Answer :
Certainly! Let's solve the problem step by step.
1. Understanding the Problem:
- We are given that the life of light bulbs follows a normal distribution.
- The mean (μ) is 750 hours.
- The standard deviation (σ) is 75 hours.
- We need to find the probability that a light bulb lasts between 525 hours and 750 hours.
2. Find the Z-scores:
The Z-score transforms a normal distribution to a standard normal distribution, allowing us to find probabilities using standard normal distribution tables or rules.
The formula for the Z-score is:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
where \(X\) is the value of our variable.
- For the lower bound \(X = 525\):
[tex]\[ Z_{\text{lower}} = \frac{525 - 750}{75} = \frac{-225}{75} = -3 \][/tex]
- For the upper bound \(X = 750\):
[tex]\[ Z_{\text{upper}} = \frac{750 - 750}{75} = \frac{0}{75} = 0 \][/tex]
3. Using the 68%-95%-99.7% Rule:
The 68%-95%-99.7% rule helps us understand the distribution of data in a normal distribution:
- About 68% of the data falls within ±1 standard deviation from the mean.
- About 95% falls within ±2 standard deviations.
- About 99.7% falls within ±3 standard deviations.
We are interested in the range from -3 to 0 standard deviations:
- From the mean to 1 standard deviation (0 to 1) covers 34% of the data (half of 68%).
- Similarly, from -1 to 0 (which is symmetric to 0 to 1) also covers 34% of the data.
- From -2 to -1 and from -3 to -2 each add successively smaller portions since the tails of the normal distribution are thin.
From our specific Z-scores:
- The probability within one standard deviation from the mean (from 675 to 825 hours) is 68%, but we're interested in the portion from 525 to 750.
- The mean (750) to one standard deviation below the mean (675) is 34%.
- Below -3 standard deviations is a much smaller portion that is commonly not included in the empirical rule approximations and negligible for practical purposes.
4. Calculating the Probability:
We are specifically interested in the range from -3 to 0 Z-scores. This portion, as we determined, corresponds to exactly half of the 68% portion (i.e., the mean to one standard deviation).
From -3 to 0 standard deviations, the probability is:
[tex]\[ \text{Probability} = 34\% \][/tex]
Thus, the probability that a light bulb lasts between 525 hours and 750 hours is 0.34, or 34%.
1. Understanding the Problem:
- We are given that the life of light bulbs follows a normal distribution.
- The mean (μ) is 750 hours.
- The standard deviation (σ) is 75 hours.
- We need to find the probability that a light bulb lasts between 525 hours and 750 hours.
2. Find the Z-scores:
The Z-score transforms a normal distribution to a standard normal distribution, allowing us to find probabilities using standard normal distribution tables or rules.
The formula for the Z-score is:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
where \(X\) is the value of our variable.
- For the lower bound \(X = 525\):
[tex]\[ Z_{\text{lower}} = \frac{525 - 750}{75} = \frac{-225}{75} = -3 \][/tex]
- For the upper bound \(X = 750\):
[tex]\[ Z_{\text{upper}} = \frac{750 - 750}{75} = \frac{0}{75} = 0 \][/tex]
3. Using the 68%-95%-99.7% Rule:
The 68%-95%-99.7% rule helps us understand the distribution of data in a normal distribution:
- About 68% of the data falls within ±1 standard deviation from the mean.
- About 95% falls within ±2 standard deviations.
- About 99.7% falls within ±3 standard deviations.
We are interested in the range from -3 to 0 standard deviations:
- From the mean to 1 standard deviation (0 to 1) covers 34% of the data (half of 68%).
- Similarly, from -1 to 0 (which is symmetric to 0 to 1) also covers 34% of the data.
- From -2 to -1 and from -3 to -2 each add successively smaller portions since the tails of the normal distribution are thin.
From our specific Z-scores:
- The probability within one standard deviation from the mean (from 675 to 825 hours) is 68%, but we're interested in the portion from 525 to 750.
- The mean (750) to one standard deviation below the mean (675) is 34%.
- Below -3 standard deviations is a much smaller portion that is commonly not included in the empirical rule approximations and negligible for practical purposes.
4. Calculating the Probability:
We are specifically interested in the range from -3 to 0 Z-scores. This portion, as we determined, corresponds to exactly half of the 68% portion (i.e., the mean to one standard deviation).
From -3 to 0 standard deviations, the probability is:
[tex]\[ \text{Probability} = 34\% \][/tex]
Thus, the probability that a light bulb lasts between 525 hours and 750 hours is 0.34, or 34%.
