UpStudy Free Solution:

Given:

- Population size \(N = 800\)

- Sample size \(n = 50\)

- Population mean \(\mu = 2.813\)

- Population standard deviation \(\sigma = 0.008\)

- Sample mean \(\bar { X} = 2.812\)

First, determine if the finite population correction (FPC) should be used. We check if the sample size is more than 5% of the population size:

\[\frac { n} { N} = \frac { 50} { 800} = 0.0625\]

Since \(0.0625\) (or 6.25%) is greater than 5%, we should use the finite population correction factor.

Calculate the finite population correction (FPC):

\[\text { FPC} = \sqrt { \frac { N - n} { N - 1} } \]

\[\text { FPC} = \sqrt { \frac { 800 - 50} { 800 - 1} } \]

\[\text { FPC} = \sqrt { \frac { 750} { 799} } \]

\[\text { FPC} \approx \sqrt { 0.9387} \]

\[\text { FPC} \approx 0.9688\]

Next, calculate the standard error without FPC:

\[\text { SE} = \frac { \sigma } { \sqrt { n} } = \frac { 0.008} { \sqrt { 50} } \]

\[\text { SE} \approx \frac { 0.008} { 7.071} \]

\[\text { SE} \approx 0.001131\]

Apply the FPC to the standard error:

\[\text { SE} _ { \text { corrected} } = \text { SE} \times \text { FPC} \]

\[\text { SE} _ { \text { corrected} } = 0.001131 \times 0.9688\]

\[\text { SE} _ { \text { corrected} } \approx 0.001095\]

Now, calculate the Z-score for the sample mean being less than 2.812:

\[Z = \frac { \bar { X} - \mu } { \text { SE} _ { \text { corrected} } } \]

\[Z = \frac { 2.812 - 2.813} { 0.001095} \]

\[Z = \frac { - 0.001} { 0.001095} \]

\[Z \approx - 0.913\]

Finally, look up the Z-score in the standard normal distribution table or use a calculator to find the probability:

\[P( Z < - 0.913) \approx 0.1809\]

Supplemental Knowledge

When dealing with sampling from a finite population, the finite population correction (FPC) factor is used to adjust the standard error of the sample mean. This adjustment is necessary when the sample size \(n\) is a significant fraction of the total population size \(N\). The FPC is particularly important when \(n/N > 0.05\).

The formula for the finite population correction factor is:

\[\text { FPC} = \sqrt { \frac { N - n} { N - 1} } \]

The standard error of the sample mean without FPC is given by:

\[\text { SE} = \frac { \sigma } { \sqrt { n} } \]

After applying the FPC, the corrected standard error becomes:

\[\text { SE} _ { \text { corrected} } = \text { SE} \times \text { FPC} \]

This corrected standard error accounts for the reduced variability due to sampling without replacement from a finite population.

Understanding statistical concepts like sampling distributions and finite population corrections can significantly strengthen your ability to analyze data accurately. At UpStudy, we offer detailed explanations and step-by-step statistical solutions for all your statistical queries - be they related to mathematics, chemistry, physics or biology! Regardless of their complexity we've got you covered!

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