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 20 and 40 seconds. p(20 < v < 40) = ?% be 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 20 and 40 seconds. p(20 < v < 40) = ?% be sure to use the 68% - 95% - 99.7% rule and do not round.

Answer

Explanation:

Step1: Identify mean and standard deviation.

The mean is $\mu = 40$ seconds and the standard deviation is $\sigma = 10$ seconds.

Step2: Determine the interval in standard deviations.

The interval is from 20 to 40 seconds. The lower bound $20 = 40 - 2 \times 10 = \mu - 2\sigma$. The upper bound $40 = \mu$. We need to find the probability $P(\mu - 2\sigma < v < \mu)$.

Step3: Apply the Empirical Rule (68-95-99.7).

The Empirical Rule states that approximately 95% of the data falls within 2 standard deviations of the mean, i.e., $P(\mu - 2\sigma < v < \mu + 2\sigma) \approx 0.95$.

Step4: Calculate the probability for the given interval.

Due to the symmetry of the normal distribution, the probability between the mean and 2 standard deviations below the mean is half of the probability within 2 standard deviations of the mean. $$P(\mu - 2\sigma < v < \mu) = \frac{1}{2} P(\mu - 2\sigma < v < \mu + 2\sigma)$$ $$P(20 < v < 40) = \frac{1}{2} \times 95% = 47.5%$$

Answer:

$47.5$