To solve this problem, we need to find the probability that a normally distributed variable falls between two given values. In this case, we need to find the probability that a light bulb lasts between 675 and 825 hours, given that the lifespan is normally distributed with a mean of 750 hours and a standard deviation of 75 hours. Here's a step-by-step solution:
### Step 1: Understand the Properties
We are given:
- Mean ([tex]\(\mu\)[/tex]) of the distribution: 750 hours
- Standard deviation ([tex]\(\sigma\)[/tex]) of the distribution: 75 hours
### Step 2: Standardize the Problem
We need to convert the given values (675 and 825 hours) to z-scores. A z-score tells us how many standard deviations away a value is from the mean.
The formula for calculating the z-score of a value [tex]\(X\)[/tex] is:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
### Step 3: Calculate the Z-Scores
1. For the lower bound (675 hours):
[tex]\[ z_{\text{lower}} = \frac{675 - 750}{75} = \frac{-75}{75} = -1 \][/tex]
2. For the upper bound (825 hours):
[tex]\[ z_{\text{upper}} = \frac{825 - 750}{75} = \frac{75}{75} = 1 \][/tex]
### Step 4: Use the 68%-95%-99.7% Rule
The 68%-95%-99.7% rule (also known as the empirical rule) helps us understand the probability distribution within certain standard deviations from the mean:
- About 68% of the data falls within 1 standard deviation from the mean ([tex]\(-1 \leq z \leq 1\)[/tex]).
Given the z-scores of -1 and 1, we can conclude that the probability of a light bulb lasting between 675 and 825 hours is the probability between these two z-scores.
### Step 5: Interpret the Result
From the 68%-95%-99.7% rule, we know that 68% of the data in a normal distribution falls within one standard deviation of the mean. So, the probability that a light bulb lasts between 675 and 825 hours is approximately 68%, or:
[tex]\[ \text{Probability}(\text{675 hours} \leq \text{Lifespan} \leq \text{825 hours}) \approx 0.6826894921370859 \][/tex]
Therefore, the probability that a given light bulb lasts between 675 and 825 hours is approximately 0.6827 (or 68.27%).
