To solve this problem, we can use the properties of the normal distribution. The lifespan of the items is normally distributed with a mean ($\mu$) of 6.2 years and a standard deviation ($\sigma$) of 2 years. We are asked to find the probability that a randomly chosen item will last longer than 8 years.

Steps:

1. Standardize the value: We first convert the raw value (8 years) to a z-score using the formula: $z = \frac{x - \mu}{\sigma}$ where:

   • $x = 8$ (the value we are interested in),
   • $\mu = 6.2$ (mean lifespan),
   • $\sigma = 2$ (standard deviation).

   Substituting the values: $z = \frac{8 - 6.2}{2} = \frac{1.8}{2} = 0.9$

2. Find the probability: We now need to find the probability that the lifespan is greater than 8 years, which corresponds to finding the probability that the z-score is greater than 0.9.

   Using standard normal distribution tables or a calculator:

   • The cumulative probability for $z = 0.9$ is approximately 0.8159.
   • Since we want the probability of the item lasting longer than 8 years, we need to subtract this from 1: $P(X > 8) = 1 - P(Z \leq 0.9) = 1 - 0.8159 = 0.1841$

Final Answer:

The probability that a randomly selected item will last longer than 8 years is approximately 0.184 (rounded to three decimal places).

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Would you like more details on any part of the solution, or do you have any questions?

Here are five related questions to extend your understanding:

1. What is the probability that the item will last less than 5 years?
2. What is the probability that an item will last between 4 and 7 years?
3. How long will an item last if you want the lifespan to be in the top 5%?
4. What is the probability that an item will last between 6 and 9 years?
5. How does the probability change if the standard deviation increases to 3 years?

Tip: The z-score allows you to compare data points from different normal distributions or the same distribution.