Statistics

Shape of a binomial and the normal approximation

A binomial count over hundreds of trials has hundreds of possible values, each with its own small probability. When does its bar chart take on a recognizable shape, and when can a single normal curve stand in for all those bars?

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Mean and standard deviation of a binomialStatistics Normal random variablesGrade 12

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Sampling distribution of a sample proportionStatistics

Where you are 4 in The binomial distribution

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Random variables11

The binomial distribution4

Mean and standard deviation of a binomial

Shape of a binomial and the normal approximation

The binomial probability formula The binomial setting

Why this exists

Ilze runs a citrus packing line. Each grapefruit that passes her station is blemished with chance 0.12, independently of the others (a made-up chance), and a crate holds 400 of them. The number of blemished grapefruit X in a crate is a binomial count (The binomial setting) with n = 400 and p = 0.12.

Janis buys crates for juice and pays less for any crate with 40 or fewer blemished fruit, since those go to the fresh market instead. He wants P(X \le 40). The binomial formula answers it, but only as a sum of 41 terms, from P(X = 0) to P(X = 40), with binomial coefficients as long as 56 digits. Technology does that sum without complaint. Still, a statistician needs to know what the distribution looks like, and for large counts there is a much simpler description: a normal curve, fixed by just the mean and standard deviation.

This lesson takes two days. Day 1 is the shape of a binomial distribution: which way it leans, and why it straightens out as n grows. Day 2 is the normal approximation and the condition that says when it can be trusted.

Definition

A binomial distribution’s shape is the pattern of its probability bar chart. It is skewed right when the bars pile up at low counts and trail off in a long tail toward high counts, symmetric when the bars on either side of the center match, and skewed left when the long tail runs toward low counts.

The normal approximation to a binomial replaces the binomial count X, with n trials and success chance p, by a normal random variable with the same mean and standard deviation:

\begin{aligned} &X \text{ is approximately} \\ &\quad N\!\left(np, \sqrt{np(1-p)}\right) \end{aligned}

and finds probabilities for X as areas under that normal curve. It is used only when the large-counts condition holds: the expected number of successes and the expected number of failures are both at least 10,

np \ge 10 \quad \text{and} \quad n(1-p) \ge 10.

The mean np and standard deviation \displaystyle \sqrt{np(1-p)} come from Mean and standard deviation of a binomial, which found that a cooking class’s no-shows balance at np and typically stray about \displaystyle \sqrt{np(1-p)} from it. Normal curves, their areas and z-scores are from Normal random variables, Normal calculations: areas and percentiles and The standard normal distribution.

Intuition

A count cannot go below 0 or above n. Those two walls decide the shape.

When p is small, the mean np sits close to the wall at 0. The bars cannot spill past 0, so they pile up against it; above the mean nothing stops them, so a thin tail stretches toward n. That is skew to the right. When p is large the same thing happens against the wall at n, and the tail runs left. At p = 0.5 the mean is exactly halfway between the walls and the two sides mirror each other.

Figure 1Ten trials, three success chances. A small p piles the count against 0 (skewed right), p = 0.5 balances it (symmetric), and a large p piles it against 10 (skewed left).

Now hold p fixed and let n grow. The mean np grows in step with n, while the standard deviation \displaystyle \sqrt{np(1-p)} grows only with \displaystyle \sqrt{n}. Measured in standard deviations, the wall at 0 moves farther and farther away, and once it is several standard deviations off, it no longer squeezes the bars. What is left is the shape of a sum of many small independent pieces: a symmetric, single-peaked hump. That hump is very close to a density curve we already know, the normal curve.

Figure 2The same made-up blemish chance, 0.12, for crates of 10, 50 and 400 grapefruit (each chart has its own height scale; dashed lines mark the means). As the crate grows, the count moves away from the wall at 0 and the lean fades.

Results

Shape. For a binomial count with n trials and success chance p:

p < 0.5: skewed right; p = 0.5: symmetric; p > 0.5: skewed left.
The distributions for p and 1 - p are mirror images: P(X = k) for chance p equals P(X = n - k) for chance 1 - p.
For fixed p, the larger n is, the more symmetric and bell-shaped the distribution.
The mean sits \displaystyle \sqrt{np/(1-p)} standard deviations above 0 and \displaystyle \sqrt{n(1-p)/p} standard deviations below n.

