Question

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

Answer 1

Answer:

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 $s = \sqrt{\frac{p(1-p)}{n}}$

3.5% of the people in the area where a large wedding is to be held are pollotarian.

This means that $p = 0.035$

300 guests

This means that $n = 300$

Mean and standard deviation:

$\mu = p = 0.035$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.035*0.965}{300}} = 0.0106$

What is the approximate probability that more than 5% of the guests are pollotarian and therefore she will not have enough pollotarian meals?

This is 1 subtracted by the pvalue of Z when X = 0.05. So

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{0.05 - 0.035}{0.0106}$

$Z = 1.41$

$Z = 1.41$ has a pvalue of 0.9207

1 - 0.9207 = 0.0793

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