Answer :
Final answer:
This mathematics question concerns probability and statistics, specifically the normal distribution involving human body temperature. Through standardized computations and reference to a normal distribution table, we can determine the associated probabilities. For instance, the probability of a randomly sampled individual having a body temperature less than 99°F is approximately 0.985.
Explanation:
The subject of this question is related to probability and statistics, specifically around the normal distribution. Let's tackle each component of the question step by step:
- In regards to what is the probability that a person’s body temperature is at most 99°F, first, we need to standardize the temperature using the formula z = (x - μ) / σ, where x = 99, μ = 97.7, and σ = 0.6. Hence, z = (99 - 97.7) / 0.6 = 2.17. Consulting a standard normal distribution table or using a z-score calculator, we find that the probability is approximately 0.985.
- For the probability that a person’s body temperature is within 1.5°F of the mean value, we calculate the z-scores for 97.7 - 1.5 = 96.2 and 97.7 + 1.5 = 99.2. We find z-scores -2.50 and 2.50. The probability is the area under the curve between these z-scores, which is approximately 0.987.
- To determine the temperature value that separates the highest 33% from the rest, we look for the z-score that corresponds to the 67th percentile (100% - 33%), which is approximately 0.43. Using the formula x = μ + z * σ, we find x = 97.7 + 0.43 * 0.6 = 98
- Lastly, the probability that all of 10 randomly sampled people have temperatures less than 99°F is simply the 10th power of the probability found in first part, which is approximately 0.85.
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