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Sagot :
Certainly! Let's break down the problem step by step using the empirical rule (68-95-99.7%).
Given:
- Mean lifespan ([tex]\(\mu\)[/tex]): 16 years
- Standard deviation ([tex]\(\sigma\)[/tex]): 1.7 years
- Lower bound ([tex]\(L\)[/tex]): 14.3 years
- Upper bound ([tex]\(U\)[/tex]): 19.4 years
1. Calculate Z-scores for the bounds:
The Z-score measures how many standard deviations an element is from the mean.
For the lower bound:
[tex]\[ Z_{\text{lower}} = \frac{L - \mu}{\sigma} = \frac{14.3 - 16}{1.7} \approx -1.00 \][/tex]
For the upper bound:
[tex]\[ Z_{\text{upper}} = \frac{U - \mu}{\sigma} = \frac{19.4 - 16}{1.7} \approx 2.00 \][/tex]
2. Interpreting the Empirical Rule:
- Approximately 68% of the data falls within 1 standard deviation of the mean: [tex]\((\mu - \sigma, \mu + \sigma)\)[/tex].
- Approximately 95% of the data falls within 2 standard deviations of the mean: [tex]\((\mu - 2\sigma, \mu + 2\sigma)\)[/tex].
- Approximately 99.7% of the data falls within 3 standard deviations of the mean: [tex]\((\mu - 3\sigma, \mu + 3\sigma)\)[/tex].
3. Relate Z-scores to the empirical rule:
- A Z-score of [tex]\(-1\)[/tex] to [tex]\(1\)[/tex] corresponds to approximately 68% of the data.
- A Z-score of [tex]\(-2\)[/tex] to [tex]\(2\)[/tex] corresponds to approximately 95% of the data.
4. Estimate the probability:
The range we are examining is from [tex]\(14.3\)[/tex] to [tex]\(19.4\)[/tex], which corresponds to Z-scores of [tex]\(-1.00\)[/tex] to [tex]\(2.00\)[/tex].
To find the probability between these two Z-scores, we can partition the empirical rule ranges:
- Since 68% of the data falls within 1 standard deviation, we know:
[tex]\[ P(-1 \leq Z \leq 1) = 68\% \][/tex]
- Since 95% of the data falls within 2 standard deviations, we know:
[tex]\[ P(-2 \leq Z \leq 2) = 95\% \][/tex]
We are interested in the area between Z-scores of [tex]\(-1.00\)[/tex] and [tex]\(2.00\)[/tex]. To find this probability, we consider the distribution within 1 to 2 standard deviations:
- The probability within exactly 2 standard deviations is 95%.
- The probability within exactly 1 standard deviation is 68%.
Thus, the probability between 1 and 2 standard deviations on one side of the mean is:
[tex]\[ \frac{95\% - 68\%}{2} = 13.5\% \][/tex]
Since the interval from [tex]\(-1.00\)[/tex] to [tex]\(2.00\)[/tex] covers from 1 standard deviation below the mean to 2 standard deviations above it, we combine:
[tex]\[ P(-1 \leq Z \leq 2) = 68\% + 13.5\% = 81.5\% \][/tex]
Therefore, based on the empirical rule, the estimated probability that a gorilla lives between 14.3 and 19.4 years is 81.5%.
Given:
- Mean lifespan ([tex]\(\mu\)[/tex]): 16 years
- Standard deviation ([tex]\(\sigma\)[/tex]): 1.7 years
- Lower bound ([tex]\(L\)[/tex]): 14.3 years
- Upper bound ([tex]\(U\)[/tex]): 19.4 years
1. Calculate Z-scores for the bounds:
The Z-score measures how many standard deviations an element is from the mean.
For the lower bound:
[tex]\[ Z_{\text{lower}} = \frac{L - \mu}{\sigma} = \frac{14.3 - 16}{1.7} \approx -1.00 \][/tex]
For the upper bound:
[tex]\[ Z_{\text{upper}} = \frac{U - \mu}{\sigma} = \frac{19.4 - 16}{1.7} \approx 2.00 \][/tex]
2. Interpreting the Empirical Rule:
- Approximately 68% of the data falls within 1 standard deviation of the mean: [tex]\((\mu - \sigma, \mu + \sigma)\)[/tex].
- Approximately 95% of the data falls within 2 standard deviations of the mean: [tex]\((\mu - 2\sigma, \mu + 2\sigma)\)[/tex].
- Approximately 99.7% of the data falls within 3 standard deviations of the mean: [tex]\((\mu - 3\sigma, \mu + 3\sigma)\)[/tex].
3. Relate Z-scores to the empirical rule:
- A Z-score of [tex]\(-1\)[/tex] to [tex]\(1\)[/tex] corresponds to approximately 68% of the data.
- A Z-score of [tex]\(-2\)[/tex] to [tex]\(2\)[/tex] corresponds to approximately 95% of the data.
4. Estimate the probability:
The range we are examining is from [tex]\(14.3\)[/tex] to [tex]\(19.4\)[/tex], which corresponds to Z-scores of [tex]\(-1.00\)[/tex] to [tex]\(2.00\)[/tex].
To find the probability between these two Z-scores, we can partition the empirical rule ranges:
- Since 68% of the data falls within 1 standard deviation, we know:
[tex]\[ P(-1 \leq Z \leq 1) = 68\% \][/tex]
- Since 95% of the data falls within 2 standard deviations, we know:
[tex]\[ P(-2 \leq Z \leq 2) = 95\% \][/tex]
We are interested in the area between Z-scores of [tex]\(-1.00\)[/tex] and [tex]\(2.00\)[/tex]. To find this probability, we consider the distribution within 1 to 2 standard deviations:
- The probability within exactly 2 standard deviations is 95%.
- The probability within exactly 1 standard deviation is 68%.
Thus, the probability between 1 and 2 standard deviations on one side of the mean is:
[tex]\[ \frac{95\% - 68\%}{2} = 13.5\% \][/tex]
Since the interval from [tex]\(-1.00\)[/tex] to [tex]\(2.00\)[/tex] covers from 1 standard deviation below the mean to 2 standard deviations above it, we combine:
[tex]\[ P(-1 \leq Z \leq 2) = 68\% + 13.5\% = 81.5\% \][/tex]
Therefore, based on the empirical rule, the estimated probability that a gorilla lives between 14.3 and 19.4 years is 81.5%.
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