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 40 seconds. p(10 < v < 40) = ?% be sure to use the 68% - 95% - 99.7% rule and do not round.

Explanation:

Step1: Recall the 68 - 95 - 99.7 rule

The 68 - 95 - 99.7 rule states that for a normal distribution:

• Approximately 68% of the data lies within \(1\) standard deviation (\(\mu\pm\sigma\)) of the mean.
• Approximately 95% of the data lies within \(2\) standard deviations (\(\mu\pm2\sigma\)) of the mean.
• Approximately 99.7% of the data lies within \(3\) standard deviations (\(\mu\pm3\sigma\)) of the mean.

Given \(\mu = 40\) (mean) and \(\sigma=10\) (standard deviation).

Step2: Calculate the number of standard deviations from the mean

For \(v = 10\), \(z=\frac{10 - 40}{10}=\frac{- 30}{10}=-3\) (where \(z=\frac{x-\mu}{\sigma}\) is the z - score formula).
For \(v = 40\), \(z=\frac{40 - 40}{10}=0\)

The interval \(10\lt v\lt40\) is from \(z=-3\) to \(z = 0\).

Since the normal distribution is symmetric about the mean (\(z = 0\)), the percentage of data from \(z=-3\) to \(z = 0\) is half of the percentage of data from \(z=-3\) to \(z = 3\).

Since the percentage of data from \(z=-3\) to \(z = 3\) is \(99.7\%\), the percentage of data from \(z=-3\) to \(z = 0\) is \(\frac{99.7\%}{2}=49.85\%\)

Answer:

\(49.85\)