A manufacturer knows that their items have a normally distributed lifespan, with a mean of 12.2 years, and standard deviation of 2.2 years. If you randomly purchase one item, what is the probability it will last longer than 15 years?
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
![## Understanding Probability in Normally Distributed Lifespans
### Problem
A manufacturer knows that their items have a normally distributed lifespan, with a mean of 12.2 years, and standard deviation of 2.2 years.
**Question:**
If you randomly purchase one item, what is the probability it will last longer than 15 years?
### Explanation
We are given:
- Mean lifespan (\(\mu\)) = 12.2 years
- Standard deviation (\(\sigma\)) = 2.2 years
To find the probability that an item will last longer than 15 years, we need to calculate the z-score:
\[ Z = \frac{X - \mu}{\sigma} \]
where \(X\) is the value for which we are finding the probability.
### Calculation
For \(X = 15\) years:
\[ Z = \frac{15 - 12.2}{2.2} = \frac{2.8}{2.2} \approx 1.27 \]
Using standard normal distribution tables or a calculator, we find the probability associated with \(Z = 1.27\). This gives the area to the left of \(Z\), but we need the area to the right of \(Z\) to find the probability of lasting longer than 15 years.
For \(Z = 1.27\), the cumulative probability is approximately 0.8980.
Therefore, the probability of the item lasting longer than 15 years is:
\[ P(X > 15) = 1 - 0.8980 = 0.1020 \]
### Conclusion
The probability that an item will last longer than 15 years is approximately **0.1020** or 10.20%.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1a3586c7-75e0-478e-8b93-fa570f1e3b86%2F9fe5e37a-3ebd-4125-bd7b-f8a9c9cb9b26%2Fw0wqwr_processed.png&w=3840&q=75)
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