Probability of duration of a phone calls like a draw made random with replacement The durations of phone calls taken by the receptionist at an office are like draws made at random with replacement from a list that has an average of $8.5$ minutes (that's $8$ minutes and $30$ seconds) and an SD of $3$ minutes. Approximately what is the chance that the total duration of the next $100$ calls is more than $15$ hours?
First convert everything into seconds. The mean is $8.5$ minutes. $8.5$ minutes is $510$ seconds for one call, 
so it's $100$ times that amount for $100$ calls. $510\times 100=51000$ seconds
The SD is $3$ minutes. $3\times 60=180$ seconds for one call, 
the variance is the SD squared. $\sigma=180^2=32,400$ seconds. 
Multiply the variance by $100$, then take the square root of it for the SD for $100$ calls. 
$SD=\sqrt{32400\times 100}=1,800$ seconds.
$x$ is the time for $100$ calls. The mean is the mean for $100$ calls, SD is the SD for $100$ calls. You'll use the $z$ score to find the probability:
$15$  hours$\times60$  minutes$\times 60$  seconds$=54,000$ seconds
$$z=\frac{x−\bar{x}}{SD}= \frac{54,000-51,000}{1,800}=1.67$$
Area under normal curve Lower end $-5$ 
and upper end $1.67=95.22\%= 0.9522$ THEREFORE $P(t\gt54,000)=1-0.9522=0.0482$
 A: So you have 100 random variables, each with average $\mu_0=8.5$ minutes and standard deviation $\sigma_0=3$ minutes. The central limit theorem thus tells you that the sum of these random variables is approximately normally distributed with $\mu=100\mu_0=850$ minutes and standard deviation $\sigma=\sqrt{100}\sigma_0 = 30$ minutes.
You're looking for the probability that such a standard distribution yields a value greater than 15 hours, or equivalently greater than $15\cdot 60 = 900$ minutes. If a random variable $X$ is normally distributed with average $\mu$ and standard deviation $\sigma$, then the probability that $X \leq a$ (or equivalently $X < a$, since the normal distribution is continuous) is $$
  \mathbb{P}(X \leq a) = \Phi\left(\frac{a - \mu}{\sigma}\right) \text{,}
$$
where $\Phi$ is the cumulative distribution function (CDF) of the standard normal distribution, i.e. the one with average 0 and standard deviation $1$. You're interested in the probability that $X = X_1+\ldots+X_{100}$ is greater than 900 minutes, and must thuse compute $$
  \mathbb{P}(X > 900) = 1 - \mathbb{P}(X \leq 900) = 1 - \Phi\left(\frac{900 - 850}{30}\right) = 1 - \Phi\left(\frac{5}{3}\right) \text{.}
$$
Now you just need to use a table (or software) to evaluate $\Phi\left(\frac{5}{3}\right) \approx 0.952$ and get $$
   \mathbb{P}(X > 900) \approx 0.048 \text{.}
$$
So you simply forgot to invert the probability you got at the end, i.e. your value (which is $\Phi\left(\frac{5}{3}\right)$) is the probability that the calls take less than 15 hours in total.
Btw, there's an easy way to see that your answer cannot be correct. You're asking from the probability that the value is greater than some value which itself is greater than the mean of the distribution. Since the normal distribution is symmetric, the mean is also the median, and thus the probability cannot be larger than 0.5, i.e. 50%. And since the distance from the mean isn't particularly small compared to the standard deviation (your distance is 50, the standard deviation is 30), it'll have to be substantially smaller than 50%.
