# Thoughts on Numerical Integration (Part 23): The normalcdf function on TI calculators

I end this series about numerical integration by returning to the most common (if hidden) application of numerical integration in the secondary mathematics curriculum: finding the area under the normal curve. This is a critically important tool for problems in both probability and statistics; however, the antiderivative of $\displaystyle \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$ cannot be expressed using finitely many elementary functions. Therefore, we must resort to numerical methods instead.

In days of old, of course, students relied on tables in the back of the textbook to find areas under the bell curve, and I suppose that such tables are still being printed. For students with access to modern scientific calculators, of course, there’s no need for tables because this is a built-in function on many calculators. For the line of TI calculators, the command is normalcdf.

Unfortunately, it’s a sad (but not well-known) fact of life that the TI-83 and TI-84 calculators are not terribly accurate at computing these areas. For example:

TI-84: $\displaystyle \int_0^1 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx \approx 0.3413447\underline{399}$

Correct answer, with Mathematica: $0.3413447\underline{467}\dots$

TI-84: $\displaystyle \int_1^2 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx \approx 0.1359051\underline{975}$

Correct answer, with Mathematica: $0.1359051\underline{219}\dots$

TI-84: $\displaystyle \int_2^3 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx \approx 0.021400\underline{0948}$

Correct answer, with Mathematica: $0.021400\underline{2339}\dots$

TI-84: $\displaystyle \int_3^4 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx \approx 0.0013182\underline{812}$

Correct answer, with Mathematica: $0.0013182\underline{267}\dots$

TI-84: $\displaystyle \int_4^5 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx \approx 0.0000313\underline{9892959}$

Correct answer, with Mathematica: $0.0000313\underline{84590261}\dots$

TI-84: $\displaystyle \int_5^6 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx \approx 2.8\underline{61148776} \times 10^{-7}$

Correct answer, with Mathematica: $2.8\underline{56649842}\dots \times 10^{-7}$

I don’t presume to know the proprietary algorithm used to implement normalcdf on TI-83 and TI-84 calculators. My honest if brutal assessment is that it’s probably not worth knowing: in the best case (when the endpoints are close to 0), the calculator provides an answer that is accurate to only 7 significant digits while presenting the illusion of a higher degree of accuracy. I can say that Simpson’s Rule with only $n = 26$ subintervals provides a better approximation to $\displaystyle \int_0^1 \frac{e^{-x^2/2}}{\sqrt{2\pi}} \, dx$ than the normalcdf function.

For what it’s worth, I also looked at the accuracy of the NORMSDIST function in Microsoft Excel. This is much better, almost always producing answers that are accurate to 11 or 12 significant digits, which is all that can be realistically expected in floating-point double-precision arithmetic (in which numbers are usually stored accurate to 13 significant digits prior to any computations).

# A Long-Sought Proof, Found and Almost Lost

I enjoyed this article from Quanta Magazine, both for its mathematical content as well as the human interest story.

# A Long-Sought Proof, Found and Almost Lost

From the opening paragraphs:

Known as the Gaussian correlation inequality (GCI), the conjecture originated in the 1950s, was posed in its most elegant form in 1972 and has held mathematicians in its thrall ever since. “I know of people who worked on it for 40 years,” said Donald Richards, a statistician at Pennsylvania State University. “I myself worked on it for 30 years.”

[Thomas] Royen hadn’t given the Gaussian correlation inequality much thought before the “raw idea” for how to prove it came to him over the bathroom sink… In July 2014, still at work on his formulas as a 67-year-old retiree, Royen found that the GCI could be extended into a statement about statistical distributions he had long specialized in. On the morning of the 17th, he saw how to calculate a key derivative for this extended GCI that unlocked the proof. “The evening of this day, my first draft of the proof was written,” he said.

Not knowing LaTeX, the word processer of choice in mathematics, he typed up his calculations in Microsoft Word, and the following month he posted his paper to the academic preprint site arxiv.org. He also sent it to Richards, who had briefly circulated his own failed attempt at a proof of the GCI a year and a half earlier. “I got this article by email from him,” Richards said. “And when I looked at it I knew instantly that it was solved” …

Proofs of obscure provenance are sometimes overlooked at first, but usually not for long: A major paper like Royen’s would normally get submitted and published somewhere like the Annals of Statistics, experts said, and then everybody would hear about it. But Royen, not having a career to advance, chose to skip the slow and often demanding peer-review process typical of top journals. He opted instead for quick publication in the Far East Journal of Theoretical Statistics, a periodical based in Allahabad, India, that was largely unknown to experts and which, on its website, rather suspiciously listed Royen as an editor. (He had agreed to join the editorial board the year before.)

With this red flag emblazoned on it, the proof continued to be ignored… No one is quite sure how, in the 21st century, news of Royen’s proof managed to travel so slowly. “It was clearly a lack of communication in an age where it’s very easy to communicate,” Klartag said.

