Question

for a family, the length of voicemails (v) is normally distributed with a mean of 40 seconds and standard deviation of 10 seconds. find the probability that a given voicemail is between 10 and 30 seconds. p(10 < v < 30) = ?% be sure to use the 68% - 95% - 99.7% rule and do not round.

Explanation:

Step1: Calculate the percentage of data within 1, 2, 3 standard deviations

According to the 68 - 95 - 99.7 rule:

• \(68\%\) of data lies within \(\mu\pm\sigma\) (i.e., \(40\pm10=(30,50)\))
• \(95\%\) of data lies within \(\mu\pm2\sigma\) (i.e., \(40\pm2\times10=(20,60)\))
• \(99.7\%\) of data lies within \(\mu\pm3\sigma\) (i.e., \(40\pm3\times10=(10,70)\))

Step2: Find the percentage of data between \(10\) and \(30\)

The total percentage of data between \(10\) and \(70\) is \(99.7\%\).
The percentage of data between \(20\) and \(60\) is \(95\%\).
The percentage of data between \(10\) and \(20\) is \(\frac{99.7\% - 95\%}{2}=2.35\%\)
The percentage of data between \(20\) and \(30\) is \(\frac{95\% - 68\%}{2}=13.5\%\)

Step3: Calculate \(P(10 < v < 30)\)

\(P(10 < v < 30)=\frac{99.7\% - 95\%}{2}+\frac{95\% - 68\%}{2}\)
\(=2.35\%+13.5\% = 15.85\%\)

Answer:

\(15.85\)