Question 1170719: The salaries of the employees in a company follow the normal distribution with mean $16000 and standard deviation $800.
(a) What is the probability that the salary of an employee is higher than $18000?
(b) There is 80% chance that the salary of an employee is less than $t. Find the value of t.
(c) If 20 employees are randomly chosen from the company and the mean salary of the employees is calculated. The manager of the company claims that over 20% of the mean salaries of the employee are higher than $16200. Do you agree?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem step by step.
**Given:**
* Mean (μ) = $16,000
* Standard Deviation (σ) = $800
* Normal distribution
**(a) Probability that salary is higher than $18,000**
1. **Calculate the z-score:**
* z = (x - μ) / σ
* z = (18000 - 16000) / 800
* z = 2000 / 800
* z = 2.5
2. **Find the probability:**
* We want to find P(X > 18000), which is the same as P(Z > 2.5).
* Using a z-table or calculator, we find P(Z < 2.5) ≈ 0.9938.
* P(Z > 2.5) = 1 - P(Z < 2.5) = 1 - 0.9938 = 0.0062
3. **Answer:**
* The probability that the salary is higher than $18,000 is approximately 0.0062 or 0.62%.
**(b) Find t such that P(X < t) = 0.80**
1. **Find the z-score:**
* We need to find the z-score that corresponds to a cumulative probability of 0.80.
* Using a z-table or calculator, we find z ≈ 0.84.
2. **Use the z-score formula:**
* z = (t - μ) / σ
* 0.84 = (t - 16000) / 800
3. **Solve for t:**
* t - 16000 = 0.84 * 800
* t - 16000 = 672
* t = 16000 + 672
* t = 16672
4. **Answer:**
* The value of t is $16,672.
**(c) Mean salary of 20 employees and claim of 20% > $16,200**
1. **Distribution of Sample Means:**
* The distribution of sample means (x̄) is also normal with:
* Mean (μ_x̄) = μ = $16,000
* Standard Deviation (σ_x̄) = σ / √n = 800 / √20 ≈ 178.89
2. **Calculate the z-score:**
* z = (x̄ - μ_x̄) / σ_x̄
* z = (16200 - 16000) / 178.89
* z = 200 / 178.89
* z ≈ 1.12
3. **Find the probability:**
* We want to find P(x̄ > 16200), which is the same as P(Z > 1.12).
* Using a z-table or calculator, we find P(Z < 1.12) ≈ 0.8686.
* P(Z > 1.12) = 1 - P(Z < 1.12) = 1 - 0.8686 = 0.1314.
4. **Convert to Percentage:**
* 0.1314 * 100% = 13.14%
5. **Compare to Manager's Claim:**
* The manager claims that over 20% of the mean salaries are higher than $16,200.
* We found that approximately 13.14% are higher than $16,200.
6. **Answer:**
* No, we do not agree with the manager's claim. The calculated percentage (13.14%) is less than 20%.
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