a catering company provides packages for weddings and for showers. the cost per person for small groups is…

Explanation:

Step1: Identify the means and standard - deviations of sample means

For weddings: $\mu_{w}=82.30$, $\sigma_{w}=18.20$, $n_{w}=9$. The mean of the sample mean for weddings is $\mu_{\bar{x}w}=\mu{w}=82.30$, and the standard deviation of the sample mean is $\sigma_{\bar{x}w}=\frac{\sigma{w}}{\sqrt{n_{w}}}=\frac{18.20}{\sqrt{9}}=\frac{18.20}{3}=6.067$. For showers: $\mu_{s}=65$, $\sigma_{s}=17.73$, $n_{s}=6$. The mean of the sample mean for showers is $\mu_{\bar{x}s}=\mu{s}=65$, and the standard deviation of the sample mean is $\sigma_{\bar{x}s}=\frac{\sigma{s}}{\sqrt{n_{s}}}=\frac{17.73}{\sqrt{6}}\approx7.23$.

Step2: Define the new random variable

Let $D = \bar{X}_w-\bar{X}s$. The mean of $D$ is $\mu_D=\mu{\bar{x}w}-\mu{\bar{x}s}=82.30 - 65=17.30$. The standard deviation of $D$ is $\sigma_D=\sqrt{\sigma{\bar{x}w}^{2}+\sigma{\bar{x}_s}^{2}}=\sqrt{6.067^{2}+7.23^{2}}=\sqrt{36.81 + 52.27}=\sqrt{89.08}\approx9.44$.

Step3: Calculate the z - score

We want to find $P(\bar{X}_w>\bar{X}_s)$, which is equivalent to $P(\bar{X}_w - \bar{X}_s>0)$ or $P(D > 0)$. The z - score is $z=\frac{0 - \mu_D}{\sigma_D}=\frac{0 - 17.30}{9.44}\approx - 1.83$.

Step4: Find the probability

Using the standard normal distribution table, $P(Z>-1.83)=1 - P(Z\leq - 1.83)$. From the standard - normal table, $P(Z\leq - 1.83)=0.0336$. So $P(Z>-1.83)=1 - 0.0336 = 0.9664\approx0.9665$.

Answer:

0.9665