Normal approximation. If np \ge 10 and n(1-p) \ge 10, then

\begin{aligned} &X \text{ is approximately} \\ &\quad N\!\left(np, \sqrt{np(1-p)}\right) \end{aligned}

so for a count a,

\begin{aligned} &P(X \le a) \\ &\quad \approx P\!\left(Z \le \frac{a - np}{\sqrt{np(1-p)}}\right) \end{aligned}

where Z is standard normal. The condition is a rule of thumb for “close enough”, not a sharp line.

Why it is true

The mirror. Swap the roles of success and failure. A count of k successes with chance p is a count of n - k failures, and each failure happens with chance 1 - p. The formula agrees: \binom{n}{k} p^k (1-p)^{n-k} is unchanged when k becomes n - k and p becomes 1 - p, because \binom{n}{k} = \binom{n}{n-k}. At p = 0.5 the two are the same distribution, so it is its own mirror image: symmetric.

How far the walls are. The distance from the wall at 0 to the mean, in standard deviations, is

\frac{np}{\sqrt{np(1-p)}} = \sqrt{\frac{np}{1-p}}

For fixed p this grows like \displaystyle \sqrt{n}, which is why the lean fades. The same steps give \displaystyle \sqrt{n(1-p)/p} for the wall at n.

Why the condition says ten. If np \ge 10, then since 1 - p \le 1,

\sqrt{\frac{np}{1-p}} \ge \sqrt{np} \ge \sqrt{10} \approx 3.16

so the wall at 0 is more than three standard deviations below the mean, and a normal curve with that center and spread puts almost no area below 0. The condition n(1-p) \ge 10 does the same for the wall at n. The cutoff 10 is a convention: a larger cutoff buys a closer match.

Why a normal curve at all. Write X = X_1 + X_2 + \cdots + X_n, where X_i is 1 if grapefruit i is blemished and 0 if not. A sum of many independent, identically distributed pieces has a distribution close to normal once there are enough pieces; this is the The central limit theorem, stated there for sample means. The large-counts condition is the binomial’s version of “enough pieces”: how many are needed depends on how lopsided each piece is, and a small p needs many more trials than p = 0.5.

Worked examples

Day 1: the shape of a binomial distribution

Example 1

A small crate leans right

Liga packs a sample crate of 10 grapefruit, each blemished with chance 0.12, independently. Find the distribution of the number blemished X for 0 to 6, describe its shape, and explain the shape.

Solution

Each probability is P(X = k) = \binom{10}{k}(0.12)^k(0.88)^{10-k}. For example, P(X = 1) = 10(0.12)(0.88)^9 \approx 0.3798. To four decimal places:

k	P(X = k)	k	P(X = k)
0	0.2785	4	0.0202
1	0.3798	5	0.0033
2	0.2330	6	0.0004
3	0.0847

Counts 7 to 10 together have probability below 0.0001. The distribution peaks at 1 and has a long, thin tail to the right: it is skewed right (Figure 1, left). The mean is np = 1.2 and the standard deviation is \displaystyle \sqrt{10(0.12)(0.88)} \approx 1.03. The mean is barely more than one standard deviation above 0, so the bars are pressed against that wall, while the other wall, at 10, is about 8.6 standard deviations above the mean.

Example 2

The good fruit lean left

In the same crate of 10, let Y be the number of unblemished grapefruit. Give P(Y = 10), P(Y = 9) and P(Y = 8), and describe the shape of Y’s distribution.

Solution

Each grapefruit is unblemished with chance 0.88, so Y is binomial with n = 10 and p = 0.88. Since Y = 10 - X, the event Y = 10 is the event X = 0, and so on:

\begin{aligned} P(Y = 10) &\approx 0.2785 \\ P(Y = 9) &\approx 0.3798 \\ P(Y = 8) &\approx 0.2330 \end{aligned}

These are Example 1’s probabilities read from the other end. The distribution is the mirror image of X’s: it peaks at 9, with a long tail toward low counts, so it is skewed left (Figure 1, right). Its mean is 10(0.88) = 8.8, and its standard deviation is the same 1.03, since a mirror image has the same spread.

