Question 1174088: Take CGPA of 05 students of your class; with n=2:
Req: (a) Develop the sampling distribution of N and n
(b) Prove that mean of the means of samples is equal to mean of population
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Alright, let's work through this problem using hypothetical CGPA data for 5 students.
**1. Hypothetical CGPA Data**
Let's assume the CGPA of 5 students are:
* Student 1: 3.2
* Student 2: 3.5
* Student 3: 3.8
* Student 4: 3.0
* Student 5: 3.7
**2. Population Calculations**
* Population (N): {3.2, 3.5, 3.8, 3.0, 3.7}
* Population Mean (μ): (3.2 + 3.5 + 3.8 + 3.0 + 3.7) / 5 = 17.2 / 5 = 3.44
**3. Sampling Distribution (n=2)**
* We need to find all possible samples of size 2 (n=2) from this population.
* The samples are:
* (3.2, 3.5)
* (3.2, 3.8)
* (3.2, 3.0)
* (3.2, 3.7)
* (3.5, 3.8)
* (3.5, 3.0)
* (3.5, 3.7)
* (3.8, 3.0)
* (3.8, 3.7)
* (3.0, 3.7)
**4. Calculate Sample Means**
* Now, calculate the mean of each sample:
* (3.2 + 3.5) / 2 = 3.35
* (3.2 + 3.8) / 2 = 3.5
* (3.2 + 3.0) / 2 = 3.1
* (3.2 + 3.7) / 2 = 3.45
* (3.5 + 3.8) / 2 = 3.65
* (3.5 + 3.0) / 2 = 3.25
* (3.5 + 3.7) / 2 = 3.6
* (3.8 + 3.0) / 2 = 3.4
* (3.8 + 3.7) / 2 = 3.75
* (3.0 + 3.7) / 2 = 3.35
**5. Sampling Distribution of the Means**
* The sampling distribution of the means is:
* {3.35, 3.5, 3.1, 3.45, 3.65, 3.25, 3.6, 3.4, 3.75, 3.35}
**(a) Develop the sampling distribution of N and n**
* N = {3.2, 3.5, 3.8, 3.0, 3.7}
* Sampling Distribution of n = {3.35, 3.5, 3.1, 3.45, 3.65, 3.25, 3.6, 3.4, 3.75, 3.35}
**(b) Prove that mean of the means of samples is equal to mean of population**
* Mean of the sample means (μ_x̄):
* (3.35 + 3.5 + 3.1 + 3.45 + 3.65 + 3.25 + 3.6 + 3.4 + 3.75 + 3.35) / 10 = 34.4 / 10 = 3.44
* Population Mean (μ): 3.44
* μ_x̄ = μ = 3.44
**Therefore, the mean of the sample means (3.44) is equal to the population mean (3.44).**
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