To find the probability that a randomly selected item will last longer than 14 years, we will use the properties of the normal distribution with the following given parameters:

• Mean ($\mu$) = 11.6 years
• Standard Deviation ($\sigma$) = 1.3 years

We want to find:

$P(X > 14)$

Step 1: Calculate the Z-score

The Z-score formula is:

$Z = \frac{X - \mu}{\sigma}$

Substitute the values:

$Z = \frac{14 - 11.6}{1.3} = \frac{2.4}{1.3} \approx 1.8462$

Step 2: Find the Probability Using the Z-score

Now, we need to find the probability that $Z$ is greater than 1.8462, which is represented as $P(Z > 1.8462)$.

Using a Z-table or a calculator, we find the cumulative probability for $Z \leq 1.8462$:

$P(Z \leq 1.8462) \approx 0.9678$

The probability of $Z > 1.8462$ is:

$P(Z > 1.8462) = 1 - 0.9678 = 0.0322$

Final Answer:

The probability that a randomly purchased item will last longer than 14 years is approximately:

$\boxed{0.0322}$

(rounded to 4 decimal places).

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Would you like more details or have any questions? Here are 5 related questions you might find interesting:

1. What is the probability that the item will last less than 10 years?
2. How long will an item last if it is at the 90th percentile of lifespans?
3. What is the mean lifespan for items if the standard deviation changes to 2 years?
4. How would the probability change if you wanted to find the likelihood of an item lasting between 11.6 and 14 years?
5. What Z-score corresponds to a lifespan of 12 years?

Tip: For probabilities involving normally distributed data, finding the Z-score helps transform the problem into one that uses the standard normal table or function.