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B. Probability Distributions · Continuous Distributions and the Normal Distribution

How Binomial and Hypergeometric Distributions Become Normal

B2.7 B2.2About 70 minutes
Learning goals
  • I can find the mean np and standard deviation √(npq) of a binomial distribution.
  • I can explain why the total area of a probability histogram is 1 and how a bar's area matches an area under a normal curve.
  • I can describe how the shapes of binomial and hypergeometric distributions approach a normal curve as the number of trials grows.
  • I can decide when a normal model is reasonable for a binomial distribution (np ≥ 5 and nq ≥ 5).

Words to know

binomial distribution
The distribution of the number of successes X in n independent trials, each with the same probability of success p. P(X = k) = C(n, k)pᵏqⁿ⁻ᵏ, where q = 1 − p.
hypergeometric distribution
The distribution of the number of successes X when n items are chosen without replacement from N items of which r are successes. P(X = k) = C(r, k)C(N − r, n − k)/C(N, n).
probability histogram
A histogram of a discrete probability distribution, with a bar of width 1 centred on each value k and height P(X = k). Its total area is 1.
standard deviation of a binomial
σ = √(npq), where q = 1 − p.
rule of thumb for normal approximation
A binomial distribution is close enough to normal when np ≥ 5 and nq ≥ 5 (some textbooks use > 5).

Bells from coins

Toss a coin once: two bars, heads or tails. Toss 10 coins and count the heads: the probability histogram already looks like a hill. Toss 100 coins: it looks like a smooth bell. In the unit Discrete Probability Distributions you studied the binomial and hypergeometric distributions as tables and histograms. In this lesson you will see that, when the number of trials is large, they take the shape of the normal distribution. That is what lets us use the z-table for counting problems in the next lesson.

Mean and standard deviation of a binomial

For X ~ Binomial(n, p) with q = 1 − p:

  • Mean (expected value): μ = np. (You met this as E(X) in Discrete Probability Distributions.)
  • Standard deviation: σ = √(npq).

The standard deviation formula comes from the same idea as in the lesson Standard Deviation: the variance is the weighted mean of the squared deviations, Σ(k − μ)²P(X = k). Doing that sum in general gives npq. Example 1 checks it for a small case.

Worked example

Example 1: Checking μ = np and σ = √(npq)

A seedling survives transplanting with probability 0.3 (a hard frost year). Four seedlings are planted. Let X be the number that survive. Use the distribution table to find the mean and variance, and compare with np and npq.

  1. Using P(X = k) = C(4, k)(0.3)ᵏ(0.7)⁴⁻ᵏ: P(0) = 0.2401, P(1) = 0.4116, P(2) = 0.2646, P(3) = 0.0756, P(4) = 0.0081.
  2. Mean: Σk·P = 0(0.2401) + 1(0.4116) + 2(0.2646) + 3(0.0756) + 4(0.0081) = 0.4116 + 0.5292 + 0.2268 + 0.0324 = 1.2.
  3. Σk²·P = 0.4116 + 4(0.2646) + 9(0.0756) + 16(0.0081) = 0.4116 + 1.0584 + 0.6804 + 0.1296 = 2.28.
  4. Variance = Σk²·P − μ² = 2.28 − 1.44 = 0.84. (This is the shortcut Σx²/N − μ² from the lesson Standard Deviation, with probabilities as weights.)
  5. Formulas: np = 4(0.3) = 1.2 and npq = 4(0.3)(0.7) = 0.84. They match.

μ = 1.2, σ² = 0.84, σ = √0.84 ≈ 0.917. The formulas np and √(npq) agree with the table.

The area of a probability histogram is 1

In a probability histogram, each value k gets a bar from k − 0.5 to k + 0.5. The width is 1 and the height is P(X = k), so

area of the bar = 1 × P(X = k) = P(X = k).

The total area is the sum of all the probabilities, which is 1. That is exactly like a density curve. So if a normal curve with the same mean and standard deviation hugs the tops of the bars, the area under the curve from k − 0.5 to k + 0.5 approximates P(X = k), and the total areas (both 1) match.