Example 3

Two lanes at even odds

Liga’s line splits at a gate that sends each grapefruit to the left or the right lane with chance 0.5 each, independently. For a group of 6 grapefruit, find the distribution of the number L sent left and describe its shape.

Solution

L is binomial with n = 6 and p = 0.5, so every one of the 2^6 = 64 ways the six fruit can split has probability (0.5)^6 = \tfrac{1}{64}, and P(L = k) = \binom{6}{k}\tfrac{1}{64}:

k	P(L = k)	k	P(L = k)
0	\tfrac{1}{64}	6	\tfrac{1}{64}
1	\tfrac{6}{64}	5	\tfrac{6}{64}
2	\tfrac{15}{64}	4	\tfrac{15}{64}
3	\tfrac{20}{64}

The distribution is exactly symmetric about its mean 6(0.5) = 3, because \binom{6}{k} = \binom{6}{6-k}: choosing which k fruit go left is the same as choosing which 6 - k go right. With p = 0.5 this holds for every n, even n = 2; symmetry at p = 0.5 does not need a large group.

Example 4

Bigger crates straighten out

Compare crates of 10, 50 and 400 grapefruit at blemish chance 0.12. For each, find the mean and standard deviation of X, and how many standard deviations the mean sits above 0. What does this say about the shapes in Figure 2?

Solution

Use \mu_X = np and \displaystyle \sigma_X = \sqrt{np(1-p)} from Mean and standard deviation of a binomial, then divide:

crate n	\mu_X	\sigma_X	\mu_X / \sigma_X
10	1.2	\displaystyle \sqrt{1.056} \approx 1.03	\approx 1.17
50	6	\displaystyle \sqrt{5.28} \approx 2.30	\approx 2.61
400	48	\displaystyle \sqrt{42.24} \approx 6.50	\approx 7.39

Each skews right, because p = 0.12 is below 0.5, but by very different amounts. For the crate of 10, the wall at 0 is about one standard deviation below the mean and squeezes the bars hard. For the crate of 50 it is about 2.6 standard deviations away and the lean is milder. For the crate of 400 it is more than seven standard deviations away, so far that it has no visible effect: the bars form a nearly symmetric hump around 48. Multiplying n by 40 (from 10 to 400) multiplied \mu_X / \sigma_X by \displaystyle \sqrt{40} \approx 6.3, as the \displaystyle \sqrt{np/(1-p)} formula predicts.

Day 2: the normal approximation

Example 5

Janis’s question, two ways

For a crate of 400 grapefruit, each blemished with chance 0.12, find P(X \le 40) with the normal approximation and exactly with technology. How far apart are they?

Solution

First the condition: np = 400(0.12) = 48 and n(1-p) = 400(0.88) = 352, both at least 10, so the approximation is justified. The normal model has

\begin{aligned} \mu &= 48 \\ \sigma &= \sqrt{400(0.12)(0.88)} \\ &= \sqrt{42.24} \approx 6.50 \end{aligned}

The count 40 has z-score

z = \frac{40 - 48}{\sqrt{42.24}} \approx -1.23

and the standard normal area to the left of it (using the unrounded z) is about 0.1092.

The exact binomial probability, the sum of P(X = 0) through P(X = 40) by technology, is P(X \le 40) \approx 0.1227. The two differ by about 0.0135: the normal answer is low by roughly one ninth of the true value. Both tell Janis the same story, that about one crate in eight or nine has 40 or fewer blemished fruit, but the normal answer is a good approximation, not a match.

Figure 3 shows where the gap comes from. The bar for X = 40 stands over the interval from 39.5 to 40.5, and cutting the curve at 40 keeps only half of it. That missing half bar, half of P(X = 40), is about 0.0149, close to the 0.0135 gap. As an aside, cutting at 40.5 instead (the continuity correction) gives \displaystyle z = \frac{40.5 - 48}{\sqrt{42.24}} \approx -1.15 and an area of about 0.1243, within 0.0016 of the exact value.

