B. Probability Distributions · Discrete Probability Distributions

The Binomial Distribution: Tables, Graphs and Expected Value

B1.4 B1.2 B1.6About 75 minutes

Learning goals

I can build the full probability distribution of a binomial random variable as a table and a histogram.
I can find cumulative probabilities such as P(X ≤ 2), P(X ≥ 3) and P(2 ≤ X ≤ 5).
I can use E(X) = np and describe the spread with σ = √(npq).
I can describe how the shape of a binomial distribution depends on n and p.

Words to know

cumulative probability
The probability that X is at most some value, P(X ≤ k): the sum of the probabilities from the smallest value up to k.

E(X) = np
The expected value (mean) of a binomial random variable X ~ B(n, p).

standard deviation σ
A measure of spread. For a binomial distribution σ = √(npq). Most of the probability lies within about 2σ of the mean.

BINOM.DIST
A spreadsheet function (Excel and Google Sheets). BINOM.DIST(k, n, p, FALSE) gives P(X = k); with TRUE it gives P(X ≤ k).

The whole distribution

In the last lesson you found single binomial probabilities. Now we build the whole distribution: every value k = 0, 1, ..., n with its probability. With it you can answer any question about X, draw its histogram and find its centre.

Let's start with the light bulb problem from the curriculum.

Worked example

Example 1: Defective light bulbs

A factory's light bulbs are defective 0.5% of the time, independently. A package holds 4 bulbs. Let X be the number of defective bulbs in a package. Build the probability distribution of X and describe its histogram.

X ~ B(4, 0.005), so q = 0.995.
P(X = 0) = 0.995⁴ ≈ 0.980 150.
P(X = 1) = C(4, 1)(0.005)(0.995)³ = 4 × 0.005 × 0.985 075 ≈ 0.019 701.
P(X = 2) = C(4, 2)(0.005)²(0.995)² = 6 × 0.000 025 × 0.990 025 ≈ 0.000 149.
P(X = 3) = 4(0.005)³(0.995) ≈ 0.000 000 5 and P(X = 4) = 0.005⁴ ≈ 0.000 000 000 6.
Check: the five probabilities add to 1 (up to rounding). ✓

About 98% of packages have no defective bulb and about 2% have exactly one. The histogram is one tall bar at x = 0, a short bar at 1, and bars at 2, 3 and 4 too small to see: extremely skewed right.

x	0	1	2	3	4
P(X = x)	0.9801495	0.0197015	0.0001485	0.0000004975	0.0000000006

The distribution of defective bulbs in a 4-pack, X ~ B(4, 0.005), to 7 decimal places (10 for x = 4).

Expected value of a binomial distribution

If a player makes 70% of free throws and shoots 10, you would expect about 7 makes. In general:

E(X) = np

Why? Each trial adds 1 success with probability p and 0 with probability q, so each trial contributes an expected p successes. Over n trials that adds up to np. (You could also prove it from Σ k·C(n, k)pᵏqⁿ⁻ᵏ with some algebra.)

For the bulbs, E(X) = 4 × 0.005 = 0.02 defective bulbs per package: in 1000 packages, expect about 20 defective bulbs.

Spread

The standard deviation measures how far values typically fall from the mean. For a binomial distribution it is

σ = √(npq).

You will compute standard deviations of data and of the normal distribution in Continuous Distributions and the Normal Distribution; for now, use it as a ruler: most of the probability lies between np − 2σ and np + 2σ.

X ~ B(8, 0.3). The mean np = 2.4 is the balance point. The mode is 2. The bars at 6, 7 and 8 are tiny, but those values are still possible.

Worked example

Example 2: Cumulative probabilities

A student takes the bus 8 times; each time, independently, there is a 30% chance it is late. Let X be the number of late buses. Find (a) P(X ≤ 2), (b) P(X ≥ 3), (c) E(X) and σ.

X ~ B(8, 0.3). From the histogram (or the formula): P(0) ≈ 0.0576, P(1) ≈ 0.1977, P(2) ≈ 0.2965.
(a) P(X ≤ 2) = 0.0576 + 0.1977 + 0.2965 ≈ 0.5518.
(b) P(X ≥ 3) = 1 − P(X ≤ 2) ≈ 1 − 0.5518 = 0.4482. (Using the complement avoids adding 6 terms.)
(c) E(X) = 8 × 0.3 = 2.4. σ = √(8 × 0.3 × 0.7) = √1.68 ≈ 1.30.

(a) ≈ 0.552; (b) ≈ 0.448; (c) E(X) = 2.4 late buses, σ ≈ 1.30

Translate the words first

"At most 2" = X ≤ 2. "At least 3" = X ≥ 3 = 1 − P(X ≤ 2). "More than 3" = X ≥ 4. "Fewer than 3" = X ≤ 2. "Between 2 and 5 inclusive" = 2 ≤ X ≤ 5. Write the list of values before you add.

How n and p shape the distribution

If p = 0.5, the distribution is symmetric about n/2.
If p < 0.5, it is skewed right (most of the probability is at small values).
If p > 0.5, it is skewed left.
As n increases (with p fixed), the histogram becomes more symmetric and bell-shaped, and it spreads out. Lesson 9 explores this, and the normal distribution in Continuous Distributions and the Normal Distribution builds on it.

