1. **Problem statement:** We have monthly rents of apartments that follow a normal distribution. A sample of 5 apartments is given with sample mean $b$. We want to find the confidence interval endpoints $a$ and $c$ around $b$ with 95% confidence. 2. **Given data:** - Sample size $n=5$ - Population standard deviation known: $\sigma=27$ - Confidence level: 95% - $Z_{0.95}=1.65$, $T_{0.95,4}=2.13$ 3. **Formulas and rules:** - When population standard deviation $\sigma$ is known, the 95% confidence interval for the mean is: $$ b \pm Z_{0.95} \times \frac{\sigma}{\sqrt{n}} $$ - When $\sigma$ is unknown, use the $t$-distribution: $$ b \pm T_{0.95, n-1} \times \frac{s}{\sqrt{n}} $$ where $s$ is the sample standard deviation. 4. **(a) Find $c$ when $\sigma$ is known:** - The margin of error is: $$ E = Z_{0.95} \times \frac{27}{\sqrt{5}} $$ - Calculate: $$ E = 1.65 \times \frac{27}{\sqrt{5}} = 1.65 \times \frac{27}{2.236} = 1.65 \times 12.07 = 19.92 $$ - So, $$ c = b + 19.92 $$ 5. **(b) Find $a$ when $\sigma$ is unknown:** - Use $t$-distribution with 4 degrees of freedom: $$ E = T_{0.95,4} \times \frac{s}{\sqrt{5}} = 2.13 \times \frac{s}{2.236} = 0.952 \times s $$ - Then, $$ a = b - 0.952 \times s $$ 6. **(c) Find $c$ for proportion of inexpensive apartments (under 1909 euro):** - This part requires proportion confidence interval, but data is incomplete here, so cannot compute $c$. **Final answers:** - (a) $$ c = b + 1.65 \times \frac{27}{\sqrt{5}} = b + 19.92 $$ - (b) $$ a = b - 2.13 \times \frac{s}{\sqrt{5}} = b - 0.952 \times s $$ Note: $b$ and $s$ are sample mean and sample standard deviation from the data file.