# When are real limits used for calculating z score?

I'm taking an introductory statistics course. At first, the textbook talks about real limits in the context of continuous variables and frequency distribution table, that is all clear. But on what conditions are real limits used for calculating the z score? Only when we are working with approximation of binomial distribution, because categories of scores are discrete (e.g. coin flip, p=1/2, can't get 2.5 heads in 4 tries)? In such case when we look for p(X > 3), we take X = 2.5, get deviation from mean, divide by std. dev. to convert to z score.

2nd contextual question. E.g. we have data of income per household. When we look for p(X > 50000), we take X = 50000, find z score of 50000, and proportion in tail from the unit normal table is the answer (e.g. if mean was 40k, std. deviation is 10k, then z = (50k - 40k) / 10k = 1. What about when we look for p(X >= 50000), do we find z score from X= 49999.5 (e.g. z = (49999.5 - 40k) / 10k = 0.99995)?

EDIT: Definition of real limits from the textbook (tl;dr continous (not discrete) variable has possible values in between, e.g. 1 and 2 can have 1.3, 1.2 in between, so measurement of score e.g. X = 2 represents an interval between 1.5 - 2.5):

• It's not clear what you're asking. Are you asking about when you take an interval instead of a point? As for your second question, z score is (x-mean)/std. To find the z score for 50000, you need to subtract the mean and then divide by the standard deviation. If you get a z score more than 5, you probably are doing something wrong. Sep 14, 2018 at 19:24
• @Acccumulation, for 2nd question: I know how to find p (X > 500): 1. find z score: (500 - mean) / std 2. find tail of this z score. How do you find p (X >= 500)? Sep 14, 2018 at 19:26
• Welcome to our site. Could you expand a little on what you might mean by "real limits"? It isn't clear what this phrase might be referring to.
– whuber
Sep 14, 2018 at 19:44
• @whuber thank you, I added a definition of real limits Sep 14, 2018 at 19:52
• That is interesting, thank you. (From what I can tell, "real limits" is a somewhat idiosyncratic term used by psychologists; it's not common among statisticians.) Although it appears relevant, I think this concept could turn out to be a little misleading, in part because values in a Binomial distribution are counts. There's no meaningful sense in which they would be conceived of as intervals. I believe you might find investigating the uses of "continuity correction" to be more fruitful. Incomes are different--but how do you even know they are reported to the nearest dollar?
– whuber
Sep 14, 2018 at 19:59

Suppose $X \sim \mathsf{Binom}(n=75,\, p=0.2)$ and you want to find $P(X \le 12).$ You have several choices.

Binomial formula: Use the formula for the binomial PDF (or PMF) to evaluate each of 13 individual terms. A somewhat tedious task. $$P(X \le 12) = P(X = 0) + P(X = 1) + \cdots + P(X = 12) = \sum_{k=0}^{12} {75 \choose k}(.2)^k(.8)^{75-k}.$$

Tables: Some books have tables for binomial probabilities. Your textbook may have a few such tables in an appendix for various (smallish) values of $n$ and a few values of $p.$ Typically, $n = 75$ and $p=0.2$ will not be found there. Before the computer era, there were entire books of binomial probabilities, but those have mainly disappeared now.

Software: Maybe you have a statistical calculator that will do such computations. But nowadays there is good software that will do the job easily. One type of software that will handle this problem is R. In R, the function pbinom is a binomial CDF function. The CDF consists of probabilities of the form $P(X \le k).$ The R code below shows that, for $X \sim \mathsf{Binom}(n=75,\, p=0.2),$ we have $P(X \le 12) = 0.2397,$ correct to four places, with more places of accuracy available if needed.

pbinom(12, 75, .2)
[1] 0.2396826


Normal approximation: For $n$ sufficiently large and $p$ not too near $0$ or $1,$ the distribution $\mathsf{Binom}(n,p)$ can be approximated by using a normal distribution with matching mean and variance: $\mathsf{Norm}(\mu, \sigma),$ where $\mu = np$ and $\sigma = \sqrt{np(1-p)}.$

An often-useful rule of thumb is that the normal approximation gives about two-place accuracy if both $np > 5$ and $n(1-p) > 5.$ Both are satisfied for $n=75$ and $p = 0.2.$ So we find $\mu = np = 15$ and $\sigma = \sqrt{np(1-p)} = 3.4641.$

Because the binomial distribution is discrete and the normal distribution is continuous, we have to be a little careful in order to get best results from the normal approximation. For $\mathsf{Binom}(75,\,0.2)$, we have $$P(X \le 12) = P(X < 12.5) = P(X < 13).$$ But for $\mathsf{Norm}(\mu = 15,\, \sigma=0.6928),$ these are three different probabilities. Briefly put, the 'continuity correction' uses the second of the three because it (usually) gives the best approximation.

$$P(X \le 12.5) = P\left(\frac{X-\mu}{\sigma} \le \frac{12.5-15}{3.461}\right) \approx P(Z \le -0.7217 ) \\ \approx P(Z \le -0.72) = 0.24,$$ where $Z \sim \mathsf{Binom}(0,1).$ The second approximation is necessary if you are using printed tables of the standard normal distribution because (without interpolation) these tables provide z-values to only two places.

pnorm(-0.7217);  pnorm(-0.72)
[1] 0.2352395
[1] 0.2357625


In the figure below, the exact binomial probability $P(X \le 12)$ is the sum of the heights of the vertical black bars to the left of the broken red line. The normal approximation is the area under the blue normal density curve to the left of the red line.

Notice that, according to the normal curve, the probability $P(X = 12)$ is represented by the area under the curve and above the interval $(11.5, 12.5].$ By using the normal approximation, we have included this entire probability--instead of just part of it.