Questions: IXL Spotlight - Precalculus RR. 5 Find confidence intervals for population means JVK Learn with an example or Watch a video A random sample of size 50 is drawn from a large population. The population standard deviation is 2.4. The sample mean is 22.8. Find a 95% confidence interval for the population mean, μ. Round your answers to the nearest tenth. <μ<

Solution Steps

To find a 95% confidence interval for the population mean, we use the formula for the confidence interval when the population standard deviation is known. The formula is:

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

where:

• \(\bar{x}\) is the sample mean,
• \(Z\) is the Z-score corresponding to the desired confidence level (for 95%, \(Z \approx 1.96\)),
• \(\sigma\) is the population standard deviation,
• \(n\) is the sample size.

We are given the following values:

• Sample mean \(\bar{x} = 22.8\)
• Population standard deviation \(\sigma = 2.4\)
• Sample size \(n = 50\)
• Confidence level = 95%

For a 95% confidence level, the Z-score is approximately \(Z = 1.960\).

The margin of error is calculated using the formula: \[ \text{Margin of Error} = Z \left(\frac{\sigma}{\sqrt{n}}\right) \] Substituting the given values: \[ \text{Margin of Error} = 1.960 \left(\frac{2.4}{\sqrt{50}}\right) \approx 0.6652 \]

The confidence interval is given by: \[ \bar{x} \pm \text{Margin of Error} \] Calculating the bounds:

• Lower bound: \(22.8 - 0.6652 = 22.1348\)
• Upper bound: \(22.8 + 0.6652 = 23.4652\)

Rounding to the nearest tenth:

• Lower bound: \(22.1\)
• Upper bound: \(23.5\)

Final Answer