0123400.20.4n = 4024681000.10.20.3n = 10010203000.050.10.15n = 30number of successes, p = 0.3
Binomial probability histograms with p = 0.3. With n = 4 the shape is lopsided (skewed right). With n = 10 it is less skewed. With n = 30 it is nearly symmetric, and the normal curve with μ = np = 9 and σ = √6.3 ≈ 2.51 fits the bars closely.
Worked example

Example 2: One bar versus the normal curve

X ~ Binomial(30, 0.3). Compare P(X = 9) with the area under the normal curve N(9, 6.3) between 8.5 and 9.5.

  1. Exact: P(X = 9) = C(30, 9)(0.3)⁹(0.7)²¹ ≈ 0.1573.
  2. Normal: μ = np = 9, σ = √(30 × 0.3 × 0.7) = √6.3 ≈ 2.5100.
  3. z-scores: (8.5 − 9)/2.51 ≈ −0.20 and (9.5 − 9)/2.51 ≈ 0.20.
  4. Area: P(Z < 0.20) − P(Z < −0.20) = 0.5793 − 0.4207 = 0.1586.

Exact 0.1573, normal estimate 0.1586. Very close: the bar from 8.5 to 9.5 and the area under the curve over the same interval are almost the same.

zP(Z < z)P(Z < −z) = 1 − P(Z < z)
0.200.57930.4207
z-table values used in this lesson (taken from the full standard normal table in the lesson Using the z-Table to Find Normal Probabilities). The middle column is the area to the left of z; the last column is the area to the left of −z.

When is the bell shape good enough?

  • If p = 0.5, the binomial is symmetric for every n and looks normal quickly.
  • If p is near 0 or 1, the distribution is squashed against one end and skewed; it needs a larger n before it looks normal.
  • A widely used rule of thumb: the normal model is reasonable when np ≥ 5 and nq ≥ 5. (Some textbooks write np > 5 and nq > 5; both are guidelines, not laws.) This makes sure the mean is far enough from 0 and from n for the curve to fit.

The same thing happens to the hypergeometric distribution. Its mean and variance are

μ = n(r/N) and σ² = n(r/N)(1 − r/N)·(N − n)/(N − 1).

When the population N and sample n are large, its histogram becomes bell-shaped too. And when the sample is a small part of the population (say n is less than about 5% of N), choosing without replacement hardly changes the probabilities, so the hypergeometric is close to a binomial with p = r/N.

0123400.250.54 chosen from 4 men, 6 women02468101200.150.312 chosen from 16 men, 24 womennumber of men on the committee
Hypergeometric distributions for the number of men on a committee. A 4-person committee from 10 people gives a rough, lopsided histogram. A 12-person committee from 40 people (still 40% men) gives a much more bell-shaped histogram centred at 12(16/40) = 4.8.
Worked example

Example 3: Checking the rule of thumb

For each binomial distribution, decide whether a normal model is reasonable, and give μ and σ.

  1. n = 20, p = 0.5
  2. n = 40, p = 0.05
  3. n = 200, p = 0.04
  1. 1. np = 10, nq = 10: both ≥ 5. Reasonable. μ = 10, σ = √(20 × 0.5 × 0.5) = √5 ≈ 2.24.
  2. 2. np = 2 < 5. Not reasonable: the distribution is piled up near 0 and skewed right. (μ = 2, σ = √1.9 ≈ 1.38.)
  3. 3. np = 8, nq = 192: both ≥ 5. Reasonable. μ = 8, σ = √(200 × 0.04 × 0.96) = √7.68 ≈ 2.77.

Normal is reasonable for 1 and 3, not for 2.

Worked example

Example 4: Mean and SD of a hypergeometric

A committee of 12 is chosen at random from 16 men and 24 women (N = 40). Let X be the number of men. Find μ and σ, and say whether a normal model is reasonable.

  1. r/N = 16/40 = 0.4. μ = 12(0.4) = 4.8.
  2. σ² = 12(0.4)(0.6)(40 − 12)/(40 − 1) = 2.88 × 28/39 ≈ 2.0677, so σ ≈ 1.438.
  3. n(r/N) = 4.8 and n(1 − r/N) = 7.2. The first is just under 5, and the histogram above is fairly bell-shaped but still a little lumpy.

μ = 4.8, σ ≈ 1.44. A normal model is borderline: it gives rough estimates, but for accurate answers use the hypergeometric formula directly.

