Question

a manufacturer knows that their items have a normally distributed lifespan, with a mean of 13.2 years, and standard deviation of 3.8 years.
if you randomly purchase one item, what is the probability it will last longer than 13 years?
round answer to three decimal places

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 13\), \(\mu=13.2\), and \(\sigma = 3.8\).

$$z=\frac{13 - 13.2}{3.8}=\frac{- 0.2}{3.8}\approx - 0.0526$$

Step2: Find the probability

We want to find \(P(X>13)\), which is equivalent to \(P(Z>-0.0526)\).
Since \(P(Z > z)=1 - P(Z\leq z)\), and looking up the value of \(P(Z\leq - 0.0526)\approx0.479\) (using a standard normal table or calculator).

$$P(Z>-0.0526)=1 - 0.479 = 0.521$$

Answer:

\(0.521\)