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 = 60 ), find the probability of a sample mean being less than 24.1 if ( mu = 24 ) and ( sigma = 1.18 ).
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 = 60 ), the probability of a sample mean being less than 24.1 if ( mu = 24 ) and ( sigma = 1.18 ) is ( square ).
(round to four decimal places as needed.)
would the given sample mean be considered unusual?
the sample mean ( square ) be considered unusual because it has a probability that is ( square ) than 5%.

Explanation:

Step1: Calculate the standard error

The standard error $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$. Given $\sigma = 1.18$ and $n = 60$, we have $\sigma_{\bar{x}}=\frac{1.18}{\sqrt{60}}\approx\frac{1.18}{7.746}\approx0.1523$.

Step2: Calculate the z - score

The z - score formula is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$. Here, $\bar{x} = 24.1$, $\mu=24$, and $\sigma_{\bar{x}}\approx0.1523$. So $z=\frac{24.1 - 24}{0.1523}=\frac{0.1}{0.1523}\approx0.66$.

Step3: Find the probability

Using the standard normal table, $P(Z\lt0.66)$ corresponds to the value in the table. Looking up $z = 0.66$ in the standard - normal table, we find $P(Z\lt0.66)=0.7454$.

Answer:

The probability of a sample mean being less than \(24.1\) is \(0.7454\). The sample mean would not be considered unusual because it has a probability that is greater than \(5\%\).