Question

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 1

Answer:

$P(Z> 3.13) = 0.000874$

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 $\mu$

According to the information given, the data have a population mean:

$\mu = 3.2$.

The standard deviation of the data is:

$\sigma = 0.7$.

Then, from the data, a sample of size $n = 30$ is taken.

We want to obtain the probability that the sample mean is greater than 3.6

If we call

$\mu_m$ to the sample mean then, we seek to find:

$P(\mu_m> 3.6)$

To find this probability we find the Z statistic.

$Z = \frac{\mu_m-\mu}{\sigma_{\mu_m}}$

Where:

Where $\sigma_{\mu_m}$ is the standard deviation of the sample

$\sigma_{\mu_m} = \frac{\sigma}{\sqrt{n}}$

$\sigma_{\mu_m} =\frac{0.7}{\sqrt{30}}\\\\\sigma_{\mu_m} = 0.1278$

$P(\frac{\mu_m-\mu}{\sigma_{\mu_m}}> \frac{3.6-3.2}{0.1278})$

Then:

$Z = \frac{3.6-3.2}{0.1278}\\\\Z = 3.13$

The probability sought is: $P(Z> 3.13)$

When looking in the standard normal probability tables for right tail we obtain:

$P(Z> 3.13) = 0.000874$