Answer: 11.5%
Explanation:
Since 1 minute = 60 seconds, we multiply 12 minutes by 60 so that 12 minutes = 720 seconds. Thus, we're looking for a probability that the auditor will spend more than 720 seconds.
Now, we get the z-score for 720 seconds by the following formula:
[tex]\text{z-score} = \frac{x - \mu}{\sigma} [/tex]
where
[tex]t = \text{time for the auditor to finish his work } = 720 \text{ seconds}
\\ \mu = \text{average time for the auditor to finish his work } = 600 \text{ seconds}
\\ \sigma = \text{standard deviation } = 100 \text{ seconds}[/tex]
So, the z-score of 720 seconds is given by:
[tex]\text{z-score} = \frac{x - \mu}{\sigma}
\\
\\ \text{z-score} = \frac{720 - 600}{100}
\\
\\ \boxed{\text{z-score} = 1.2}[/tex]
Let
t = time for the auditor to finish his work
z = z-score of time t
Since the time is normally distributed, the probability for t > 720 is the same as the probability for z > 1.2. In terms of equation:
[tex]P(t \ \textgreater \ 720)
\\ = P(z \ \textgreater \ 1.2)
\\ = 1 - P(z \leq 1.2)
\\ = 1 - 0.885
\\ \boxed{P(t \ \textgreater \ 720) = 0.115}[/tex]
Hence, there is 11.5% chance that the auditor will spend more than 12 minutes in an invoice.
