To find the probability that a voicemail duration is between 20 and 50 seconds, we can use the properties of the normal distribution, specifically the concept of z-scores and the cumulative distribution function (CDF).
1. Identify the given values:
- Mean ([tex]\(\mu\)[/tex]) = 40 seconds
- Standard deviation ([tex]\(\sigma\)[/tex]) = 10 seconds
- Lower bound = 20 seconds
- Upper bound = 50 seconds
2. Calculate the z-scores:
The z-score is a measure of how many standard deviations an element is from the mean. It is calculated using the formula:
[tex]\[
Z = \frac{X - \mu}{\sigma}
\][/tex]
where [tex]\(X\)[/tex] is the value we are converting to a z-score.
- For the lower bound (20 seconds):
[tex]\[
z_{lower} = \frac{20 - 40}{10} = \frac{-20}{10} = -2.0
\][/tex]
- For the upper bound (50 seconds):
[tex]\[
z_{upper} = \frac{50 - 40}{10} = \frac{10}{10} = 1.0
\][/tex]
3. Use the cumulative distribution function (CDF) of the standard normal distribution:
The CDF gives the probability that a standard normal variable is less than or equal to a given value. We need to find the CDF values for the calculated z-scores.
- CDF(-2.0) = The probability that a value is less than or equal to -2.0
- CDF(1.0) = The probability that a value is less than or equal to 1.0
4. Calculate the required probability:
The probability that the voicemail duration is between 20 and 50 seconds is the difference between the CDF values for the upper and lower z-scores:
[tex]\[
P(20 < v < 50) = \text{CDF}(1.0) - \text{CDF}(-2.0)
\][/tex]
From standard normal distribution tables or similar references, we obtain:
- CDF(1.0) ≈ 0.8413
- CDF(-2.0) ≈ 0.0228
Thus,
[tex]\[
P(20 < v < 50) = 0.8413 - 0.0228 = 0.8185
\][/tex]
To summarize, the probability that a given voicemail is between 20 and 50 seconds is approximately [tex]\(0.8185\)[/tex].
