To find the probability that a randomly selected item lasts longer than 10 years, we can use the properties of the normal distribution.

The lifespan of the items is normally distributed with:

• Mean ($\mu$) = 13.3 years
• Standard deviation ($\sigma$) = 1.1 years

We are looking for the probability that the lifespan ($X$) is greater than 10 years: $P(X > 10)$.

Steps:

1. Convert the raw score to a z-score using the formula: $z = \frac{X - \mu}{\sigma}$ where $X$ is the value we're interested in (10 years), $\mu$ is the mean, and $\sigma$ is the standard deviation.

   $z = \frac{10 - 13.3}{1.1} = \frac{-3.3}{1.1} = -3.00$

2. Now, we look up the z-score of -3.00 in the standard normal distribution table or use a calculator. The cumulative probability corresponding to a z-score of -3.00 is approximately 0.00135.

3. The value from the z-table gives the probability of $X$ being less than 10 years, but we need the probability of $X > 10$. Therefore, subtract this value from 1:

   $P(X > 10) = 1 - P(X \leq 10) = 1 - 0.00135 = 0.99865$

Final Answer:

The probability that the item will last longer than 10 years is approximately $0.999$ (rounded to three decimal places).

Would you like more details, or do you have any other questions?

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Here are 5 related questions:

1. What is the probability the item lasts less than 12 years?
2. What is the probability the item lasts between 12 and 14 years?
3. How would the probability change if the standard deviation was 2 years instead of 1.1 years?
4. What is the 90th percentile of the lifespan of these items?
5. If you purchase 3 items, what is the probability that all will last longer than 10 years?

Tip: The z-score helps standardize any normal distribution, making it easier to use standard tables or calculators for probability.