the distribution of the number of words in text messages between employees at a large company is skewed…

Explanation:

Step1: Identify the mean and standard deviation of the sampling - distribution

The mean of the sampling - distribution of the sample mean $\mu_{\bar{x}}$ is equal to the population mean $\mu$. So, $\mu_{\bar{x}}=\mu = 8.6$. The standard deviation of the sampling - distribution of the sample mean (also known as the standard error) is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard deviation and $n$ is the sample size. Given $\sigma = 4.3$ and $n = 39$, then $\sigma_{\bar{x}}=\frac{4.3}{\sqrt{39}}\approx\frac{4.3}{6.245}\approx0.689$.

Step2: Calculate the z - score

The z - score is calculated using the formula $z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}$. We want to find $P(\bar{X}>10)$. Substitute $\bar{x} = 10$, $\mu_{\bar{x}}=8.6$, and $\sigma_{\bar{x}}\approx0.689$ into the z - score formula: $z=\frac{10 - 8.6}{0.689}=\frac{1.4}{0.689}\approx2.03$.

Step3: Find the probability

We want $P(Z>2.03)$. Since the total area under the standard normal curve is 1, and $P(Z>z)=1 - P(Z\leq z)$. Looking up $P(Z\leq2.03)$ in the standard - normal table, we find that $P(Z\leq2.03)=0.9788$. So, $P(Z>2.03)=1 - 0.9788 = 0.0212\approx0.0210$.

Answer:

0.0210