Sure! Let's solve this problem step-by-step.
### Given:
- Mean life of the light bulb ([tex]\(\mu\)[/tex]) = 800 hours
- Standard deviation ([tex]\(\sigma\)[/tex]) = 40 hours
- Lower bound = 750 hours
- Upper bound = 900 hours
### Step-by-Step Solution:
1. Calculate the z-scores for the lower and upper bounds using the formula for the z-score:
[tex]\[
z = \frac{X - \mu}{\sigma}
\][/tex]
- For the lower bound (750 hours):
[tex]\[
z_{\text{lower}} = \frac{750 - 800}{40} = \frac{-50}{40} = -1.25
\][/tex]
- For the upper bound (900 hours):
[tex]\[
z_{\text{upper}} = \frac{900 - 800}{40} = \frac{100}{40} = 2.5
\][/tex]
2. Find the cumulative distribution function (CDF) values for these z-scores:
- [tex]\( P(Z \leq -1.25) \approx 0.10564977366685535 \)[/tex]
- [tex]\( P(Z \leq 2.5) \approx 0.9937903346742238 \)[/tex]
3. Calculate the probability that the light bulb lasts between 750 and 900 hours:
[tex]\[
P(750 \leq X \leq 900) = P(Z \leq 2.5) - P(Z \leq -1.25) \approx 0.9937903346742238 - 0.10564977366685535 = 0.8881405610073685
\][/tex]
### Conclusion:
The probability that a given light bulb lasts between 750 and 900 hours is approximately [tex]\(0.8881\)[/tex], or 88.81%.
This solution includes using the z-scores and the cumulative distribution function to find the probability accurately without rounding. This ensures the precision of our result.
