Questions: A manufacturer knows that their items have a normally distributed lifespan, with a mean of 7 years, and standard deviation of 1.2 years. If you randomly purchase one item, what is the probability it will last longer than 5 years?

Solution Steps

The lifespan of the items is normally distributed with a mean (\( \mu \)) of 7 years and a standard deviation (\( \sigma \)) of 1.2 years. We are interested in finding the probability that a randomly purchased item lasts longer than 5 years.

To find the probability, we first calculate the Z-score for the value of 5 years using the formula:

\[ Z = \frac{X - \mu}{\sigma} \]

Substituting the values:

\[ Z_{start} = \frac{5 - 7}{1.2} = \frac{-2}{1.2} \approx -1.6667 \]

The probability that the item lasts longer than 5 years can be expressed as:

\[ P(X > 5) = 1 - P(X \leq 5) = 1 - \Phi(Z_{start}) \]

Using the Z-score calculated:

\[ P(X > 5) = 1 - \Phi(-1.6667) \]

From the output, we find:

\[ P(X > 5) = \Phi(inf) - \Phi(-1.6667) \approx 0.9522 \]

Final Answer