for a family, the length of voicemails (v) is normally distributed with a mean of 40 seconds and standard…

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.\np(10 < v < 40) = ?%\nbe sure to use the 68% - 95% - 99.7% rule and do not round.

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.\np(10 < v < 40) = ?%\nbe sure to use the 68% - 95% - 99.7% rule and do not round.

Answer

Explanation:

Step1: Recall the empirical rule

The empirical rule (68 - 95 - 99.7 rule) for a normal distribution states that about 68% of the data lies within 1 - standard - deviation of the mean, 95% within 2 - standard - deviations, and 99.7% within 3 - standard - deviations. The mean $\mu = 40$ seconds and the standard deviation $\sigma=10$ seconds.

Step2: Calculate the z - scores

The z - score is calculated as $z=\frac{x-\mu}{\sigma}$. For $x = 10$, $z_1=\frac{10 - 40}{10}=\frac{- 30}{10}=-3$. For $x = 40$, $z_2=\frac{40 - 40}{10}=0$.

Step3: Apply the empirical rule

The interval from $z=-3$ to $z = 0$ represents half of the data within 3 - standard - deviations of the mean. Since the normal distribution is symmetric about the mean, and the data within 3 - standard - deviations of the mean is 99.7%, the probability of data between $z=-3$ and $z = 0$ is $\frac{99.7%}{2}=49.85%$.

Answer:

49.85%