An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 40 hours. If a sample of 30 bulbs has an average life of 780 hours, find a 96% confidence interval for the population mean of all bulbs produced by this firm.

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4. Please help me answer this including the z-table.

**Problem 2.**

An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 40 hours. If a sample of 30 bulbs has an average life of 780 hours, find a 96% confidence interval for the population mean of all bulbs produced by this firm.

**Solution:**

To find a 96% confidence interval for the population mean, we use the formula for the confidence interval of the mean when the population standard deviation is known:

\[ \bar{X} \pm Z \left( \frac{\sigma}{\sqrt{n}} \right) \]

Where:
- \(\bar{X}\) is the sample mean
- \(Z\) is the Z-value from the standard normal distribution corresponding to the desired confidence level
- \(\sigma\) is the population standard deviation
- \(n\) is the sample size

Given:
- Sample mean (\(\bar{X}\)) = 780 hours
- Population standard deviation (\(\sigma\)) = 40 hours
- Sample size (\(n\)) = 30
- Confidence level = 96%

First, we need to find the Z-value for a 96% confidence level. The 96% confidence level corresponds to a Z-value of approximately 2.05 (You can find this value in standard Z-tables or using software/tools for more accuracy).

Now, we calculate the margin of error:

\[ \text{Margin of Error} = Z \left( \frac{\sigma}{\sqrt{n}} \right) = 2.05 \left( \frac{40}{\sqrt{30}} \right) \approx 14.97 \]

Next, we calculate the confidence interval:

\[ \text{Confidence Interval} = \bar{X} \pm \text{Margin of Error} \]
\[ \text{Confidence Interval} = 780 \pm 14.97 \]

Therefore, the 96% confidence interval for the population mean is:

\[ 780 - 14.97 \leq \mu \leq 780 + 14.97 \]
\[ 765.03 \leq \mu \leq 794.97 \]

So, we are 96% confident that the population mean life of all light bulbs produced by this firm is between 765.03 hours and 794.97 hours.
Transcribed Image Text:**Problem 2.** An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 40 hours. If a sample of 30 bulbs has an average life of 780 hours, find a 96% confidence interval for the population mean of all bulbs produced by this firm. **Solution:** To find a 96% confidence interval for the population mean, we use the formula for the confidence interval of the mean when the population standard deviation is known: \[ \bar{X} \pm Z \left( \frac{\sigma}{\sqrt{n}} \right) \] Where: - \(\bar{X}\) is the sample mean - \(Z\) is the Z-value from the standard normal distribution corresponding to the desired confidence level - \(\sigma\) is the population standard deviation - \(n\) is the sample size Given: - Sample mean (\(\bar{X}\)) = 780 hours - Population standard deviation (\(\sigma\)) = 40 hours - Sample size (\(n\)) = 30 - Confidence level = 96% First, we need to find the Z-value for a 96% confidence level. The 96% confidence level corresponds to a Z-value of approximately 2.05 (You can find this value in standard Z-tables or using software/tools for more accuracy). Now, we calculate the margin of error: \[ \text{Margin of Error} = Z \left( \frac{\sigma}{\sqrt{n}} \right) = 2.05 \left( \frac{40}{\sqrt{30}} \right) \approx 14.97 \] Next, we calculate the confidence interval: \[ \text{Confidence Interval} = \bar{X} \pm \text{Margin of Error} \] \[ \text{Confidence Interval} = 780 \pm 14.97 \] Therefore, the 96% confidence interval for the population mean is: \[ 780 - 14.97 \leq \mu \leq 780 + 14.97 \] \[ 765.03 \leq \mu \leq 794.97 \] So, we are 96% confident that the population mean life of all light bulbs produced by this firm is between 765.03 hours and 794.97 hours.
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