SOLUTION: 40% of the employees at a large corporation are female a sample of 50 employees is taken and gender is recorded what is the probability of sample proportion of females in sample be

Question 1192318: 40% of the employees at a large corporation are female a sample of 50 employees is taken and gender is recorded what is the probability of sample proportion of females in sample between 0.35 and 0.45?
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
**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.**

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