Using central limit theorem for normal distribution So i have the following question:

The times that patients spend in a doctor’s surgery have mean 5 minutes, and standard
deviation 2 minutes. On one particular day, the doctor sees 30 patients during his surgery
which starts at 4.30pm. Find an approximate probability that she finishes with her last
patient before 6.50pm. State clearly any assumptions you need to make.

I have so far got to the following:
Let X be the time (in minutes) spent at the surgery. From the question we have $\text{E}(X) = 5$
and $\text{Var}(X) = 4$. 
Using a random samply 30 from $X_1, X_2, \ldots , X_{30}$.

Let $$Y =
\sum_{i=1}^{30}
X_i$$
By the Central Limit Theorem $Y ∼ N(\mu = 150, \sigma^2 = 120)$ approximately.
As surgery starts at 4 : 30, and the doctor desires to finish surgery at 6 : 50, we desire
$P(Y &lt; 140)$ 

However I am not sure where to go from here? In the answer it says $P(Y &lt; 140)=0.1807$ could someone explain how this was calculated? Thanks
Perhaps a statistician might be able to offer some insight, but I believe your application of the CLT as stated in the question is wrong. This question defines the CLT as follows:


  If $X_{1}, ... , X_{n}$ are i.i.d continuous random variables with
  mean $\mu$ and variance $\sigma^2$, then as $n \rightarrow \infty$,
  $$ \sqrt{n} \frac{\bar{X} - \mu}{\sigma} $$ will have the standard
  normal distribution


It seems like you're off by a factor of $\sqrt{n}$ somewhere. Let's break it down. First, the empirical mean of your sample, $\bar{X}$, is defined as:

$$\bar{X} = \frac{\sum_{i=1}^{30}
X_i}{30}$$

Applying the CLT, as above:

$$\sqrt{n} \frac{\bar{X} - \mu}{\sigma} = \sqrt{30} \frac{\frac{\sum_{i=1}^{30}
X_i}{30} - 5}{2} \approx N(\mu = 0, \sigma^2 = 1)$$

Now, we're interested in $P(Y&lt;140)$, where $Y$ is defined as:

$Y = \sum_{i=1}^{30}{X_i} = 30\bar{X}$

So:
$$P(Y&lt;140) = P(\bar{X}&lt;\frac{14}{3})$$

Now, rearranging our normal approximation identity from above, with $n = 30$

$$\sqrt{30}(\bar{X} - \mu) \approx 2 N(\mu = 0, \sigma^2 = 1)$$
$$(\bar{X} - \mu) \approx \dfrac{2 N(\mu = 0, \sigma^2 = 1)}{\sqrt{30}}$$
$$\bar{X} \approx \dfrac{2 N(\mu = 0, \sigma^2 = 1)}{\sqrt{30}} + 5$$

So:

$$P(\bar{X}&lt;\frac{14}{3}) \approx P(\dfrac{2 N(\mu = 0, \sigma^2 = 1)}{\sqrt{30}} + 5 &lt; \frac{14}{3})$$

Rearranging again we have:

$$P(N(\mu = 0, \sigma^2 = 1) &lt; \frac{\sqrt{30}(\frac{14}{3} - 5))}{2})$$

Crunching out the RHS, this boils down to:

$$P(N(\mu = 0, \sigma^2 = 1) &lt; -0.91287092917)$$

Now that the LHS is your standard normal distribution, you can plug the RHS number into any bog-standard cumulative probability calculator such as https://stattrek.com/online-calculator/normal.aspx and you get a cumulative probability of $0.181$, which is the same as that quoted in the question up to a rounding error.

Assumptions

The Q asks to state assumptions. The assumptions we are making are:


In order to apply the CLT as above, we have to assume that the variables $X_i$ are indeed all independent and are indeed all identically distributed. This is a big assumption. Because in reality, if one patient takes longer, a doctor might rush the next one, for example.
We assume that a sample of $30$ patients is sufficiently large to apply the CLT, but in theory the CLT only guarantees us a normal distribution in the infinite limit. We are assuming that our underlying distributions for patient wait time are sufficiently "well behaved" that the convergence guaranteed by the CLT is enough to give us a good approximation with sample size $30$. However, in general, we can't always make that assumption. Some distributions are so skewed that we need many more than $30$ terms in our sum in order for the resulting distribution to start looking normal.


Common Sense Check

Does that answer make sense? Well, yeah. If everyone on average takes 5 mins, and you have 30 people, you'd expect them to take $150$ mins. $140$ mins is ten mins quicker than that, and this calculation gives that an $18\%$ chance of that happening, which seems pretty reasonable to me.

Explanation of Strategy

In general in these types of problems, you want applly the CLT to allow you to restate the problem in terms of the standard normal distribution, which you can then use to get your probabilities easily using standard tables or a calculator like I did.
$P(Y&lt;140)=P(Z&lt;\frac{140-150}{\sqrt{120}}$)