Figure 3A crate of 400 (made-up chance 0.12). The shaded bars add to the exact P(X \le 40) \approx 0.1227; the normal area left of the dashed line is 0.1092. The cut at 40 keeps only half of the bar for 40; that missing half bar, about 0.0149, is close to the 0.0135 gap.

Example 6

When the condition fails

A smaller crate holds 50 grapefruit, blemish chance 0.12. Check the large-counts condition, then compare the normal approximation of P(X \le 3) with the exact value 0.1345. What is the smallest crate for which the condition holds?

Solution

Here np = 50(0.12) = 6, below 10, so the condition fails (even though n(1-p) = 44 is fine). Ignoring that for the moment, the normal model would have \mu = 6 and \displaystyle \sigma = \sqrt{5.28} \approx 2.30, so

z = \frac{3 - 6}{\sqrt{5.28}} \approx -1.31

with area about 0.0958 to its left. The exact probability is 0.1345, so the normal answer is off by about 0.0387, close to 30\% of the true value. The crate of 50 is still visibly skewed right (Figure 2, middle), and a symmetric curve cannot follow a lopsided distribution.

For the condition, both 0.12n \ge 10 and 0.88n \ge 10 must hold. The first needs \displaystyle n \ge \frac{10}{0.12} \approx 83.3, the second only \displaystyle n \ge \frac{10}{0.88} \approx 11.4. So the smallest crate is n = 84, where np = 10.08. The rarer outcome, blemished fruit, sets the size.

Example 7

Is this crate unusual?

Ilze opens one crate of 400 and finds 66 blemished grapefruit. If the chance really is 0.12, how likely is a count of 66 or more? What should she conclude?

Solution

The condition holds (Example 5), so use N(48, 6.50). The count 66 sits

z = \frac{66 - 48}{\sqrt{42.24}} \approx 2.77

standard deviations above the mean. The standard normal area to the right of 2.77 is about 0.0028. The exact binomial tail, by technology, is P(X \ge 66) \approx 0.0048. Far out in a tail, the exact value is about 1.7 times the normal one, yet both say the same thing: at chance 0.12, a crate this bad turns up only a few times in a thousand.

So either Ilze saw a rare crate or the chance for this crate is higher than 0.12, perhaps from a bad batch of fruit. A count almost three standard deviations from the mean is reason to inspect that batch. This reasoning, measuring how surprising a count is under an assumed p, is the start of testing a claim about a proportion.

Boundaries

The shape results hold for binomial counts: fixed n, independent trials, the same p on each. If the trials are not independent (blemishes that spread from fruit to fruit), the count is not binomial and neither its shape nor the approximation follows.
The normal approximation needs np \ge 10 and n(1-p) \ge 10. A large n alone is not enough: with p tiny, the count stays pressed against 0 for any reasonable n.
Even with the condition met, the approximation is only approximate. Example 5 was off by about 0.014, and in a far tail (Example 7) the relative error can be large. When technology can give the exact binomial probability, the exact value is the answer; the normal curve explains it.
A normal curve gives every single value probability 0, while P(X = 48) is about 0.06. To approximate one count, take the area over the bar from 47.5 to 48.5, never the area at a point (Exercise 22).
The approximation is for the count X. The share of blemished fruit, X/n, has its own version, in Sampling distribution of a sample proportion.

Where this goes next

Dividing the count by the crate size turns X into the sample proportion \hat{p} = X/n. The same large-counts condition then makes the distribution of \hat{p} approximately normal with mean p and standard deviation \displaystyle \sqrt{p(1-p)/n}; that is Sampling distribution of a sample proportion, the base of confidence intervals and tests for a proportion. Example 7’s question, how surprising a count is if the claimed p is right, is the shape every one of those tests takes. The normal curve’s appearance here is one case of the The central limit theorem: sums of many independent pieces become normal.

Exercises

Every blemish chance is made up, and grapefruit are independent unless the exercise says otherwise. Give probabilities to four decimal places and standard deviations to two. For a normal area, use the unrounded z.

