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A manufacturer knows that their items have a normally distributed lifespan, with a mean of 13.3 years, and standard deviation of 1.1 years.If you randomly purchase one item, what is the probability it will last longer than 10 years?Use the normal table and round answer to four decimal places

Respuesta :

We want to find the following probability:

[tex]P(X>10)[/tex]

where X is a normal random variable with mean 13.3 and standard deviation 1.1. To find this probability let's normalize the random variable; to do this we use the z-score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

Then, in this case, we have:

[tex]P(X>10)=P(z>\frac{10-13.3}{1.1})=P(z>-3)[/tex]

Using the standard normal table, we have:

[tex]P(X\gt10)=P(z\gt(10-13.3)\/1.1)=P(z\gt-3)=0.9987[/tex]

Therefore, the probability of purchasing an item with a lifespan greater than 10 years is 0.9987

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