Math Problem Statement

The mean and standard deviation of the wages of 6,000 workers engaged in a factory are Rs. 1,200 and Rs. 400 respectively. Assuming the wages are normally distributed, estimate: a. Percentage of workers getting wages above Rs. 1,600. b. No. of workers getting wages between Rs. 600 and Rs. 900. c. No. of workers getting wages between Rs. 1,100 and Rs. 1,500.

Solution

The problem you're asking involves finding percentages and counts of workers within certain wage ranges, assuming a normal distribution. Here's how to approach each part:

Given:

  • Mean wage (μ) = Rs. 1200
  • Standard deviation (σ) = Rs. 400
  • Total number of workers (N) = 6000

a. Percentage of workers getting wages above Rs. 1600

To solve this, we use the z-score formula: z=xμσz = \frac{x - \mu}{\sigma} where:

  • xx = 1600 (the wage we're interested in)
  • μ\mu = 1200 (mean)
  • σ\sigma = 400 (standard deviation)

Substituting values: z=16001200400=400400=1z = \frac{1600 - 1200}{400} = \frac{400}{400} = 1

From standard normal distribution tables, the area to the left of z=1z = 1 is approximately 0.8413. Therefore, the area (or percentage) to the right (wages above Rs. 1600) is: 10.8413=0.1587 or 15.87%1 - 0.8413 = 0.1587 \text{ or } 15.87\%

b. Number of workers getting wages between Rs. 600 and Rs. 900

We calculate the z-scores for Rs. 600 and Rs. 900.

For Rs. 600: z=6001200400=600400=1.5z = \frac{600 - 1200}{400} = \frac{-600}{400} = -1.5 For Rs. 900: z=9001200400=300400=0.75z = \frac{900 - 1200}{400} = \frac{-300}{400} = -0.75

Using the z-table:

  • The area to the left of z=1.5z = -1.5 is approximately 0.0668.
  • The area to the left of z=0.75z = -0.75 is approximately 0.2266.

The percentage of workers earning between Rs. 600 and Rs. 900 is the difference between these two areas: 0.22660.0668=0.1598 or 15.98%0.2266 - 0.0668 = 0.1598 \text{ or } 15.98\% The number of workers is: 0.1598×6000=959 workers0.1598 \times 6000 = 959 \text{ workers}

c. Number of workers getting wages between Rs. 1100 and Rs. 1500

For Rs. 1100: z=11001200400=100400=0.25z = \frac{1100 - 1200}{400} = \frac{-100}{400} = -0.25 For Rs. 1500: z=15001200400=300400=0.75z = \frac{1500 - 1200}{400} = \frac{300}{400} = 0.75

Using the z-table:

  • The area to the left of z=0.25z = -0.25 is approximately 0.4013.
  • The area to the left of z=0.75z = 0.75 is approximately 0.7734.

The percentage of workers earning between Rs. 1100 and Rs. 1500 is: 0.77340.4013=0.3721 or 37.21%0.7734 - 0.4013 = 0.3721 \text{ or } 37.21\% The number of workers is: 0.3721×6000=2233 workers0.3721 \times 6000 = 2233 \text{ workers}

Summary:

  • a. Percentage of workers getting wages above Rs. 1600: 15.87%
  • b. Number of workers getting wages between Rs. 600 and Rs. 900: 959 workers
  • c. Number of workers getting wages between Rs. 1100 and Rs. 1500: 2233 workers

Would you like more details or have any questions?

Here are five related questions to explore further:

  1. How would the percentages change if the standard deviation was larger or smaller?
  2. How can you calculate wages for a different percentile, like the 90th percentile?
  3. What if wages followed a skewed distribution rather than a normal one?
  4. How can you estimate the probability for any arbitrary wage using the z-score?
  5. How do different confidence intervals affect the precision of wage estimates?

Tip: When using z-scores, always remember the direction you're interested in: above or below the given wage level!

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Math Problem Analysis

Mathematical Concepts

Normal Distribution
Z-Score Calculation
Probability

Formulas

Z = (X - μ) / σ
Area under the curve (Standard Normal Distribution Table)

Theorems

68-95-99.7 Rule (Empirical Rule)

Suitable Grade Level

Grades 10-12