Part 1: the shape of a binomial distribution

Liga packs a gift box of 8 grapefruit, each blemished with chance 0.12. Find P(X = k) for k = 0, 1, 2, 3, name the most likely count, and describe the shape.
For the same gift box, let Y be the number unblemished. Find P(Y = 8) and P(Y = 7), the mean of Y, and the shape of its distribution.
Compare two binomial counts with n = 20: one with p = 0.12, one with p = 0.3. For each find \mu_X, \sigma_X and \mu_X / \sigma_X. Which is more strongly skewed, and why?
Liga’s gate sends each of 8 grapefruit left or right with chance 0.5. Find the probability that exactly 3 go left and the probability that exactly 5 go left. Why are they equal?
For blemish chance 0.12, find \mu_X, \sigma_X and \mu_X / \sigma_X for a box of 25 and a box of 100. Which distribution is closer to symmetric?
For a tray of 30 grapefruit, blemish chance 0.12, find P(X = 2), P(X = 3) and P(X = 4). Which count is most likely?
Three binomial counts each have n = 15, with p = 0.1, p = 0.5 and p = 0.9. Describe the shape of each without computing, and say which two are mirror images.

Part 2: the large-counts condition

With blemish chance 0.12, check the large-counts condition for crates of 60, 90 and 200 grapefruit.
At a second orchard each grapefruit is graded first class with chance 0.95. For the number graded first class in a box of n, find the smallest n that meets the large-counts condition.
Find the smallest n that meets the condition when p = 0.5.

Part 2: using the normal approximation

A bin holds 250 grapefruit, blemish chance 0.12. Check the condition, find \mu_X and \sigma_X, and approximate P(X \le 25). Technology gives the exact value 0.1920; how far off is the approximation?
For the same bin of 250, approximate P(X \ge 40) and compare with the exact 0.0361.
A truckload holds 1{,}000 grapefruit, blemish chance 0.12. Approximate the probability that the number blemished is between 100 and 140 inclusive.
A crate of 500 grapefruit, blemish chance 0.12. Let Y be the number unblemished. Approximate P(Y \le 430).
Janis flags any crate of 600 whose blemished count is so high that, at chance 0.12, a count that high happens only about 5\% of the time. Using the normal model and z \approx 1.645, find the cutoff count.
A tray of 30 grapefruit, blemish chance 0.12. Find the normal approximation of P(X \le 1) and compare with the exact value 0.1100. Explain the gap using the condition.
Janis finds 50 blemished grapefruit in a crate of 300. Approximate P(X \ge 50) at chance 0.12, and say whether the count is unusual.

Reasoning

Explain, using the wall at 0, why a binomial count with p = 0.12 and n = 10 is skewed right.
Explain why the binomial distributions for p and 1 - p (same n) are mirror images.
Why does the large-counts condition check both np and n(1-p), rather than n alone?
Janis says that with n = 1{,}000{,}000 trials the normal approximation works for any p. Give a value of p that proves him wrong, and explain.
With a crate of 400 and chance 0.12, the normal curve gives P(X = 48) = 0. Explain why, then approximate P(X = 48) by the area from 47.5 to 48.5 and compare with the exact 0.0613.

