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------------------------------------------------ A population of values has a normal distribution with μ=229.6 and σ=97.4. If a random sample of size n=15 is selected, find the probability that a single randomly selected value is less than 297.5. Round your answer to four decimals.

To find the probability that a single randomly selected value is less than 297.5 from a normally distributed population, we'll use the properties of the normal distribution.

Given data:

Mean ([tex]\mu[/tex]) = 229.6
Standard deviation ([tex]\sigma[/tex]) = 97.4

We want to find [tex]P(X < 297.5)[/tex].

Step-by-step Solution

Calculate the Z-score:
The Z-score helps us understand how many standard deviations a specific value (in this case, 297.5) is away from the mean. The formula for calculating the Z-score is:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Plug in the values:
[tex]Z = \frac{297.5 - 229.6}{97.4} = \frac{67.9}{97.4} \approx 0.6971[/tex]
Use the Z-score to find the probability:
To find the probability [tex]P(X < 297.5)[/tex], we'll use the standard normal distribution table (Z-table) or a calculator with normal distribution functions.
Find the cumulative probability corresponding to the Z-score of 0.6971.
According to the Z-table or calculator, [tex]P(Z < 0.6971) \approx 0.7570.[/tex]
Interpret the result:
The probability that a randomly selected value from this population is less than 297.5 is approximately 0.7570.
Therefore, there is a 75.70% chance that a value randomly chosen from this population will be less than 297.5.