Question

select the correct answer. julie manages a franchise restaurant and is analyzing the tips left at her location. from corporate informational material, she knows the overall population mean is $6.85 with a standard deviation of $1.25. julie has a sample of 180 tips for her franchise. by the central limit theorem, which interval can julie be 95% certain that the sample mean will fall within? a. $6.76 and $6.94 b. $6.83 and $6.87 c. $6.57 and $7.13 d. $6.66 and $7.04

Explanation:

Step1: Recall the formula for confidence interval

For a 95% confidence interval with large - sample (by central limit theorem, $n = 180$ is large), the formula is $\bar{x}\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$. The $z -$ value for a 95% confidence interval, $z_{\alpha/2}=1.96$. The population mean $\mu = 6.85$, population standard deviation $\sigma = 1.25$, and sample size $n = 180$.

Step2: Calculate the margin of error

The margin of error $E=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$. Substitute $z_{\alpha/2}=1.96$, $\sigma = 1.25$, and $n = 180$ into the formula. $\frac{\sigma}{\sqrt{n}}=\frac{1.25}{\sqrt{180}}\approx\frac{1.25}{13.42}\approx0.0931$. Then $E = 1.96\times0.0931\approx0.1825$.

Step3: Calculate the confidence interval

The lower - bound of the confidence interval is $\mu - E=6.85−0.1825 = 6.6675\approx6.66$ and the upper - bound is $\mu + E=6.85 + 0.1825=7.0325\approx7.04$.

Answer:

D. $6.66$ and $7.04$