Question
You manufacture car batteries and need to determine how long you will provide a warranty under which you will replace a battery for free if it fails before the warranty runs out. You know from testing that a battery’s life span is normally distributed and has a mean of 4.3 years with a standard deviation of 0.92. You want to ensure that you replace a maximum of 33% of batteries for free. For how long will you make the limited warranty on the batteries last? Why? What role does customer satisfaction play in making this decision? write a short response to this post
Answers
GPT-4o mini
To determine the warranty period for the car batteries, we need to find the time frame in which only 33% of the batteries are expected to fail. Since the lifespan follows a normal distribution with a mean of 4.3 years and a standard deviation of 0.92 years, we can use the z-score formula to find the corresponding time period.
1. **Find the z-score for 33%**: The z-score that corresponds to the top 33% (or the 67th percentile since we want to cover only 33% of failures) can be found using a z-table. The z-score for the 67th percentile is approximately 0.44.
2. **Calculate the warranty length**: We can use the following formula to determine the warranty length:
\[
X = \mu + Z \sigma
\]
where \( \mu \) is the mean, \( Z \) is the z-score, and \( \sigma \) is the standard deviation.
Plugging in the values:
\[
X = 4.3 + (0.44 \times 0.92) \approx 4.3 + 0.4048 \approx 4.70 \text{ years}
\]
Thus, the limited warranty should last approximately **4.7 years**.
**Customer Satisfaction**: Offering a warranty that ensures only 33% of batteries are replaced for free balances cost control with customer satisfaction. A longer warranty might enhance customer trust and satisfaction, leading to increased loyalty and positive brand perception. However, too generous a policy could lead to excessive costs. Therefore, the 4.7-year warranty effectively limits replacements, while still providing reasonable assurance to clients about the reliability and durability of the product, thereby positively influencing customer satisfaction.
1. **Find the z-score for 33%**: The z-score that corresponds to the top 33% (or the 67th percentile since we want to cover only 33% of failures) can be found using a z-table. The z-score for the 67th percentile is approximately 0.44.
2. **Calculate the warranty length**: We can use the following formula to determine the warranty length:
\[
X = \mu + Z \sigma
\]
where \( \mu \) is the mean, \( Z \) is the z-score, and \( \sigma \) is the standard deviation.
Plugging in the values:
\[
X = 4.3 + (0.44 \times 0.92) \approx 4.3 + 0.4048 \approx 4.70 \text{ years}
\]
Thus, the limited warranty should last approximately **4.7 years**.
**Customer Satisfaction**: Offering a warranty that ensures only 33% of batteries are replaced for free balances cost control with customer satisfaction. A longer warranty might enhance customer trust and satisfaction, leading to increased loyalty and positive brand perception. However, too generous a policy could lead to excessive costs. Therefore, the 4.7-year warranty effectively limits replacements, while still providing reasonable assurance to clients about the reliability and durability of the product, thereby positively influencing customer satisfaction.