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 sizeGiven:- 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.

Name: 4. Please help me answer this including the z-table. **Problem 2.**An
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Description: 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

Given information: n = 30 Mean = 780 S.D = 40 PZ<z=1-0.042=1-0.02=0.98 Use the z- cumulative…

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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.

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.

MATLAB: An Introduction with Applications

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ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

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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.

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