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Answer :
Final answer:
Using the z-score formula and the standard normal distribution table, the probability that a single randomly selected value is between 100.2 and 103.2, for a population with a mean of 97.7 and a standard deviation of 46.1, is 0.0259.
Explanation:
The subject of your question pertains to probability in statistics, specifically about normal distribution and the Central Limit Theorem. Given that the population is normally distributed with a mean (μ) of 97.7 and a standard deviation (σ) of 46.1, we can calculate the z-score for the range 100.2 to 103.2. The z-score can be calculated using the formula (X - μ)/σ, where X is the value we want to estimate.
For X=100.2, the z-score is (100.2-97.7)/46.1 = 0.054. For X=103.2, the z-score is (103.2 - 97.7)/46.1 = 0.119. Then, we find the probabilities that correspond to these z-scores from the standard normal distribution table which are around 0.5219 and 0.5478 respectively. The difference between these two probabilities will give the probability that a randomly selected value is between 100.2 and 103.2. So, P(100.20.0259.
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