Exercise
The human resources department of a company with 5,000 employees wants to estimate the average gross monthly salary of their employees. A sample of 400 employees was randomly selected, and it was estimated that the monthly salary follows a normal distribution with a mean of 1,500 euros and a standard deviation of 300 euros. Answer the following questions:
1. What is the probability that a randomly selected employee has a monthly salary of less than 1,000 euros?
2. What is the probability that a randomly selected employee has a monthly salary greater than 1,500 euros?
3. What percentage of employees have a monthly salary between 1,000 and 2,000 euros?
4. What salary should an employee have to be in the top 10% of earners in the company?
5. If 10 employees are randomly selected, what is the probability that 5 of them have a monthly salary greater than 1,500 euros?
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Solution
Question 1: Probability that a randomly selected employee has a monthly salary of less than 1,000 euros.
We are asked to find \( P(X < 1000) \), where \( X \) follows a normal distribution with \( \mu = 1500 \) and \( \sigma = 300 \).
First, calculate the z-score:
\[
z = \frac{1000 – 1500}{300} = \frac{-500}{300} \approx -1.67
\]
Using the standard normal distribution table, the probability corresponding to a z-score of -1.67 is approximately 0.0475.
So, the probability that a randomly selected employee has a salary less than 1,000 euros is 0.0475, or 4.75%.
Question 2: Probability that a randomly selected employee has a monthly salary greater than 1,500 euros.
We are asked to find \( P(X > 1500) \).
Since 1,500 is the mean of the distribution, the probability that \( X > 1500 \) is 0.5 (since the normal distribution is symmetric).
Thus, the probability that a randomly selected employee has a salary greater than 1,500 euros is 0.5, or 50%.
Question 3: Percentage of employees with a salary between 1,000 and 2,000 euros.
We are asked to find \( P(1000 < X < 2000) \).
First, calculate the z-scores for 1,000 and 2,000:
\[
z_1 = \frac{1000 – 1500}{300} = -1.67
\]
\[
z_2 = \frac{2000 – 1500}{300} = 1.67
\]
Using the standard normal distribution table:
– The probability for \( z_1 = -1.67 \) is 0.0475.
– The probability for \( z_2 = 1.67 \) is 0.9525.
Thus, the probability that an employee has a salary between 1,000 and 2,000 euros is:
\[
P(1000 < X < 2000) = 0.9525 – 0.0475 = 0.905
\]
So, 90.5%Â of employees have a salary between 1,000 and 2,000 euros.
Question 4: Salary needed to be in the top 10% of earners.
We need to find the salary corresponding to the top 10%, or the 90th percentile.
Using the standard normal distribution table, the z-score corresponding to the 90th percentile is approximately 1.28.
Now, calculate the corresponding salary:
\[
X = \mu + z \cdot \sigma = 1500 + 1.28 \times 300 = 1500 + 384 = 1884
\]
So, an employee must have a salary of at least 1,884 euros to be in the top 10% of earners.
Question 5: Probability that 5 out of 10 randomly selected employees have a salary greater than 1,500 euros.
This is a binomial probability problem where:
– \( n = 10 \) (number of trials),
– \( p = 0.5 \) (probability of success, as calculated in Question 2),
– \( k = 5 \) (number of successes).
The binomial probability formula is:
\[
P(X = k) = \binom{n}{k} p^k (1 – p)^{n – k}
\]
Substitute the values:
\[
P(X = 5) = \binom{10}{5} (0.5)^5 (0.5)^{5} = \binom{10}{5} (0.5)^{10}
\]
\[
P(X = 5) = \frac{10!}{5!5!} \times (0.5)^{10} = 252 \times 0.0009765625 = 0.246
\]
So, the probability that 5 out of 10 randomly selected employees have a salary greater than 1,500 euros is approximately 0.246, or 24.6%.
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Summary of Answers
1. The probability of a salary less than 1,000 euros is 4.75%.
2. The probability of a salary greater than 1,500 euros is 50%.
3. The percentage of employees with a salary between 1,000 and 2,000 euros is 90.5%.
4. To be in the top 10% of earners, an employee needs to earn at least 1,884 euros.
5. The probability that 5 out of 10 employees have a salary greater than 1,500 euros is 24.6%.
