UPSC MainsMANAGEMENT-PAPER-II201710 Marks
Q2.

Question 2

A new company in 1000 paraffin-bulbs, with an average life of 120 days, have been installed. Their life is normally distributed with a standard deviation of 20 days. How many bulbs will expire in less than 90 days? If all the bulbs are to be replaced at a time, what interval should be allowed between group replacements so that not more than 10% should expire before replacement?

How to Approach

This question tests the application of statistical concepts – specifically, the normal distribution – to a practical problem of inventory management and replacement policies. The approach should involve calculating the probability of bulb failure before 90 days, then using this probability to determine the optimal replacement interval to limit failures to 10%. The solution requires understanding Z-scores, standard normal distribution tables, and applying them to a real-world scenario. The answer should be structured into two parts, addressing each sub-question separately.

Model Answer

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Introduction

In operations management, understanding the lifespan of components and implementing effective replacement strategies is crucial for minimizing downtime and costs. The reliability of products is often modeled using probability distributions, with the normal distribution being a common choice due to its mathematical properties and applicability to many real-world phenomena. This question presents a scenario involving the lifespan of paraffin bulbs, requiring the application of normal distribution principles to determine the number of bulbs likely to fail prematurely and the optimal replacement interval to maintain a desired level of operational reliability.

Part 1: Number of Bulbs Expiring in Less Than 90 Days

We are given that the average life of a bulb is 120 days (μ = 120) with a standard deviation of 20 days (σ = 20). We want to find the number of bulbs that will expire in less than 90 days. This requires calculating the probability of a bulb failing before 90 days and then multiplying that probability by the total number of bulbs (1000).

Step 1: Calculate the Z-score

The Z-score represents the number of standard deviations a particular value is away from the mean. The formula for the Z-score is:

Z = (X - μ) / σ

Where:

  • X = The value we are interested in (90 days)
  • μ = The mean (120 days)
  • σ = The standard deviation (20 days)

Z = (90 - 120) / 20 = -1.5

Step 2: Find the Probability from the Z-table

Using a standard normal distribution table (or a statistical calculator), we find the probability associated with a Z-score of -1.5. This probability represents the proportion of bulbs that will expire in less than 90 days. P(Z < -1.5) ≈ 0.0668

Step 3: Calculate the Number of Bulbs

Number of bulbs expiring in less than 90 days = Total number of bulbs * Probability

Number of bulbs = 1000 * 0.0668 = 66.8

Since we can't have a fraction of a bulb, we round to the nearest whole number. Approximately 67 bulbs will expire in less than 90 days.

Part 2: Replacement Interval for 10% Failure Rate

We want to find the interval between replacements such that no more than 10% of the bulbs expire before replacement. This means we want to find a time 't' such that P(life < t) = 0.10.

Step 1: Find the Z-score for 10% Probability

We need to find the Z-score corresponding to a cumulative probability of 0.10. Using a standard normal distribution table (or a statistical calculator), we find that the Z-score is approximately -1.28.

Step 2: Calculate the Time 't'

We can use the Z-score formula to solve for 't':

Z = (t - μ) / σ

-1.28 = (t - 120) / 20

t - 120 = -1.28 * 20

t - 120 = -25.6

t = 120 - 25.6 = 94.4 days

Therefore, the replacement interval should be approximately 94.4 days to ensure that not more than 10% of the bulbs expire before replacement.

Conclusion

In conclusion, approximately 67 bulbs are expected to expire within the first 90 days of operation. To maintain a failure rate of no more than 10%, the bulbs should be replaced every 94.4 days. These calculations demonstrate the practical application of normal distribution principles in inventory management and reliability engineering. Regular monitoring of bulb lifespan and adjustments to the replacement interval may be necessary to account for variations in manufacturing quality or operating conditions.

Answer Length

This is a comprehensive model answer for learning purposes and may exceed the word limit. In the exam, always adhere to the prescribed word count.

Additional Resources

Key Definitions

Normal Distribution
A probability distribution that is symmetrical about the mean, representing many natural phenomena. It is characterized by its bell-shaped curve and defined by its mean (μ) and standard deviation (σ).
Z-score
A statistical measurement that describes a value's relationship to the mean of a group of values. It indicates how many standard deviations an element is from the mean.

Key Statistics

According to a report by the U.S. Department of Energy, the average lifespan of an incandescent light bulb is approximately 750-1000 hours (as of 2023).

Source: U.S. Department of Energy

The global lighting market was valued at USD 108.4 billion in 2022 and is expected to grow at a CAGR of 6.5% from 2023 to 2030 (Source: Grand View Research, 2023).

Source: Grand View Research

Examples

Quality Control in Manufacturing

Manufacturers often use normal distribution to assess the quality of their products. If a product's characteristic (e.g., weight, length) deviates significantly from the mean, it may be considered defective and rejected.

Frequently Asked Questions

What if the bulb lifespan doesn't perfectly follow a normal distribution?

While the normal distribution is a useful approximation, real-world data may deviate. Other distributions (e.g., exponential, Weibull) might be more appropriate depending on the specific failure pattern. Statistical tests can be used to assess the goodness of fit.