To find the probability that a given light bulb lasts between 750 and 900 hours, given that the lifetimes are normally distributed with a mean of 750 hours and a standard deviation of 75 hours, follow these steps:
1. Identify key parameters and values in the problem:
- Mean (μ) = 750 hours
- Standard deviation (σ) = 75 hours
- Lower bound of the interval (L) = 750 hours
- Upper bound of the interval (U) = 900 hours
2. Convert the raw scores to z-scores:
The z-score for a value in a normal distribution is found using the formula:
[tex]\[
z = \frac{x - \mu}{\sigma}
\][/tex]
where [tex]\(x\)[/tex] is the value, [tex]\(\mu\)[/tex] is the mean, and [tex]\(\sigma\)[/tex] is the standard deviation.
- For the lower bound (750 hours):
[tex]\[
z_{\text{lower}} = \frac{750 - 750}{75} = \frac{0}{75} = 0.0
\][/tex]
- For the upper bound (900 hours):
[tex]\[
z_{\text{upper}} = \frac{900 - 750}{75} = \frac{150}{75} = 2.0
\][/tex]
3. Use the z-scores to find the corresponding probabilities:
In a standard normal distribution, we can find the area (probability) to the left of a given z-score using standard normal distribution tables or a cumulative distribution function (CDF).
Since the z-scores are:
[tex]\[
z_{\text{lower}} = 0.0 \quad \text{and} \quad z_{\text{upper}} = 2.0
\][/tex]
- The cumulative probability for [tex]\( z = 0.0 \)[/tex]:
[tex]\[
P(Z \leq 0.0) = 0.5
\][/tex]
- The cumulative probability for [tex]\( z = 2.0 \)[/tex]:
[tex]\[
P(Z \leq 2.0) \approx 0.9772498680518208
\][/tex]
4. Find the probability that the bulb's lifetime falls between 750 and 900 hours:
This is the difference between the cumulative probabilities:
[tex]\[
P(750 \leq b \leq 900) = P(Z \leq 2.0) - P(Z \leq 0.0) = 0.9772498680518208 - 0.5 = 0.4772498680518208
\][/tex]
Thus, the probability that a given light bulb lasts between 750 and 900 hours is approximately [tex]\(0.4772498680518208\)[/tex].
