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Andrei [34K]
3 years ago
11

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.
Mathematics
1 answer:
jenyasd209 [6]3 years ago
8 0

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

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