Question

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 14.2 years, andstandard deviation of 2.7 years.If you randonvily purchase one item, what is the probability it will last longer than 13 years?Round answer to three decimal places

Answer 1

Probability for Z score is given by the formula

`\begin{gathered} Z=(x-\mu)/(\sigma) \\ \\ \text{Where x =sample mean} \\ \mu=\text{population mean and } \\ \sigma=\text{standard deviation } \\ Z=(x-\mu)/(\sigma) \\ \\ Z=(13-14.2)/(2.7) \\ \\ Z=\text{ -0.44} \\ P(x>Z)\text{ = 0.67} \end{gathered}`

So from the Z score calculator, the Probability of it lasting more than 13 years = 0.67