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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

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SOLUTION

Probability for Z score is given by the formula

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

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