Question

monitors manufactured by tsi electronics have life spans that have a normal distribution with a variance of 3,610,000 and a mean life span of 15,000 hours. if a monitor is selected at random, find the probability that the life span of the monitor will be more than 17,090 hours. round your answer to four decimal places.

Explanation:

Step1: Calculate the standard deviation

The variance $\sigma^{2}=3610000$, so the standard deviation $\sigma = \sqrt{3610000}=1900$.

Step2: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $\mu = 15000$, $x = 17090$, and $\sigma=1900$.
Substitute the values: $z=\frac{17090 - 15000}{1900}=\frac{2090}{1900}=1.1$.

Step3: Find the probability

We want to find $P(X>17090)$, which is equivalent to $P(Z > 1.1)$.
Since $P(Z>z)=1 - P(Z\leq z)$, and from the standard normal table $P(Z\leq1.1)=0.8643$.
So $P(Z > 1.1)=1 - 0.8643=0.1357$.

Answer:

$0.1357$