Question

A catering company provides packages for weddings and for showers. The cost per person for small groups

Answer 1

Using the normal distribution and the central limit theorem, it is found that the probability the mean cost of the weddings is more than the mean cost of the showers is of 0.9665.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

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

• It 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, which is the percentile of X.

• By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

• When two variables are subtracted, the mean is the subtraction of the means, while the standard error is the square root of the sum of the variances.

What is the mean and the standard error of the distribution of differences?

For each sample, they are given by:

$\mu_W = 82.3, s_W = \frac{18.2}{\sqrt{9}} = 6.0667$

$\mu_S = 65, s_S = \frac{17.73}{\sqrt{6}} = 7.2382$

For the distribution of differences, we have that:

$\mu = \mu_W - \mu_S = 82.3 - 65 = 17.3$

$s = \sqrt{s_W^2 + s_S^2} = \sqrt{6.0667^2 + 7.2382^2} = 9.4444$

The probability the mean cost of the weddings is more than the mean cost of the showers is P(X > 0), that is, one subtracted by the p-value of Z when X = 0, hence:

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

By the Central Limit Theorem

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

$Z = \frac{0 - 17.3}{9.4444}$

$Z = -1.83$

$Z = -1.83$ has a p-value of 0.0335.

1 - 0.0335 = 0.9665.

More can be learned about the normal distribution and the central limit theorem at brainly.com/question/24663213