Just for fun

In the 1800s Francis Galton built a board of pins where little balls bounce left or right at each pin. Each ball makes a string of yes-no choices, which is a binomial experiment, and the balls pile up at the bottom in the shape of a bell. Toy versions are still sold as "Galton boards".

Hands-on activity

Ten coins, fifty times

You need: 10 coins (or a spreadsheet: =BINOM.INV(10, 0.5, RAND()) gives the number of heads in 10 tosses), paper and a pencil.

  1. Toss all 10 coins and count the heads. Repeat 50 times.
  2. Make a frequency table for 0 to 10 heads and draw a histogram with bars of width 1 centred on each value.
  3. On the same axes, plot the expected frequencies 50 × P(X = k) using the binomial probabilities 0.0010, 0.0098, 0.0439, 0.1172, 0.2051, 0.2461, 0.2051, 0.1172, 0.0439, 0.0098, 0.0010 (for k = 0 to 10).
  4. Sketch a normal curve with μ = 5 and σ = √2.5 ≈ 1.58 over your histogram.

What to look for: a hill centred near 5, roughly symmetric, with very few results of 0, 1, 9 or 10 heads. Your 50 results will be bumpier than the expected frequencies; with a whole class's results (or 1000 spreadsheet trials) they would be much closer.

Your turn

1

X ~ Binomial(50, 0.2). Find μ and σ.

2

Is a normal model reasonable for Binomial(50, 0.2)? Explain.

3

Is a normal model reasonable for Binomial(30, 0.1)? Explain.

4

What is the total area of any probability histogram? Why?

5

Which bar interval of a probability histogram corresponds to P(X = 12)?

6

Find μ and σ for the number of hearts in a 13-card hand dealt from a standard deck (hypergeometric: N = 52, r = 13, n = 13).

7

Why does Binomial(n, 0.5) look symmetric even for small n?

More practice: warm-up

1

Write the formula for the mean of a binomial distribution.

2

Write the formula for the standard deviation of a binomial distribution.

3

X ~ Binomial(100, 0.5). Find μ and σ.

4

In a probability histogram, how wide is each bar?

5

State the rule of thumb for using a normal model for a binomial.

More practice: on your own

1

A multiple-choice test has 40 questions with 4 choices each. A student guesses every answer. Find μ and σ for the number correct, and decide whether a normal model is reasonable.

2

What is the smallest n for which Binomial(n, 0.1) satisfies np ≥ 5?

3

For Binomial(4, 0.3) (Example 1), what is the area of the bar for k = 1? What is the area of all bars together?

4

A box of 200 bulbs has 30 defective ones. You test 10 without replacement. Is a binomial with p = 0.15 a good approximation? Explain.

5

Explain in your own words why the binomial distribution with large n looks like a normal curve.

Challenge

1

Show that for a single trial (n = 1), the variance of the number of successes is pq. Then explain why n independent trials give npq.

2

For Binomial(30, 0.3), use the empirical rule to estimate the interval that contains about 95% of outcomes. Compare with the exact probability that 4 ≤ X ≤ 14, which is about 0.97.

3

For a hypergeometric distribution, the factor (N − n)/(N − 1) is called the finite population correction. What is it when N = 1000 and n = 20? What does that tell you?

Check your understanding

Quiz

1.

X ~ Binomial(80, 0.25). What is σ?

2.

For which binomial distribution is a normal model reasonable?

3.

In a probability histogram, what is the area of the bar for X = k?

4.

As n increases with p fixed at 0.3, the binomial probability histogram...

5.

Which interval of the normal curve best matches the binomial probability P(X = 20)?

6.

A hypergeometric distribution has N = 50, r = 20 and n = 10. What is its mean?

For parents and teachers

Big idea (B2.7): binomial (and hypergeometric) distributions look more and more like a normal curve as the number of trials grows. Because each probability-histogram bar has width 1 and area P(X = k), the area under a matching normal curve N(np, npq) over the same interval approximates that probability.

How to help: The 10-coin activity is quick and convincing. Ask your teen to predict the centre (5) and spread (about 1.6) before tossing. Then ask: "Why would p = 0.05 with only 20 trials not look like a bell?"

Common difficulties:

  • Using npq as the standard deviation (it is the variance; σ = √(npq)).
  • Forgetting the rule of thumb np ≥ 5 and nq ≥ 5.
  • Thinking the binomial becomes continuous. It stays discrete; the normal curve is only an approximation.