Question

the population mean and standard deviation are given below. find the indicated probability and determine whether the given sample mean would be considered unusual. for a sample of n = 40, find the probability of a sample mean being less than 12,750 or greater than 12,753 when μ = 12,750 and σ = 1.4. for the given sample, the probability of a sample mean being less than 12,750 or greater than 12,753 is (round to four decimal places as needed.)

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$. Given $\sigma = 1.4$ and $n = 40$, we have $\sigma_{\bar{x}}=\frac{1.4}{\sqrt{40}}\approx\frac{1.4}{6.3246}\approx0.2214$.

Step2: Calculate the z - scores

For $\bar{x}_1 = 12750$, the z - score $z_1=\frac{\bar{x}_1-\mu}{\sigma_{\bar{x}}}=\frac{12750 - 12750}{0.2214}=0$.
For $\bar{x}_2 = 12753$, the z - score $z_2=\frac{\bar{x}_2-\mu}{\sigma_{\bar{x}}}=\frac{12753 - 12750}{0.2214}=\frac{3}{0.2214}\approx13.55$.

Step3: Find the probabilities

We want $P(\bar{X}<12750\ or\ \bar{X}>12753)$. Since $P(\bar{X}<12750)=\Phi(0)$ and $P(\bar{X}>12753)=1 - \Phi(13.55)$.
The standard normal distribution $\Phi(0)=0.5$ and $\Phi(13.55)\approx1$. So $P(\bar{X}<12750\ or\ \bar{X}>12753)=0.5+(1 - 1)=0.5000$.

Answer:

$0.5000$