Using the <em>normal distribution and the central limit theorem</em>, 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.

<h3>Normal Probability Distribution</h3>

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.

<h3>What is the mean and the standard error of the distribution of differences?</h3>

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, <u>one subtracted by the p-value of Z when X = 0</u>, hence:

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

By the Central Limit Theorem

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