College

Proportion of body temperature scores between 96.9 and 98.1:

1. First z-score for [tex]x = 96.9[/tex] is:
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2. Second z-score for [tex]x = 98.1[/tex] is:
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3. First proportion for [tex]x = 96.9[/tex] is:
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4. Second proportion for [tex]x = 98.1[/tex] is:
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5. Final total proportions:
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Answer :

Sure! Let's go through the steps to find the proportion of body temperature scores between 96.9 and 98.1 using a statistical approach:

1. Identify the Mean and Standard Deviation:
- Typically, the mean body temperature is around 98.6 degrees Fahrenheit.
- The standard deviation is approximately 0.7 degrees Fahrenheit.

2. Calculate the Z-Scores:
- The z-score is a way to measure how many standard deviations an element is from the mean.
- For a temperature of 96.9 degrees:
- First z-score: Calculate using the formula [tex]\( z = \frac{(x - \text{mean})}{\text{standard deviation}} \)[/tex].
- Substituting the values, we get a first z-score: [tex]\(-2.43\)[/tex].
- For a temperature of 98.1 degrees:
- Second z-score: Using the same formula, the second z-score is [tex]\(-0.71\)[/tex].

3. Find the Proportions Using Z-Scores:
- We use the z-scores to find the cumulative probabilities from the standard normal distribution.
- First proportion (for 96.9 degrees): The cumulative probability is approximately [tex]\(0.0076\)[/tex].
- Second proportion (for 98.1 degrees): The cumulative probability is approximately [tex]\(0.2375\)[/tex].

4. Calculate the Final Proportion:
- To find the proportion of scores between these two temperatures, subtract the first proportion from the second.
- So, the final total proportion is [tex]\(0.2375 - 0.0076 = 0.2299\)[/tex].

This means that approximately 22.99% of body temperature scores fall between 96.9 and 98.1 degrees Fahrenheit.