Question

Suppose the length of voicemails (in seconds) 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.

Answer 1

0.1587, or about 15.87%

Explanation:

If you want to find the probability that a given voicemail is between 10 and 30 seconds, we can use the properties of the normal distribution. We'll need to standardize the values and then use a standard normal distribution table or calculator. Here are the steps:

Standardize the lower and upper bounds of the time interval using the z-score formula:

For 10 seconds:

Z = (X - μ) / σ

Z = (10 - 40) / 10

Z = -3

For 30 seconds:

Z = (X - μ) / σ

Z = (30 - 40) / 10

Z = -1

Now, we need to find the probabilities associated with these z-scores.

P(Z ≤ -3) represents the probability that a voicemail is less than or equal to 10 seconds.

P(Z ≤ -1) represents the probability that a voicemail is less than or equal to 30 seconds.

We can use a standard normal distribution table or calculator to find these probabilities:

P(Z ≤ -3) is the probability that a standard normal random variable is less than or equal to -3. This probability is very close to 0.

P(Z ≤ -1) is the probability that a standard normal random variable is less than or equal to -1. This probability is approximately 0.1587.

To find the probability that a given voicemail is between 10 and 30 seconds, we can subtract the probability that it's less than or equal to 10 seconds from the probability that it's less than or equal to 30 seconds:

P(10 ≤ X ≤ 30) = P(Z ≤ -1) - P(Z ≤ -3)

P(10 ≤ X ≤ 30) ≈ 0.1587 - 0 ≈ 0.1587

So, the probability that a given voicemail is between 10 and 30 seconds is approximately 0.1587, or about 15.87%.