Answer:
81.5%
Step-by-step explanation:
The 68%-95%-99.7% rule (also known is the Empirical rule) states that nearly all of the data within a normal distribution will fall within three standard deviations of the mean:
- Approximately 68% of the data will fall within one standard deviation of the mean.
- Approximately 95% of the data will fall within two standard deviations of the mean.
- Approximately 99.7% of the data will fall within three standard deviations of the mean.
We are given that the length of voicemails (v) for a family is normally distributed with a mean of 40 seconds and standard deviation of 10 seconds.
According to the 68%-95%-99.7% rule, this means that:
- P(30 < v < 50) = 68%
- P(20 < v < 60) = 95%
- P(10 < v < 70) = 99.7%
To find the probability that a given voicemail is between 20 and 50 seconds, we can subtract 68% from 95%, divide this by 2, then add the result to 68%:
[tex]\sf P(20 < v < 50) =P(30 < v < 50)+ \dfrac{P(20 < v < 60) - P(30 < v < 50)}{2}[/tex]
[tex]\sf P(20 < v < 50) =68\%+ \dfrac{95\% - 68\%}{2}[/tex]
[tex]\sf P(20 < v < 50) =68\%+ \dfrac{27\%}{2}[/tex]
[tex]\sf P(20 < v < 50) =68\%+ 13.5\%[/tex]
[tex]\sf P(20 < v < 50) =81.5\%[/tex]
Therefore, the the probability that a given voicemail is between 20 and 50 seconds using the 68%-95%-99.7% is:
[tex]\Large\textsf{$\sf P(20 < v < 50)=\boxed{81.5}$\;\%}[/tex]
