# How do companies decide the warranty period for a product?

$$\text{Question:}$$ A firm produces machines with a lifespan, whose distribution has a mean of $$\mu$$ months and standard deviation of $$\sigma$$ months. The firm wishes to introduce a warranty scheme in which it would like to replace all the dysfunctional machines with new ones within warranty period. But they do not wish to do so for more than $$m\%$$ of the machines they produce. If the lifespan of the machine is assumed to follow a normal distribution (or maybe some other distribution), how long a guarantee period should be offered?

I don’t want to do this question just to get the answer. I want to do it to really understand how companies decide on their warranty period, I want to know it deeply. Also, can we somehow minimise this $$m\%$$ and maximise the warranty period for the benefit of the company?

• If you truly want to understand how to find warranty periods, then you need a more realistic question, because in real life nobody truly knows $\mu,$ $\sigma,$ or even the true distribution -- but estimates can be made. This textbook problem is just an elementary exercise in understanding the definition of a quantile; but the real problem is usually framed as a tolerance limit.
– whuber
Commented Jun 28 at 19:27
• In the real world, the firm would set the price jointly with the warranty length since consumers are willing to pay more for a product with a generous warranty. A solution would have to handle that and make some assumptions about market structure (monopoly, duopoly, monopolistic competition, potential for competitor entry) and whether the warranty might attract high-risk customers or change behavior (a generous warranty might reduce the incentive to maintain the machine). I'm not sure if this is the sort of thing you want to delve into. Commented Jun 28 at 23:48
• @whuber Could you tell me what $P(X \le m)$ or $P(X = m)$ represent here? (where $X$ is a random variable representing lifespan of a machine). Commented Jun 29 at 13:32
• @whuber And I tried to understand Tolerance Intervals but unable to get it. (you can see my response in the comment of answer provided by @jginestet). I will definitely study it after I complete all pre-requisite knowledge. Commented Jun 29 at 13:37

A TI gives you an interval which is likely to contain a certain percentage of the population (coverage), with a certain confidence, e.g. 95% confidence of at least 95% of the population. In your case (survival), one is only interested in a 1-sided interval (lower bound); we want to ensure that at least $$p%$$ of the product will last a certain time.
The form of a normal TI is $$\mu - k_1*\sigma$$; k is a so-called k-factor, available from tables, or various software packages. I am using $$k_1$$ to indicate it is for a one-sided TI. For example, for a sample size of 30, coverage of 95%, and confidence of 95%, $$k_1=2.2198$$. Note that, even for the normal distribution, there are various methods for computing the "k-factors", some being more or less approximative; so not all sources will give you exactly the same values for k; the differences become very small when N gets large (a few hundreds).
So, with your values of $$\mu$$ and $$\sigma$$, it becomes very simple to find the low bound for a commercially acceptable warranty period.