Math Problem Statement

The salaries of the employees of a corporation are normally distributed with a mean of €25,000 and a standard deviation of €5,000. Answer the following: A. What is the probability that a randomly selected employee will have a starting salary of at least €31,000? B. What percentage of employees has salaries of less than €12,200? C. What are the minimum and the maximum salaries of the middle 95% of the employees? D. What percentage will earn between €20,000 and €35,000?

Solution

The questions from the image are related to the properties of a normal distribution. Let’s solve them systematically:

The salaries of employees follow a normal distribution with:

  • Mean (μ) = €25,000
  • Standard deviation (σ) = €5,000

A. Probability of a salary of at least €31,000

To find the probability, we calculate the z-score: z=xμσz = \frac{x - \mu}{\sigma} Here, x=31,000x = 31,000, μ=25,000\mu = 25,000, and σ=5,000\sigma = 5,000: z=31,00025,0005,000=6,0005,000=1.2z = \frac{31,000 - 25,000}{5,000} = \frac{6,000}{5,000} = 1.2

Using the standard normal distribution table, the cumulative probability for z=1.2z = 1.2 is approximately 0.8849.

The probability of x31,000x \geq 31,000 is: P(x31,000)=1P(z1.2)=10.8849=0.1151P(x \geq 31,000) = 1 - P(z \leq 1.2) = 1 - 0.8849 = 0.1151

Answer: The probability is 11.51%.


B. Percentage of employees earning less than €12,200

Similarly, calculate the z-score for x=12,200x = 12,200: z=12,20025,0005,000=12,8005,000=2.56z = \frac{12,200 - 25,000}{5,000} = \frac{-12,800}{5,000} = -2.56

Using the z-table, the cumulative probability for z=2.56z = -2.56 is approximately 0.0052.

Answer: The percentage is 0.52%.


C. Minimum and maximum salaries for the middle 95%

For the middle 95% of a normal distribution, the z-scores are approximately ±1.96.

The corresponding values are calculated as: x=μ+zσx = \mu + z \cdot \sigma For the minimum: xmin=25,000+(1.96)5,000=25,0009,800=15,200x_{\text{min}} = 25,000 + (-1.96) \cdot 5,000 = 25,000 - 9,800 = 15,200 For the maximum: xmax=25,000+(1.96)5,000=25,000+9,800=34,800x_{\text{max}} = 25,000 + (1.96) \cdot 5,000 = 25,000 + 9,800 = 34,800

Answer: The salaries range from €15,200 to €34,800.


D. Percentage of employees earning between €20,000 and €35,000

Calculate the z-scores for both values:

  1. For x=20,000x = 20,000: z=20,00025,0005,000=5,0005,000=1z = \frac{20,000 - 25,000}{5,000} = \frac{-5,000}{5,000} = -1 The cumulative probability for z=1z = -1 is approximately 0.1587.

  2. For x=35,000x = 35,000: z=35,00025,0005,000=10,0005,000=2z = \frac{35,000 - 25,000}{5,000} = \frac{10,000}{5,000} = 2 The cumulative probability for z=2z = 2 is approximately 0.9772.

The probability of being between x=20,000x = 20,000 and x=35,000x = 35,000 is: P(20,000x35,000)=P(z2)P(z1)=0.97720.1587=0.8185P(20,000 \leq x \leq 35,000) = P(z \leq 2) - P(z \leq -1) = 0.9772 - 0.1587 = 0.8185

Answer: The percentage is 81.85%.


Any Questions or Clarifications?

Would you like detailed calculations or more context?


5 Related Questions:

  1. What is the probability of earning between €15,000 and €30,000?
  2. How does the standard deviation affect the probability calculations?
  3. What percentage of employees earn above €40,000?
  4. How would the range for the middle 99% of salaries differ from the 95%?
  5. Can a salary of €50,000 be considered an outlier in this distribution?

Tip:

The z-score is key in solving normal distribution problems. Memorizing common z-values (e.g., ±1.96 for 95%) can save time!

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Z-scores
Probability
Percentiles

Formulas

z = (x - μ) / σ
P(x ≤ x₀) = cumulative probability from z-tables
Middle k% range: μ ± z * σ

Theorems

Empirical Rule
Standard Normal Distribution

Suitable Grade Level

Undergraduate (Statistics/Probability Course)