Chapter 6: Problem 17
The lifetime of a type of battery is normally distributed with mean value \(10 \mathrm{~h}\) and standard deviation \(1 \mathrm{~h}\). There are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only \(5 \%\) of all packages?
Short Answer
Step by step solution
Understand the Problem
Calculate the Total Mean and Standard Deviation
Identify the Appropriate Z-Score
Calculate the Desired Lifetime
Compute the Total Lifetime
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Battery Lifetime
Understanding the normal distribution is significant because it allows us to predict variations in battery lifetime and make probabilistic conclusions about their performance and reliability.
- This means we can estimate not only the expected lifetime but also the likelihood of variations from the mean.
- For example, some batteries might last longer than 10 hours, while others might last less.
- These variations are measured using the standard deviation.
Z-Score
To find a lifetime value "such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages," we have to identify the Z-Score corresponding to the 95th percentile. This statistical value helps us understand where an observation stands in relation to a normal distribution posited with known means and standard deviations.
- The Z-Score for the 95th percentile is approximately 1.645.
- This means that the value is 1.645 standard deviations above the mean.
- Knowing the Z-Score is crucial for further calculations involving the standard normal distribution.
Standard Deviation
In our exercise, the standard deviation of the battery's lifetime is given as 1 hour. For the package of four batteries, we use a formula that calculates the total standard deviation of a combined set of data:
\ The total standard deviation = \( \sqrt{n} \times \sigma \), where \( n \) is the number of items, and \( \sigma \) is the standard deviation for one item.
- In our case, the total standard deviation for four batteries is \( \sqrt{4} \times 1 = 2 \, \mathrm{h} \).
- This tells us how much variation can be expected in the total lifetime of all four batteries combined.
Percentile Calculation
We wanted to find the total lifetime that is higher than 95% of all packages, which means we were looking for the 95th percentile of the distribution. Using a Z-Score table, we can translate this percentile into a Z-Score, as explained before, which is crucial for finding the desired value in a normally distributed set.
Using the formula:
- \( X = \mu_{total} + Z \times \sigma_{total} \)
- where \( \mu_{total} = 40 \), \( Z = 1.645 \), and \( \sigma_{total} = 2 \)