A study reports that only $12 \%$ of people stick to their New Year's resolutions. Suppose you sample $100$ people about whether they stuck to their resolutions. What is the probability that less than $6$ out of $100$ people respond that they did(stuck to the resolution)?
It is pretty clear that $X$ is a binomial random variable with success probability $p=0.12$
The mean is $\mu=12$ and $\sigma^2=10.56$, so the standard deviation is $\sigma=3.249$
Now we need to find $P(X<6)$ which is same as $P(X \leq 5)$ since $X$ is discrete.Now here should I do $$P\left(\frac{x-\mu}{\sigma}<\frac{6-\mu}{\sigma}\right)=P(Z<-1.84)=0.0329$$ or should I do $$P\left(\frac{x-\mu}{\sigma} \leq \frac{5-\mu}{\sigma}\right)=P(Z \leq -2.15)=0.0158 $$
Here I have plotted the PMF of the binomial distribution with parameters $n = 100$, $p = 0.12$, which is represented as the blue shaded region, and the normal approximation with mean $\mu = np = 12$ and variance $\sigma^2 = np(1-p) = 10.56$, shown as the red curve.
While in actuality the binomial PMF should be represented as discrete points on the plot rather than bars of unit width, the reason for presenting it this way is to illustrate the reason for continuity correction. When we want to compute $\Pr[X < 6]$ for instance, what we want is the sum of the areas of the shaded bars corresponding to $X \in \{0, 1, 2, 3, 4, 5\}$. But if $Y$ is the aforementioned normal approximation, then computing $\Pr[Y < 5]$ will not capture half of the probability mass at $X = 5$:
And if we calculate $\Pr[Y < 6]$, we also have a problem since it captures half of the probability mass at $X = 6$:
Thus, the continuity-corrected calculation has us choose $\Pr[Y < 5.5]$ in order to approximate $\Pr[X < 6]$: