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 = 70, find the probability of a sample mean being greater than 211 if μ = 210 and σ = 5.7.

for a sample of n = 70, the probability of a sample mean being greater than 211 if μ = 210 and σ = 5.7 is . (round to four decimal places as needed.)

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard - deviation and $n$ is the sample size.
$\sigma_{\bar{x}}=\frac{5.7}{\sqrt{70}}\approx\frac{5.7}{8.3666}\approx0.6813$

Step2: Calculate the z - score

The z - score formula for the sample mean is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean, and $\sigma_{\bar{x}}$ is the standard error of the mean.
$z=\frac{211 - 210}{0.6813}=\frac{1}{0.6813}\approx1.47$

Step3: Find the probability

We want to find $P(\bar{X}>211)$, which is equivalent to $P(Z > 1.47)$ in the standard normal distribution.
Since the total area under the standard - normal curve is 1, $P(Z>1.47)=1 - P(Z\leq1.47)$.
Looking up $P(Z\leq1.47)$ in the standard - normal table, we find that $P(Z\leq1.47) = 0.9292$.
So, $P(Z>1.47)=1 - 0.9292=0.0708$

Answer:

$0.0708$