Question 1197679: 9. A population has a mean of 200 and a standard deviation of 50. A simple random sample of
size 100 will be taken and the sample mean 𝑥 will be used to estimate the population mean.
(a) What is the expected value of 𝑥?
(b) What is the standard deviation of 𝑥?
(c) Show the sampling distribution of 𝑥?
(d) What does the sampling distribution of 𝑥 show?
(e) What is the probability that the sample mean will be within ±5 of the population mean?
(f) What is the probability that the sample mean will be within ±10 of the population mean
Answer by onyulee(41)   (Show Source):
You can put this solution on YOUR website! **a) Expected Value of 𝑥**
* The expected value of the sample mean (𝑥) is equal to the population mean (μ).
* **E(𝑥) = μ = 200**
**b) Standard Deviation of 𝑥 (Standard Error)**
* The standard deviation of the sample mean is given by:
* σ_𝑥 = σ / √n
* where:
* σ is the population standard deviation (50)
* n is the sample size (100)
* σ_𝑥 = 50 / √100 = 50 / 10 = 5
**c) Sampling Distribution of 𝑥**
* **According to the Central Limit Theorem:**
* If the sample size (n) is sufficiently large (generally considered to be n ≥ 30), the sampling distribution of the sample mean (𝑥) will be approximately normally distributed, regardless of the shape of the population distribution.
* **In this case:**
* The sample size (n = 100) is large.
* Therefore, the sampling distribution of 𝑥 will be approximately normally distributed.
**d) What does the sampling distribution of 𝑥 show?**
* The sampling distribution of 𝑥 shows the probability distribution of all possible sample means that could be obtained from repeated random samples of size 100 from the population.
* It illustrates how the sample means are likely to vary around the population mean.
**e) Probability that the sample mean will be within ±5 of the population mean**
1. **Calculate the z-scores:**
* z1 = (μ - 5 - μ) / σ_𝑥 = -5 / 5 = -1
* z2 = (μ + 5 - μ) / σ_𝑥 = 5 / 5 = 1
2. **Find the probability using a standard normal distribution table (z-table):**
* P(-1 < z < 1) = P(z < 1) - P(z < -1)
* P(-1 < z < 1) = 0.8413 - 0.1587 = 0.6826
* **The probability that the sample mean will be within ±5 of the population mean is approximately 0.6826 or 68.26%.**
**f) Probability that the sample mean will be within ±10 of the population mean**
1. **Calculate the z-scores:**
* z1 = (μ - 10 - μ) / σ_𝑥 = -10 / 5 = -2
* z2 = (μ + 10 - μ) / σ_𝑥 = 10 / 5 = 2
2. **Find the probability using a standard normal distribution table (z-table):**
* P(-2 < z < 2) = P(z < 2) - P(z < -2)
* P(-2 < z < 2) = 0.9772 - 0.0228 = 0.9544
* **The probability that the sample mean will be within ±10 of the population mean is approximately 0.9544 or 95.44%.**
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