Question 1207924
**a) Find a 90% lower confidence bound for the population mean 𝜇.**

* **Understand the Concept:**
    * A lower confidence bound provides a minimum value for the population mean with a certain level of confidence. 
    * For a 90% lower bound, we're essentially saying that we are 90% confident that the population mean is greater than or equal to this value.

* **Calculate the Lower Bound:**
    * Since we're given the 90% confidence interval (27.658 to 29.742), the lower bound of this interval directly represents the 90% lower confidence bound for the population mean.

* **Answer:**
    * The 90% lower confidence bound for the population mean 𝜇 is **27.66**.

**b) Determine the sample size needed to estimate 𝜇 to within 0.6 with a 95% probability.**

* **Understand the Goal:**
    * We want to find the sample size (n) that will ensure the margin of error (E) is 0.6 with a 95% confidence level.

* **Formula:**
    * The margin of error (E) for a confidence interval for the mean is given by:
        * E = zα/2 * (σ / √n) 
        * where:
            * zα/2 is the critical value from the standard normal distribution for the desired confidence level (95% in this case)
            * σ is the population standard deviation (we'll use the sample standard deviation 's' as an estimate)
            * n is the sample size

* **Rearrange the formula to solve for n:**
    * n = (zα/2 * σ / E)²

* **Find the critical value (zα/2):**
    * For a 95% confidence level, zα/2 = 1.96 (from the standard normal distribution table)

* **Plug in the values:**
    * n = (1.96 * 3.8 / 0.6)² 
    * n = (12.32)² 
    * n ≈ 151.75

* **Round up to the nearest whole number:**
    * n = 152

* **Answer:**
    * You need **152 observations** to estimate 𝜇 to within 0.6 with a probability of 0.95.