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
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
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
and standard deviation ![s = \sqrt{\frac{p(1-p)}{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D)
1000 people were given assurance of a room.
This means that ![n = 1000](https://tex.z-dn.net/?f=n%20%3D%201000)
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](https://tex.z-dn.net/?f=p%20%3D%200.9)
Probability more than 900 still keeps their booking:
![n = 1000, p = 0.9](https://tex.z-dn.net/?f=n%20%3D%201000%2C%20p%20%3D%200.9)
So
![\mu = 0.9, s = \sqrt{\frac{0.9*0.1}{1000}} = 0.0095](https://tex.z-dn.net/?f=%5Cmu%20%3D%200.9%2C%20s%20%3D%20%5Csqrt%7B%5Cfrac%7B0.9%2A0.1%7D%7B1000%7D%7D%20%3D%200.0095)
901/1000 = 0.91
So this is 1 subtracted by the pvalue of Z when X = 0.91.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{0.91 - 0.9}{0.0095}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B0.91%20-%200.9%7D%7B0.0095%7D)
![Z = 1.05](https://tex.z-dn.net/?f=Z%20%3D%201.05)
has a pvalue of 0.8531
1 - 0.8531 = 0.1469
14.69% probability that this happens