Question 1195144: 3. (a) The starting salary among 1999 graduates of the Evangelican Business School has a standard deviation of $17,000. If you randomly survey 40 students and average their starting salaries, what is the probability that the average among these students will greaterthan the average among all students?
(b) What is the probability that the average in the sample will exceed the average amo all students by more than $5,000?
(c) What is the probability that the average in the sample will differ from the avera among all students by more than $5,000?
Answer by ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website! **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!
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