To solve this problem, we need to find the probability that a randomly selected item will last longer than 5 years. This is a normal distribution problem, so we will use the Z-score formula:

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

Where:

• $X = 5$ (the value we are interested in),
• $\mu = 6.8$ (the mean lifespan),
• $\sigma = 1.4$ (the standard deviation).

Step 1: Calculate the Z-score

$Z = \frac{5 - 6.8}{1.4} = \frac{-1.8}{1.4} \approx -1.286$

Step 2: Find the probability corresponding to the Z-score

We now need to look up the probability associated with a Z-score of -1.286 in a standard normal distribution table, or use a calculator. The Z-table gives the cumulative probability for values less than a given Z-score.

The cumulative probability for $Z = -1.286$ is approximately 0.0996. This means the probability that the item lasts less than 5 years is 0.0996.

Step 3: Find the probability of lasting longer than 5 years

To find the probability that the item lasts longer than 5 years, we subtract the cumulative probability from 1:

$P(X > 5) = 1 - 0.0996 = 0.9004$

Final Answer:

The probability that a randomly purchased item will last longer than 5 years is approximately $\mathbf{0.900}$.

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Would you like more details or have any questions?

Here are 5 related questions for deeper exploration:

1. What is the probability that an item will last between 5 and 7 years?
2. If the manufacturer guarantees items for 4 years, what percentage of items will last beyond the warranty?
3. What lifespan corresponds to the 95th percentile?
4. How does the probability change if the standard deviation increases to 2 years?
5. What is the Z-score for an item lasting exactly 8 years?

Tip: For normally distributed data, the empirical rule states that approximately 68% of values fall within one standard deviation of the mean, 95% within two, and 99.7% within three.