac{60}{35}[/tex]

Now, you would look up the Z-score in the standard normal distribution table (or use a calculator) to find the corresponding probability.

Let's calculate the Z-score:

Z ≈ 1.7143

Now, find the probability associated with this Z-score. You can use a standard normal distribution table or a calculator. The probability that the hotel will sell out all 790 rooms is the area to the right of this Z-score.

For a Z-score of 1.7143, the probability is approximately 0.9564.

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