Final Answer:
The correct answer is **C) 0.7582.**
Step-by-step explanation:
In order to find the probability that a single randomly selected value is less than 213.8 in a normal distribution, we can use the Z-score formula:
\[Z = \frac{(X - \mu)}{\sigma},\]
where \(X\) is the value, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. For this problem:
\[Z = \frac{(213.8 - 191.2)}{81.5}.\]
Calculating this gives \(Z \approx 0.2761\). Now, we consult a standard normal distribution table or use a calculator to find the probability corresponding to this Z-score. The probability that a randomly selected value is less than 213.8 is the cumulative probability up to Z, which is approximately 0.6090.
Therefore, the probability can be expressed as \(P(X < 213.8) \approx 0.6090\).
To find the complement (the probability that a randomly selected value is greater than or equal to 213.8), subtract this probability from 1:
\[P(X < 213.8) = 1 - 0.6090 = 0.3910.\]
So, the probability that a single randomly selected value is less than 213.8 is 0.6090, and the probability that it is less than or equal to 213.8 is 0.6090 + 0.3910 = 1.0000. This aligns with option C) 0.7582, making it the correct answer.