A manufacturer knows that their items have a normally distributed lifespan, with a mean of 2.2 years, and standard deviation of 0.6 years. If you randomly purchase one item, what is the probability it will last longer than 3 years? Round answer to three decimal places

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To find the probability that a randomly purchased item will last longer than 3 years, we can use the properties of the normal distribution. Given: - Mean ($$\mu$$) = 2.2 years - Standard deviation ($$\sigma$$) = 0.6 years We want to find $$P(X > 3)$$, where $$X$$ is the lifespan of the item. First, we need to convert the lifespan of 3 years into a z-score using the formula: $$ z = \frac{X - \mu}{\sigma} $$ Substituting the values: $$ z = \frac{3 - 2.2}{0.6} = \frac{0.8}{0.6} \approx 1.3333 $$ Next, we will look up the z-score in the standard normal distribution table or use a calculator to find the probability associated with $$z = 1.3333$$. Using a standard normal distribution table or calculator, we find: $$ P(Z < 1.3333) \approx 0.9082 $$ This value represents the probability that a randomly selected item lasts less than 3 years. To find the probability that it lasts longer than 3 years, we subtract this value from 1: $$ P(X > 3) = 1 - P(Z < 1.3333) \approx 1 - 0.9082 = 0.0918 $$ Rounding to three decimal places, the probability that a randomly purchased item will last longer than 3 years is: $$ \boxed{0.092} $$ The probability that a randomly purchased item will last longer than 3 years is 0.092.

A manufacturer knows that their items have a normally distributed lifespan, with a mean of 2.2 years, and standard deviation of 0.6 years. If you randomly purchase one item, what is the probability it will last longer than 3 years? Round answer to three decimal places

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