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 = 65 ), find the probability of a sample mean being less than ( 20.1 ) if ( mu = 20 ) and ( sigma = 1.2 ).
click the icon to view page 1 of the standard normal table.
click the icon to view page 2 of the standard normal table.
for a sample of ( n = 65 ), the probability of a sample mean being less than ( 20.1 ) if ( mu = 20 ) and ( sigma = 1.2 ) is (round to four decimal places as needed.)
would the given sample mean be considered unusual?

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$.
Given $\sigma = 1.2$ and $n = 65$, we have $\sigma_{\bar{x}}=\frac{1.2}{\sqrt{65}}\approx\frac{1.2}{8.0623}\approx0.149$.

Step2: Calculate the z - score

The formula for the z - score is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$.
Here, $\bar{x} = 20.1$, $\mu = 20$, and $\sigma_{\bar{x}}\approx0.149$. So $z=\frac{20.1 - 20}{0.149}=\frac{0.1}{0.149}\approx0.67$.

Step3: Find the probability using the standard normal table

We want to find $P(\bar{X}<20.1)$, which is equivalent to $P(Z < 0.67)$.
Looking up the value in the standard normal table, $P(Z < 0.67)=0.7486$.

Answer:

The probability is $0.7486$. Since the probability ($0.7486>0.05$), the sample mean is not unusual.