Question

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 8 years, and standard deviation of 2.4 years.

If you randomly purchase one item, what is the probability it will last longer than 5 years?

Round answer to three decimal places

Answer 1

To find the probability that an item will last longer than 5 years, we first need to standardize the value of 5 years using the Z-score formula for a normal distribution:

\[
Z = \frac{(X - \mu)}{\sigma}
\]

where:
- \( X \) is the value we are comparing (5 years),
- \( \mu \) is the mean (8 years),
- \( \sigma \) is the standard deviation (2.4 years).

Substituting the values into the formula gives us:

\[
Z = \frac{(5 - 8)}{2.4} = \frac{-3}{2.4} \approx -1.25
\]

Next, we need to find the probability that \( Z \) is greater than -1.25. We can use the standard normal distribution table (or a calculator) to find the probability associated with \( Z \leq -1.25 \).

Looking up \( Z = -1.25 \) in the Z-table, we find that:

\[
P(Z \leq -1.25) \approx 0.1056
\]

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

\[
P(Z > -1.25) = 1 - P(Z \leq -1.25) \approx 1 - 0.1056 = 0.8944
\]

Rounding to three decimal places, the probability that a randomly purchased item will last longer than 5 years is:

\[
\boxed{0.894}
\]