Question 1192318
**1. Define Variables**

* **Population Proportion (p):** 0.40 (40% of employees are female)
* **Sample Size (n):** 50
* **Sample Proportion (p-hat):** Proportion of females in the sample

**2. Check Conditions for Normal Approximation**

* **Independence:** We assume that the employees are randomly selected and that the sample size is less than 10% of the total population.
* **Success-Failure Condition:**
    * np = 0.40 * 50 = 20 (Number of female employees in the sample)
    * n(1-p) = 50 * 0.60 = 30 (Number of male employees in the sample)
    * Both np and n(1-p) are greater than 10, so the condition is met.

Since the conditions are met, we can use the normal approximation to the binomial distribution.

**3. Calculate Standard Error**

* Standard Error (SE) = √[p * (1-p) / n] 
* SE = √[0.40 * 0.60 / 50] 
* SE = √[0.24 / 50] 
* SE = √0.0048 
* SE ≈ 0.0693

**4. Standardize the Values**

* **For p-hat = 0.35:**
    * z = (0.35 - 0.40) / 0.0693 ≈ -0.72
* **For p-hat = 0.45:**
    * z = (0.45 - 0.40) / 0.0693 ≈ 0.72

**5. Find the Probability**

* We want to find P(0.35 < p-hat < 0.45) 
* This is equivalent to finding P(-0.72 < z < 0.72) 
* Using a standard normal table or calculator, we find:
    * P(z < 0.72) ≈ 0.7642
    * P(z < -0.72) ≈ 0.2358
* P(-0.72 < z < 0.72) = P(z < 0.72) - P(z < -0.72) 
* P(-0.72 < z < 0.72) ≈ 0.7642 - 0.2358 
* P(-0.72 < z < 0.72) ≈ 0.5284

**Therefore, the probability that the sample proportion of females in the sample is between 0.35 and 0.45 is approximately 0.5284.**