To find the probability that an item will last longer than 3 years, we first need to standardize the value using the z-score formula for a normal distribution:
\[ z = \frac{(X - \mu)}{\sigma} \]
Where:
- \(X\) is the value we are interested in (3 years),
- \(\mu\) is the mean (3.5 years),
- \(\sigma\) is the standard deviation (0.8 years).
Now, substitute in the values:
\[ z = \frac{(3 - 3.5)}{0.8} = \frac{-0.5}{0.8} = -0.625 \]
Next, we need to find the probability that the lifespan is greater than 3 years, which corresponds to the area to the right of \(z = -0.625\).
We can find this using the standard normal distribution table, or a calculator. The table gives us the area to the left of the z value. For \(z = -0.625\):
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Look up the value in the z-table, or use a standard normal distribution calculator. The approximate area to the left of \(z = -0.625\) is about \(0.2659\).
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Since we want the area to the right, we subtract this value from 1:
\[ P(X > 3) = 1 - P(Z < -0.625) = 1 - 0.2659 = 0.7341 \]
Therefore, the probability that a randomly purchased item will last longer than 3 years is approximately:
\[ \boxed{0.7341} \]