# My Favorite One-Liners: Part 62

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

This is a story that I’ll tell after doing a couple of back-to-back central limit theorem problems. Here’s the first:

The chances of winning a column bet in roulette is 12/38. The bet pays 2 to 1, meaning that if you lose, you lose $1. However, if you win, you get your$1 back and $2 more. If this bet is made 1000 times, what is the probability of winning at least$0?

With my class, we solve this problem using standard techniques with the normal approximation:

$\mu = E(X) = 2 \times \displaystyle \frac{12}{38} + (-1) \frac{26}{38} = - \displaystyle \frac{1}{19}$

$E(X^2) = 2^2 \times \displaystyle \frac{12}{38} + (-1)^2 \frac{26}{38} = \displaystyle \frac{37}{19}$

$\sigma = SD(X) = \sqrt{ \displaystyle \frac{37}{19} - \left( - \displaystyle \frac{1}{19} \right)^2} = \displaystyle \frac{\sqrt{702}}{19}$

$E(T_0) = n\mu = 1000 \left( -\displaystyle \frac{1}{19} \right) \approx -52.63$

$\hbox{SD}(T_0) = \sigma \sqrt{n} = \displaystyle \frac{\sqrt{702}}{19} \sqrt{1000} \approx 44.10$

$P(T_0 > 0) \approx P\left(Z > \displaystyle \frac{0-(-52.63)}{44.10} \right) \approx P(Z > 1.193) \approx 0.1163$.

Next, I’ll repeat the problem, except playing the game 10,000 times.

The chances of winning a column bet in roulette is 12/38. The bet pays 2 to 1, meaning that if you lose, you lose $1. However, if you win, you get your$1 back and $2 more. If this bet is made 10,000 times, what is the probability of winning at least$0?

The last three lines of the above calculation have to be changed:

$E(T_0) = n\mu = 10,000 \left( -\displaystyle \frac{1}{19} \right) \approx -526.32$

$\hbox{SD}(T_0) = \sigma \sqrt{n} = \displaystyle \frac{\sqrt{702}}{19} \sqrt{10,000} \approx 139.45$

$P(T_0 > 0) \approx P\left(Z > \displaystyle \frac{0-(-526.32)}{139.45} \right) \approx P(Z > 3.774) \approx 0.00008$.

In other words, the chance of winning drops dramatically. This is an example of the Law of Large Numbers: if you do something often enough, then what ought to happen eventually does happen.

As a corollary, if you’re going to bet at roulette, you should only bet a few times. And, I’ll tell my students, one Englishman took this to the (somewhat) logical extreme by going to Las Vegas and making the ultimate double-or-nothing bet, betting his entire life savings on one bet. After all, his odds of coming out ahead by making one bet were a whole lot higher than by making a sequence of bets.

# My Favorite One-Liners: Part 40

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

In some classes, the Greek letter $\phi$ or $\Phi$ naturally appears. Sometimes, it’s an angle in a triangle or a displacement when graphing a sinusoidal function. Other times, it represents the cumulative distribution function of a standard normal distribution.

Which begs the question, how should a student pronounce this symbol?

I tell my students that this is the Greek letter “phi,” pronounced “fee”. However, other mathematicians may pronounce it as “fie,” rhyming with “high”. Continuing,

Other mathematicians pronounce it as “foe.” Others, as “fum.”

# In-class demo: The binomial distribution and the bell curve

Many years ago, the only available in-class technology at my university was the Microsoft Office suite — probably Office 95 or 98. This placed severe restrictions on what I could demonstrate in my statistics class, especially when I wanted to have an interactive demonstration of how the binomial distribution gets closer and closer to the bell curve as the number of trials increases (as long as both $np$ and $n(1-p)$ are also decently large).

The spreadsheet in the link below is what I developed. It shows

• The probability histogram of the binomial distribution for $n \le 150$
• The bell curve with mean $\mu = np$ and standard deviation $\sigma = \sqrt{np(1-p)}$
• Also, the minimum and maximum values on the $x-$axis can be adjusted. For example, if $n = 100$ and $p = 0.01$, it doesn’t make much sense to show the full histogram; it suffices to have a maximum value around 5 or so.

In class, I take about 3-5 minutes to demonstrate the following ideas with the spreadsheet:

• If $n$ is large and both $np$ and $n(1-p)$ are greater than 10, then the normal curve provides a decent approximation to the binomial distribution.
• The probability distribution provides exact answers to probability questions, while the normal curve provides approximate answers.
• If $n$ is small, then the normal approximation isn’t very good.
• If $n$ is large but $p$ is small, then the normal approximation isn’t very good. I’ll say in words that there is a decent approximation under this limit, namely the Poisson distribution, but (for a class in statistics) I won’t say much more than that.

Doubtlessly, there are equally good pedagogical tools for this purpose. However, at the time I was limited to Microsoft products, and it took me untold hours to figure out how to get Excel to draw the probability histogram. So I continue to use this spreadsheet in my classes to demonstrate to students this application of the Central Limit Theorem.