Normal Distribution Support cal per minute how to find Z-score?

Question

I know how to calculate question which are phrased like so

A study of data collected at a company manufacturing flashlight batteries shows that a batch of 8000 batteries have a mean life of 250 minutes with a standard deviation of 20 minutes. Assuming a Normal Distribution, estimate:

(i) How many batteries will fail before 220 minutes?

Answer = 534.4

But I can not figure questions phrased like this:

Support call times at a technical support center are Normally distributed with a mean time of 8 minutes and 45 seconds and a standard deviation of 1 minute and 5 seconds. On a particular day, a total of 500 calls are taken at the centre. How many of these calls are likely to last more than 10 minutes

I dont understand how to find the z-score in this question as its to do with time?

Answer 1

To calculate the z-score you have to standardize the random variable. The support call time is distributed as $T\sim\mathcal N\left(8.75, (1\frac1{12})^2 \right)$

Reasoning: $45$ seconds are $0.75$ minutes. And 5 seconds are $\frac1{12}$ minutes.

Therefore $Z=\frac{T-8.75}{1\frac1{12}}=\frac{T-8.75}{\frac{13}{12}}$. Then it is asked for

$$P(T> 10)=1-P(T\leq 10)=1-\Phi\left(\frac{10-8.75}{\frac{13}{12}}\right)=1-\Phi\left(\frac{\frac54}{\frac{13}{12}}\right)=1-\Phi\left(\frac{15}{13}\right)$$

This is the probability that one arbitrary call last more than 10 minutes.

Answer 2

The Z score is how many standard deviations above the mean 10 minutes is. It is $5/4$ minutes more than the mean of eight minutes and forty five seconds, and the standard deviation is $13/12$ minutes, so the Z score is $15/13$.