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A population of values has a normal distribution with $\mu=225$ and $\sigma=22.7$. If a random sample of size $n=12$ is selected,
a. Find the probability that a single randomly selected value is greater than 211.2 . Round your answer to four decimals.
\[
P(X> 211.2)=.9824
\]

Step 1 ：The problem is asking for the probability that a single randomly selected value from a normally distributed population is greater than 211.2. The population has a mean (μ) of 225 and a standard deviation (σ) of 22.7.

Step 2 ：To solve this, we need to convert the raw score (211.2) to a z-score. The z-score is a measure of how many standard deviations an element is from the mean. The formula to calculate the z-score is: \[ z = \frac{(X - μ)}{σ} \] where: X is the raw score (211.2 in this case), μ is the population mean (225), and σ is the standard deviation (22.7).

Step 3 ：After calculating the z-score, we can use a z-table or a statistical software to find the probability that a randomly selected value is greater than the z-score. However, the z-table gives the probability that the value is less than the z-score. So, we need to subtract the result from 1 to get the probability that the value is greater than the z-score.

Step 4 ：The calculated z-score is -0.6079295154185027 and the corresponding probability from the z-table is 0.7284.

Step 5 ：Final Answer: The probability that a single randomly selected value is greater than 211.2 is \(\boxed{0.7284}\).