Question

the population mean and standard deviation are given below. find the required probability and determine whether the given sample mean would be considered unusual. for a sample of n = 75, find the probability of a sample mean being greater than 212 if μ = 211 and σ = 6.1. for a sample of n = 75, the probability of a sample mean being greater than 212 if μ = 211 and σ = 6.1 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 $\sigma = 6.1$ and $n = 75$, we have $\sigma_{\bar{x}}=\frac{6.1}{\sqrt{75}}\approx\frac{6.1}{8.66}\approx0.7044$.

Step2: Calculate the z - score

The z - score formula for the sample mean is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$. Here, $\bar{x}=212$, $\mu = 211$ and $\sigma_{\bar{x}}\approx0.7044$. So $z=\frac{212 - 211}{0.7044}\approx1.42$.

Step3: Find the probability

We want $P(\bar{X}>212)$, which is equivalent to $P(Z > 1.42)$ in the standard normal distribution. Since $P(Z>z)=1 - P(Z\leq z)$ and from the standard - normal table $P(Z\leq1.42)=0.9222$, then $P(Z > 1.42)=1 - 0.9222 = 0.0778$.

Answer:

$0.0778$