Question 1195144
**a) Probability that the average among these students will be greater than the average among all students**

* **Assumptions:**
    * We assume that the starting salaries of all 1999 graduates are normally distributed.
    * We are comparing the sample mean to the population mean.

* **Central Limit Theorem:** 
    * The sampling distribution of the sample mean will be approximately normally distributed with:
        * Mean (μ_x̄) = μ (population mean)
        * Standard Deviation (σ_x̄) = σ / √n 
            * where:
                * σ = population standard deviation ($17,000)
                * n = sample size (40)
            * σ_x̄ = $17,000 / √40 ≈ $2690.87

* **Probability of Sample Mean Exceeding Population Mean:**
    * Since we're comparing the sample mean to the population mean, we are essentially asking for the probability that the sample mean is greater than the population mean. 
    * In a normal distribution, 50% of the data lies above the mean. 
    * Therefore, the probability that the average among the 40 students will be greater than the average among all students is **50%**.

**b) Probability that the average in the sample will exceed the average among all students by more than $5,000**

* **Calculate the z-score:**
    * z = (X - μ_x̄) / σ_x̄
        * where:
            * X = Difference in means ($5,000)
            * μ_x̄ = 0 (since we're comparing the sample mean to the population mean)
            * σ_x̄ = $2690.87

    * z = ($5,000 - $0) / $2690.87 
    * z ≈ 1.86

* **Find the Probability:**
    * Use a standard normal distribution table (z-table) to find the probability that z is greater than 1.86. 
    * P(z > 1.86) ≈ 0.0314

* **Therefore, the probability that the average in the sample will exceed the average among all students by more than $5,000 is approximately 3.14%.**

**c) Probability that the average in the sample will differ from the average among all students by more than $5,000**

* This includes both cases: 
    * Sample mean is greater than population mean by more than $5,000 (calculated in part b)
    * Sample mean is less than population mean by more than $5,000

* **Due to the symmetry of the normal distribution:**
    * Probability of differing by more than $5,000 = 2 * P(z > 1.86) 
    * Probability = 2 * 0.0314 = 0.0628

* **Therefore, the probability that the average in the sample will differ from the average among all students by more than $5,000 is approximately 6.28%.**

**Key Assumptions:**

* The starting salaries of all 1999 graduates are normally distributed.
* The sample of 40 students is a random sample from the population of all 1999 graduates.

I hope this comprehensive explanation is helpful!