Question 1197679
**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%.**