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DaniilM [7]
3 years ago
6

Help with this please

Mathematics
1 answer:
Gwar [14]3 years ago
4 0

Answer:1

Step-by-step explanation:

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a) A large hotel in Miami has 900 rooms (all rooms are equivalent). During Christmas, the hotel is usually fully booked. However
Olegator [25]

Answer:

14.69% probability that this happens

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

1000 people were given assurance of a room.

This means that n = 1000

Let us assume that each customer cancels their reservation with a probability of 0.1.

So 0.9 probability that they still keep their booking, which means that p = 0.9

Probability more than 900 still keeps their booking:

n = 1000, p = 0.9

So

\mu = 0.9, s = \sqrt{\frac{0.9*0.1}{1000}} = 0.0095

901/1000 = 0.91

So this is 1 subtracted by the pvalue of Z when X = 0.91.

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{0.91 - 0.9}{0.0095}

Z = 1.05

Z = 1.05 has a pvalue of 0.8531

1 - 0.8531 = 0.1469

14.69% probability that this happens

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