Answer :
Certainly! Let's explore the given problem step-by-step.
The length of voicemails for a family is normally distributed with:
- Mean [tex]$(μ) = 40$[/tex] seconds
- Standard Deviation [tex]$(σ) = 10$[/tex] seconds
We need to find the probability that a given voicemail's length [tex]$(v)$[/tex] falls between 10 and 40 seconds: [tex]$P(10 < v < 40)$[/tex].
### Step 1: Determine the Z-scores
To standardize our values, we need to compute the Z-scores for the boundaries of this interval:
- For [tex]$v = 10$[/tex] seconds:
[tex]\[ z_{10} = \frac{10 - 40}{10} = -3 \][/tex]
- For [tex]$v = 40$[/tex] seconds:
[tex]\[ z_{40} = \frac{40 - 40}{10} = 0 \][/tex]
### Step 2: Apply the Empirical (68%-95%-99.7%) Rule
The Empirical Rule tells us about the distribution of data in a normal distribution:
- Approximately 68% of the data lies within 1 standard deviation of the mean [tex]$(μ ± σ)$[/tex].
- Approximately 95% of the data lies within 2 standard deviations of the mean [tex]$(μ ± 2σ)$[/tex].
- Approximately 99.7% of the data lies within 3 standard deviations of the mean [tex]$(μ ± 3σ)$[/tex].
Next, let's identify the ranges for these standard deviations:
- 1 standard deviation: [tex]$[30, 50]$[/tex]
- 2 standard deviations: [tex]$[20, 60]$[/tex]
- 3 standard deviations: [tex]$[10, 70]$[/tex]
### Step 3: Probability Calculations
Since [tex]$v = 40$[/tex] is exactly the mean, it is right at the median of the distribution. Thus, we understand that the area to the left of [tex]$v = 40$[/tex] is exactly 50% ([tex]$0.5$[/tex]) of the total area under the normal distribution curve.
Next, let's focus on [tex]$v = 10$[/tex] which is [tex]$3$[/tex] standard deviations below the mean. According to the Empirical Rule, 99.7% of the data falls within [tex]$3$[/tex] standard deviations of the mean [tex]$(10 < v < 70)$[/tex]. Therefore, the remaining portion outside this range is:
[tex]\[ (1 - 0.997) = 0.003 = 0.3\% \][/tex]
This remaining portion is divided equally between both tails (left and right of the distribution), hence:
[tex]\[ \text{Probability of } v < 10 \text{ seconds} = \frac{0.003}{2} = 0.0015 \][/tex]
### Step 4: Calculate Desired Probability
Finally, the probability that the voicemail falls between [tex]$10$[/tex] and [tex]$40$[/tex] seconds is the area under the curve from [tex]$v = 10$[/tex] to [tex]$v = 40$[/tex]:
[tex]\[ P(10 < v < 40) = \text{Total area up to } v = 40 \text{ (0.5)} - \text{Probability of } v < 10 \text{ (0.0015)} \][/tex]
[tex]\[ P(10 < v < 40) = 0.5 - 0.0015 = 0.4985 \][/tex]
Thus, the probability is 0.4985 or 49.85%.
### Summary and Additional Information
- Z-scores: [tex]$z_{10} = -3$[/tex], [tex]$z_{40} = 0$[/tex]
- Range within 1 standard deviation: [tex]$30$[/tex] to [tex]$50$[/tex] seconds
- Range within 2 standard deviations: [tex]$20$[/tex] to [tex]$60$[/tex] seconds
- Range within 3 standard deviations: [tex]$10$[/tex] to [tex]$70$[/tex] seconds
- Probability [tex]$(10 < v < 40)$[/tex]: [tex]$0.4985$[/tex]
The final answer is that the probability of a voicemail length being between 10 and 40 seconds is [tex]$0.4985$[/tex] or 49.85%.
The length of voicemails for a family is normally distributed with:
- Mean [tex]$(μ) = 40$[/tex] seconds
- Standard Deviation [tex]$(σ) = 10$[/tex] seconds
We need to find the probability that a given voicemail's length [tex]$(v)$[/tex] falls between 10 and 40 seconds: [tex]$P(10 < v < 40)$[/tex].
### Step 1: Determine the Z-scores
To standardize our values, we need to compute the Z-scores for the boundaries of this interval:
- For [tex]$v = 10$[/tex] seconds:
[tex]\[ z_{10} = \frac{10 - 40}{10} = -3 \][/tex]
- For [tex]$v = 40$[/tex] seconds:
[tex]\[ z_{40} = \frac{40 - 40}{10} = 0 \][/tex]
### Step 2: Apply the Empirical (68%-95%-99.7%) Rule
The Empirical Rule tells us about the distribution of data in a normal distribution:
- Approximately 68% of the data lies within 1 standard deviation of the mean [tex]$(μ ± σ)$[/tex].
- Approximately 95% of the data lies within 2 standard deviations of the mean [tex]$(μ ± 2σ)$[/tex].
- Approximately 99.7% of the data lies within 3 standard deviations of the mean [tex]$(μ ± 3σ)$[/tex].
Next, let's identify the ranges for these standard deviations:
- 1 standard deviation: [tex]$[30, 50]$[/tex]
- 2 standard deviations: [tex]$[20, 60]$[/tex]
- 3 standard deviations: [tex]$[10, 70]$[/tex]
### Step 3: Probability Calculations
Since [tex]$v = 40$[/tex] is exactly the mean, it is right at the median of the distribution. Thus, we understand that the area to the left of [tex]$v = 40$[/tex] is exactly 50% ([tex]$0.5$[/tex]) of the total area under the normal distribution curve.
Next, let's focus on [tex]$v = 10$[/tex] which is [tex]$3$[/tex] standard deviations below the mean. According to the Empirical Rule, 99.7% of the data falls within [tex]$3$[/tex] standard deviations of the mean [tex]$(10 < v < 70)$[/tex]. Therefore, the remaining portion outside this range is:
[tex]\[ (1 - 0.997) = 0.003 = 0.3\% \][/tex]
This remaining portion is divided equally between both tails (left and right of the distribution), hence:
[tex]\[ \text{Probability of } v < 10 \text{ seconds} = \frac{0.003}{2} = 0.0015 \][/tex]
### Step 4: Calculate Desired Probability
Finally, the probability that the voicemail falls between [tex]$10$[/tex] and [tex]$40$[/tex] seconds is the area under the curve from [tex]$v = 10$[/tex] to [tex]$v = 40$[/tex]:
[tex]\[ P(10 < v < 40) = \text{Total area up to } v = 40 \text{ (0.5)} - \text{Probability of } v < 10 \text{ (0.0015)} \][/tex]
[tex]\[ P(10 < v < 40) = 0.5 - 0.0015 = 0.4985 \][/tex]
Thus, the probability is 0.4985 or 49.85%.
### Summary and Additional Information
- Z-scores: [tex]$z_{10} = -3$[/tex], [tex]$z_{40} = 0$[/tex]
- Range within 1 standard deviation: [tex]$30$[/tex] to [tex]$50$[/tex] seconds
- Range within 2 standard deviations: [tex]$20$[/tex] to [tex]$60$[/tex] seconds
- Range within 3 standard deviations: [tex]$10$[/tex] to [tex]$70$[/tex] seconds
- Probability [tex]$(10 < v < 40)$[/tex]: [tex]$0.4985$[/tex]
The final answer is that the probability of a voicemail length being between 10 and 40 seconds is [tex]$0.4985$[/tex] or 49.85%.
