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Sagot :
Certainly! Let's solve this step-by-step:
### Problem Statement
We have the duration of voicemails which are normally distributed with the following parameters:
- Mean duration ([tex]\(\mu\)[/tex]) = 30 seconds
- Standard deviation ([tex]\(\sigma\)[/tex]) = 10 seconds
We need to find the probability that a given voicemail duration is between 20 and 50 seconds.
### 1. Understanding the Normal Distribution
The normal distribution is characterized by the mean (average) and standard deviation, and the [tex]\(68\%-95\%-99.7\%\)[/tex] rule helps us understand the spread of data in a normal distribution:
- About 68% of the data falls within 1 standard deviation from the mean.
- About 95% of the data falls within 2 standard deviations from the mean.
- About 99.7% of the data falls within 3 standard deviations from the mean.
### 2. Finding [tex]\(z\)[/tex]-Scores
To find the probability, we first need to convert our boundaries (20 seconds and 50 seconds) into [tex]\(z\)[/tex]-scores. The [tex]\(z\)[/tex]-score tells us how many standard deviations a particular value is from the mean.
The [tex]\(z\)[/tex]-score is calculated as:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
#### - For the lower bound (20 seconds):
[tex]\[ z_{\text{lower}} = \frac{20 - 30}{10} = \frac{-10}{10} = -1.0 \][/tex]
#### - For the upper bound (50 seconds):
[tex]\[ z_{\text{upper}} = \frac{50 - 30}{10} = \frac{20}{10} = 2.0 \][/tex]
### 3. Applying the [tex]\(68\%-95\%-99.7\%\)[/tex] Rule
Now that we have the [tex]\(z\)[/tex]-scores:
- The lower bound [tex]\(z\)[/tex]-score is [tex]\(-1.0\)[/tex]
- The upper bound [tex]\(z\)[/tex]-score is [tex]\(2.0\)[/tex]
Using the [tex]\(68\%-95\%-99.7\%\)[/tex] rule:
- A [tex]\(z\)[/tex]-score between [tex]\(-1.0\)[/tex] and [tex]\(1.0\)[/tex] covers 68% of the data approximately.
- A [tex]\(z\)[/tex]-score between [tex]\(-2.0\)[/tex] and [tex]\(2.0\)[/tex] covers 95% of the data approximately.
- A [tex]\(z\)[/tex]-score between [tex]\(-3.0\)[/tex] and [tex]\(3.0\)[/tex] covers 99.7% of the data approximately.
Since our range of [tex]\(z\)[/tex]-scores is [tex]\(-1.0\)[/tex] to [tex]\(2.0\)[/tex], we fall into the 95% range. This means the probability that a given voicemail is between 20 and 50 seconds is approximately 95%.
### Conclusion
Therefore, the probability that a given voicemail has a duration between 20 and 50 seconds is 95%.
### Problem Statement
We have the duration of voicemails which are normally distributed with the following parameters:
- Mean duration ([tex]\(\mu\)[/tex]) = 30 seconds
- Standard deviation ([tex]\(\sigma\)[/tex]) = 10 seconds
We need to find the probability that a given voicemail duration is between 20 and 50 seconds.
### 1. Understanding the Normal Distribution
The normal distribution is characterized by the mean (average) and standard deviation, and the [tex]\(68\%-95\%-99.7\%\)[/tex] rule helps us understand the spread of data in a normal distribution:
- About 68% of the data falls within 1 standard deviation from the mean.
- About 95% of the data falls within 2 standard deviations from the mean.
- About 99.7% of the data falls within 3 standard deviations from the mean.
### 2. Finding [tex]\(z\)[/tex]-Scores
To find the probability, we first need to convert our boundaries (20 seconds and 50 seconds) into [tex]\(z\)[/tex]-scores. The [tex]\(z\)[/tex]-score tells us how many standard deviations a particular value is from the mean.
The [tex]\(z\)[/tex]-score is calculated as:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
#### - For the lower bound (20 seconds):
[tex]\[ z_{\text{lower}} = \frac{20 - 30}{10} = \frac{-10}{10} = -1.0 \][/tex]
#### - For the upper bound (50 seconds):
[tex]\[ z_{\text{upper}} = \frac{50 - 30}{10} = \frac{20}{10} = 2.0 \][/tex]
### 3. Applying the [tex]\(68\%-95\%-99.7\%\)[/tex] Rule
Now that we have the [tex]\(z\)[/tex]-scores:
- The lower bound [tex]\(z\)[/tex]-score is [tex]\(-1.0\)[/tex]
- The upper bound [tex]\(z\)[/tex]-score is [tex]\(2.0\)[/tex]
Using the [tex]\(68\%-95\%-99.7\%\)[/tex] rule:
- A [tex]\(z\)[/tex]-score between [tex]\(-1.0\)[/tex] and [tex]\(1.0\)[/tex] covers 68% of the data approximately.
- A [tex]\(z\)[/tex]-score between [tex]\(-2.0\)[/tex] and [tex]\(2.0\)[/tex] covers 95% of the data approximately.
- A [tex]\(z\)[/tex]-score between [tex]\(-3.0\)[/tex] and [tex]\(3.0\)[/tex] covers 99.7% of the data approximately.
Since our range of [tex]\(z\)[/tex]-scores is [tex]\(-1.0\)[/tex] to [tex]\(2.0\)[/tex], we fall into the 95% range. This means the probability that a given voicemail is between 20 and 50 seconds is approximately 95%.
### Conclusion
Therefore, the probability that a given voicemail has a duration between 20 and 50 seconds is 95%.
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