Question

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 3.5 years, and standard deviation of 0.8 years.

If you randomly purchase one item, what is the probability it will last longer than 3 years?

Answer 1

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 \(z = -0.625\):

1. 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 \(0.2659\).

2. 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:

\[
\boxed{0.7341}
\]