To find the probability that a given voicemail length [tex]\( v \)[/tex] is between 20 and 50 seconds for a normally distributed variable with a mean ([tex]\(\mu\)[/tex]) of 40 seconds and a standard deviation ([tex]\(\sigma\)[/tex]) of 10 seconds, follow these steps:
1. Identify Mean and Standard Deviation:
[tex]\[
\mu = 40 \text{ seconds}, \quad \sigma = 10 \text{ seconds}
\][/tex]
2. Calculate Z-scores:
Convert the raw scores (20 seconds and 50 seconds) into their respective z-scores, which show how many standard deviations away from the mean these values are.
The z-score formula is:
[tex]\[
z = \frac{x - \mu}{\sigma}
\][/tex]
- For the lower bound, 20 seconds:
[tex]\[
z_{\text{lower}} = \frac{20 - 40}{10} = \frac{-20}{10} = -2
\][/tex]
- For the upper bound, 50 seconds:
[tex]\[
z_{\text{upper}} = \frac{50 - 40}{10} = \frac{10}{10} = 1
\][/tex]
3. Interpret using the Empirical Rule:
The empirical rule (68%-95%-99.7%) helps to determine the probability in a normal distribution:
- [tex]\(68\%\)[/tex] of the data falls within 1 standard deviation of the mean.
- [tex]\(95\%\)[/tex] of the data falls within 2 standard deviations of the mean.
- [tex]\(99.7\%\)[/tex] of the data falls within 3 standard deviations of the mean.
4. Determine Probability:
- The z-scores [tex]\( -2 \)[/tex] and [tex]\( 1 \)[/tex] indicate we are dealing with data spanning from [tex]\(-2\sigma\)[/tex] to [tex]\( +1\sigma\)[/tex].
- According to the empirical rule, approximately [tex]\(95\%\)[/tex] of the data lies within ±2 standard deviations from the mean.
- Specifically, between [tex]\(-2\sigma\)[/tex] and [tex]\( +1\sigma\)[/tex], you can reference the cumulative probabilities for each region.
Therefore, for voicemails between 20 and 50 seconds, the probability is approximately [tex]\( 95\% \)[/tex].
Thus, the calculated probability that a given voicemail is between 20 and 50 seconds is:
[tex]\[
\boxed{0.95}
\][/tex]
