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 3.6
If 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]
