Question: Question 9
A manufacturer knows that their items have a normally distributed lifespan, with a mean of $8.4$ years, and standard deviation of $1.7$ years.
If you randomly purchase one item, what is the probability it will last longer than $11$ years?
Round answer to three decimal places
1. **State the problem:** We want to find the probability that a randomly purchased item lasts longer than $11$ years, given the lifespan is normally distributed with mean $\mu = 8.4$ years and standard deviation $\sigma = 1.7$ years.
2. **Formula and rules:** For a normal distribution, the probability that a value $X$ is greater than some value $x$ is given by:
$$P(X > x) = 1 - P(X \leq x) = 1 - \Phi\left(\frac{x - \mu}{\sigma}\right)$$
where $\Phi$ is the cumulative distribution function (CDF) of the standard normal distribution.
3. **Calculate the z-score:**
$$z = \frac{11 - 8.4}{1.7} = \frac{2.6}{1.7} = 1.5294117647 \approx 1.53$$
4. **Find the cumulative probability for $z=1.53$:**
Using standard normal tables or a calculator,
$$\Phi(1.53) \approx 0.9370$$
5. **Calculate the probability of lasting longer than 11 years:**
$$P(X > 11) = 1 - 0.9370 = 0.0630$$
6. **Round the answer:**
The probability rounded to three decimal places is $0.063$.
**Final answer:**
$$\boxed{0.063}$$