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A manufacturer knows that their items have a normally distributed lifespan, with a mean of 6.9 years, and standard deviation of 1.6 years. If you randomly purchase one item, what is the probability it will last longer than 10 years ?

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ANSWER

0.9738 or 97.38%

Step-by-step explanation

Given:


\begin{gathered} mean(\mu)=6.9 \\ Standard\text{ }Deviation(\sigma)=1.6 \end{gathered}

Desired Outcome:

Probability that it will last 10 years

z-score for the sample:


\begin{gathered} z-score=(X-\mu)/(\sigma) \\ z-score=(10-6.9)/(1.6) \\ z-score=1.9376 \end{gathered}

p-value

For the z-score of 1.9376, the p-value is 0.9738 or 97.38%

Hence, the probability that it will last longer than 10 years if you randonly purchase one item is 97.38%

User Eroironico
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