Question 1199774
**a) Describe the sampling distribution of p**

* **Shape:** Since the sample size (n = 200) is large and the population size (N = 25,000) is much larger, the sampling distribution of the sample proportion (p̂) can be approximated by a normal distribution. This is due to the Central Limit Theorem.

* **Mean:** The mean of the sampling distribution of p̂ is equal to the population proportion (p): 
   * μ<sub>p̂</sub> = p = 0.65

* **Standard Deviation:** The standard deviation of the sampling distribution of p̂ is given by:
   * σ<sub>p̂</sub> = √[p * (1 - p) / n] 
   * σ<sub>p̂</sub> = √[0.65 * (1 - 0.65) / 200] 
   * σ<sub>p̂</sub> = √[0.2275 / 200] 
   * σ<sub>p̂</sub> ≈ 0.0337

**Therefore, the sampling distribution of p̂ is approximately normal with mean 0.65 and standard deviation 0.0337.**

**b) Probability of obtaining x = 136 or fewer individuals with the characteristic (P(p̂ >= 0.68))**

1. **Calculate the z-score for p̂ = 0.68:**
   * z = (p̂ - μ<sub>p̂</sub>) / σ<sub>p̂</sub> 
   * z = (0.68 - 0.65) / 0.0337 
   * z ≈ 0.89

2. **Find the probability:**
   * P(p̂ >= 0.68) = P(Z >= 0.89) 
   * Using a standard normal table or calculator, find the area to the right of z = 0.89. 
   * P(Z >= 0.89) ≈ 0.1867

**Therefore, the probability of obtaining 136 or fewer individuals with the characteristic is approximately 0.1867.**

**c) Probability of obtaining x = 118 or fewer individuals with the characteristic (P(p̂ <= 0.59))**

1. **Calculate the z-score for p̂ = 0.59:**
   * z = (0.59 - 0.65) / 0.0337 
   * z ≈ -1.78

2. **Find the probability:**
   * P(p̂ <= 0.59) = P(Z <= -1.78)
   * Using a standard normal table or calculator, find the area to the left of z = -1.78. 
   * P(Z <= -1.78) ≈ 0.0375

**Therefore, the probability of obtaining 118 or fewer individuals with the characteristic is approximately 0.0375.**