Answers

P(0) = 0.88^8 \approx 0.3596, P(1) = 8(0.12)(0.88)^7 \approx 0.3923, P(2) \approx 0.1872, P(3) \approx 0.0511. The most likely count is 1; the distribution is skewed right, piled against 0.
Y = 8 - X, so P(Y = 8) = P(X = 0) \approx 0.3596 and P(Y = 7) = P(X = 1) \approx 0.3923. Mean 8(0.88) = 7.04. Skewed left, the mirror image of Exercise 1.
p = 0.12: \mu_X = 2.4, \displaystyle \sigma_X = \sqrt{2.112} \approx 1.45, ratio \approx 1.65. p = 0.3: \mu_X = 6, \displaystyle \sigma_X = \sqrt{4.2} \approx 2.05, ratio \approx 2.93. The p = 0.12 count is more skewed: its mean is closer to the wall at 0 in standard deviations.
Both are \binom{8}{3}(0.5)^8 = \binom{8}{5}(0.5)^8 = \tfrac{56}{256} \approx 0.2188. Three going left is five going right, and with p = 0.5 left and right are equally likely.
Box of 25: \mu_X = 3, \displaystyle \sigma_X = \sqrt{2.64} \approx 1.62, ratio \approx 1.85. Box of 100: \mu_X = 12, \displaystyle \sigma_X = \sqrt{10.56} \approx 3.25, ratio \approx 3.69. The box of 100 is closer to symmetric.
P(2) \approx 0.1747, P(3) \approx 0.2224, P(4) \approx 0.2047. The most likely count is 3.
p = 0.1: skewed right. p = 0.5: symmetric. p = 0.9: skewed left. The p = 0.1 and p = 0.9 distributions are mirror images, since 0.9 = 1 - 0.1.
n = 60: np = 7.2, fails. n = 90: np = 10.8 and n(1-p) = 79.2, holds. n = 200: np = 24 and n(1-p) = 176, holds.
The failures set it: 0.05n \ge 10 gives n \ge 200, and then np = 190. Smallest n = 200.
0.5n \ge 10 gives n = 20.
np = 30, n(1-p) = 220, holds. \mu_X = 30, \displaystyle \sigma_X = \sqrt{26.4} \approx 5.14. \displaystyle z = \frac{25 - 30}{\sqrt{26.4}} \approx -0.97, area \approx 0.1652. Low by about 0.0268.
\displaystyle z = \frac{40 - 30}{\sqrt{26.4}} \approx 1.95, area to the right \approx 0.0258, low by about 0.0103 against the exact 0.0361.
\mu_X = 120, \displaystyle \sigma_X = \sqrt{105.6} \approx 10.28. The z-scores are about -1.95 and 1.95, so the area between is about 0.9484 (exact value by technology: 0.9542).
Y is binomial with p = 0.88: \mu_Y = 440, \displaystyle \sigma_Y = \sqrt{52.8} \approx 7.27, and both counts (440 and 60) are at least 10. \displaystyle z = \frac{430 - 440}{\sqrt{52.8}} \approx -1.38, area \approx 0.0844.
\mu_X = 72, \displaystyle \sigma_X = \sqrt{63.36} \approx 7.96. Cutoff 72 + 1.645(7.96) \approx 85.1, so a crate with more than 85.1 blemished fruit, that is 86 or more, is flagged.
\mu_X = 3.6, \displaystyle \sigma_X = \sqrt{3.168} \approx 1.78, \displaystyle z = \frac{1 - 3.6}{\sqrt{3.168}} \approx -1.46, area \approx 0.0720, low by about 0.0379. The condition fails (np = 3.6 < 10), so the count is still strongly skewed and the curve does not fit.
\mu_X = 36, \displaystyle \sigma_X = \sqrt{31.68} \approx 5.63, \displaystyle z = \frac{50 - 36}{\sqrt{31.68}} \approx 2.49, area to the right \approx 0.0064. Yes: about 6 crates in 1{,}000, so the count is unusual at chance 0.12.
The mean is 1.2, only about one standard deviation (1.03) above 0. The count cannot fall below 0, so the bars pile up there, while above the mean they can run all the way to 10, leaving a long right tail.
A count of k successes with chance p is a count of n - k failures with chance 1 - p, so P(X = k) under p equals P(X = n - k) under 1 - p, which is reflection about n/2. In the formula, \binom{n}{k} = \binom{n}{n-k} and the powers swap.
Each wall needs to be far from the mean in standard deviations. np \ge 10 keeps the wall at 0 more than \displaystyle \sqrt{10} standard deviations away and n(1-p) \ge 10 does the same for the wall at n. A large n with an extreme p can still have the mean next to a wall.
Take p = 0.000001. Then np = 1: about one success expected, the count piled against 0 and strongly skewed right, so the condition fails and the normal curve does not fit.
A normal model is continuous, and the area above a single point is 0. The bar for 48 stands from 47.5 to 48.5; the normal area there is about 0.0613, matching the exact 0.0613 to four places.