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:7.93%

Explanation:

Answer 2

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

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 = (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 = (\sigma)/(√(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{(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{(p(1-p))/(n)} = \sqrt{(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 = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (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