Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

Find the probability that a given voicemail is between 20 and 50 seconds, assuming a mean of 30 seconds and a standard deviation of 10 seconds.

Use the [tex]$68\%-95\%-99.7\%$[/tex] rule and do not round.


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%.
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.