A set of normally distributed data has a mean of 3.2 and a standard deviation of 0.7. Find the probability of randomly selecting 30 values and getting a mean greater than 3.6.

Answer:[tex]P(Z> 3.13) = 0.000874[/tex]Step-by-step explanation:A set of normally distributed data has a mean of 3.2 and a standard deviation of 0.7. Find the probability of randomly selecting 30 values and obtaining an average greater than 3.6.We can denote the population mean with the symbol [tex]\mu[/tex]According to the information given, the data have a population mean:[tex]\mu = 3.2[/tex].The standard deviation of the data is:[tex]\sigma = 0.7[/tex].Then, from the data, a sample of size [tex]n = 30[/tex] is taken.We want to obtain the probability that the sample mean is greater than 3.6If we call[tex]\mu_m[/tex] to the sample mean then, we seek to find:[tex]P(\mu_m> 3.6)[/tex]To find this probability we find the Z statistic.[tex]Z = \frac{\mu_m-\mu}{\sigma_{\mu_m}}[/tex]Where:Where [tex]\sigma_{\mu_m}[/tex] is the standard deviation of the sample[tex]\sigma_{\mu_m} = \frac{\sigma}{\sqrt{n}}[/tex][tex]\sigma_{\mu_m} =\frac{0.7}{\sqrt{30}}\\\\\sigma_{\mu_m} = 0.1278[/tex][tex]P(\frac{\mu_m-\mu}{\sigma_{\mu_m}}> \frac{3.6-3.2}{0.1278})[/tex]Then:[tex]Z = \frac{3.6-3.2}{0.1278}\\\\Z = 3.13[/tex]The probability sought is: [tex]P(Z> 3.13)[/tex]When looking in the standard normal probability tables for right tail we obtain:[tex]P(Z> 3.13) = 0.000874[/tex]