Mathematics, 08.04.2021 22:00, loxtin495

wedding planner does some research and finds that approximately 3.5% of the people in the area where a large wedding is to be held are pollotarian. Treat the 300 guests expected at the wedding as a simple random sample from the local population of about 2,000,000. 18) Suppose the wedding planner assumes that 5% of the guests will be pollotarian so she orders 15 pollotarian meals. What is the approximate probability that more than 5% of the guests are pollotarian and therefore she will not have enough pollotarian meals

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0.0793 = 7.93% probability that more than 5% of the guests are pollotarian and therefore she will not have enough pollotarian meals

Step-by-step explanation:

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

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, 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 $s = \sqrt{\frac{p(1-p)}{n}}$