Let's solve the problem of finding the probability that the battery life of a phone is between 18 and 22 hours, assuming we have a normally distributed battery life with a mean of 20 hours and a standard deviation of 2 hours.
1. Define the problem parameters:
- Mean ([tex]\(\mu\)[/tex]) = 20 hours
- Standard deviation ([tex]\(\sigma\)[/tex]) = 2 hours
- Lower bound = 18 hours
- Upper bound = 22 hours
2. Calculate the z-scores:
Z-scores help us convert our normal distribution into a standard normal distribution (mean = 0, standard deviation = 1).
- The z-score for the lower bound (18 hours) is calculated using the formula:
[tex]\[
z_{lower} = \frac{18 - 20}{2} = \frac{-2}{2} = -1.0
\][/tex]
- The z-score for the upper bound (22 hours) is calculated using the formula:
[tex]\[
z_{upper} = \frac{22 - 20}{2} = \frac{2}{2} = 1.0
\][/tex]
3. Find the cumulative probability for each z-score:
Using the cumulative distribution function (CDF) for the standard normal distribution, we find the probabilities corresponding to [tex]\(-1.0\)[/tex] and [tex]\(1.0\)[/tex].
- The cumulative probability for [tex]\(z_{lower} = -1.0\)[/tex] is approximately 0.1587.
- The cumulative probability for [tex]\(z_{upper} = 1.0\)[/tex] is approximately 0.8413.
4. Calculate the probability that the battery life is between 18 and 22 hours:
The desired probability is the difference between the two cumulative probabilities.
[tex]\[
\text{Probability} = \text{CDF}(1.0) - \text{CDF}(-1.0) = 0.8413 - 0.1587 = 0.6826
\][/tex]
Therefore, the probability that a phone's battery life will fall between 18 and 22 hours is approximately 0.6826 or 68.26%.
In summary:
- The z-score for the lower bound of 18 hours is [tex]\(-1.0\)[/tex].
- The z-score for the upper bound of 22 hours is [tex]\(1.0\)[/tex].
- The probability that the battery life is between 18 and 22 hours is about 68.27%.
