Question 1170244: In a certain stats class, the marks obtained by students on a class test followed a normal distribution with a mean of 68% and a standard deviation of 10%. What is the probability that the mean test mark from a sample of 25 students from the class was more than 72%?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem step by step.
**Understanding the Problem**
We are dealing with the sampling distribution of the mean. We're given:
* Population mean (μ) = 68%
* Population standard deviation (σ) = 10%
* Sample size (n) = 25
* We want to find the probability that the sample mean (x̄) > 72%
**Key Concepts**
* **Sampling Distribution of the Mean:** When we take multiple samples from a population, the means of those samples form a distribution.
* **Central Limit Theorem:** If the population is normally distributed or if the sample size is large enough (n ≥ 30), the sampling distribution of the mean is approximately normal.
* **Standard Error of the Mean (SEM):** The standard deviation of the sampling distribution of the mean is given by σ / √n.
**Calculations**
1. **Calculate the Standard Error of the Mean (SEM):**
SEM = σ / √n = 10% / √25 = 10% / 5 = 2%
2. **Calculate the Z-score:**
The z-score tells us how many standard errors the sample mean is away from the population mean.
z = (x̄ - μ) / SEM = (72% - 68%) / 2% = 4% / 2% = 2
3. **Find the Probability:**
We want to find P(x̄ > 72%), which is equivalent to finding P(Z > 2).
Using a standard normal distribution table or a calculator, we find that:
P(Z < 2) ≈ 0.9772
Therefore, P(Z > 2) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228
**Answer**
The probability that the mean test mark from a sample of 25 students was more than 72% is approximately 0.0228.
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