Ten trials with three different values of p. B(10, 0.2) and B(10, 0.8) are mirror images of each other; B(10, 0.5) is symmetric about 5. In each case the mean is np: 2, 5 and 8.

Worked example

Example 3: Using technology

A spreadsheet can produce a binomial table quickly. Show how to get the distribution of X ~ B(20, 0.15) and use it to find P(X ≤ 2) and P(X ≥ 5).

In cells A1 to A21, enter 0 to 20. In B1 enter =BINOM.DIST(A1, 20, 0.15, FALSE) and fill down: this gives P(X = k).
In C1 enter =BINOM.DIST(A1, 20, 0.15, TRUE) and fill down: this gives P(X ≤ k).
P(X ≤ 2) is the TRUE value for k = 2: about 0.0388 + 0.1368 + 0.2293 = 0.4049.
P(X ≥ 5) = 1 − P(X ≤ 4). P(X ≤ 4) ≈ 0.0388 + 0.1368 + 0.2293 + 0.2428 + 0.1821 = 0.8298, so P(X ≥ 5) ≈ 0.1702.

P(X ≤ 2) ≈ 0.405, P(X ≥ 5) ≈ 0.170. By hand you would need the formula 5 times; the spreadsheet (or a graphing calculator's binompdf and binomcdf) does all of them at once.

Worked example

Example 4: Is the result surprising?

A coin is tossed 100 times. Find E(X) and σ for the number of heads. Would 62 heads be surprising?

X ~ B(100, 0.5). E(X) = 50. σ = √(100 × 0.5 × 0.5) = √25 = 5.
62 is 12 above the mean, which is 12/5 = 2.4 standard deviations.
Most results fall within 2σ of the mean (40 to 60). 62 is outside that range.

E(X) = 50, σ = 5. 62 heads is unusual (more than 2σ above the mean) and might make you question whether the coin is fair, though it does happen by chance now and then.

Just for fun

Why is B(10, 0.5) so easy to get along with?

It always sees both sides equally: it is perfectly symmetric.

Hands-on activity

Paper-clip spinner: B(5, 0.25)

You need: a paper clip, a pencil, paper with a circle divided into 4 equal sectors (one sector shaded), and a calculator.

Put the pencil tip through the paper clip at the centre of the circle and flick the clip. Landing in the shaded sector is a success (p = 0.25).
Spin 5 times and record X = number of successes. Repeat for 40 sets of 5 spins.
Use the formula to make the theoretical table for B(5, 0.25): P(0) ≈ 0.237, P(1) ≈ 0.396, P(2) ≈ 0.264, P(3) ≈ 0.088, P(4) ≈ 0.015, P(5) ≈ 0.001.
Draw both histograms (relative frequency and probability) on the same axes.
Find the mean of your 40 values of X and compare with np = 1.25.

What to look for: both histograms should be skewed right with mode 1. Your mean should be roughly between 0.9 and 1.6. If your spinner sticks or the sectors are uneven, the results may drift; that is a good discussion about whether the model's conditions really hold.

Your turn

X ~ B(5, 0.4). Build the distribution table.

For X ~ B(5, 0.4), find P(X ≤ 1) and E(X).

X ~ B(6, 0.5). Find P(2 ≤ X ≤ 4).

X ~ B(50, 0.1). Find E(X) and σ.

Describe the shape of B(12, 0.9).

X ~ B(12, 0.25). Find P(X ≥ 3) using the complement. (P(0) ≈ 0.0317, P(1) ≈ 0.1267, P(2) ≈ 0.2323.)

Write the spreadsheet formula for P(X ≤ 7) when X ~ B(30, 0.2).

More practice: warm-up

X ~ B(20, 0.5). Find E(X).

X ~ B(10, 0.3). Find E(X).

What does "at most 3" mean in symbols?

What does "at least 2" equal in terms of a complement?

Is B(10, 0.5) symmetric?

More practice: on your own

About 12% of a large maple seedling batch fail to sprout. You plant 25. Find the expected number that fail and σ.

X ~ B(10, 0.2). Find P(2 ≤ X ≤ 4). (P(2) ≈ 0.3020, P(3) ≈ 0.2013, P(4) ≈ 0.0881.)

A binomial distribution has mean 6 and n = 15. Find p and σ.

In 400 tosses of a coin, between what two numbers of heads do about 95% of results fall (use mean ± 2σ)?

Why do B(10, 0.2) and B(10, 0.8) have mirror-image histograms?

Challenge

A binomial distribution has E(X) = 12 and σ = 3. Find n and p.

A multiple-choice test has 40 questions with 4 choices each. A student who guesses on every question gets 18 right. Is that surprising? Use mean and σ.

Show that for X ~ B(n, p), the ratio P(X = k + 1)/P(X = k) = [(n − k)/(k + 1)](p/q). Use it to find the mode of B(8, 0.3).

Check your understanding

Quiz

X ~ B(40, 0.25). What is E(X)?

X ~ B(100, 0.5). What is σ?

Which expression equals P(X ≥ 4) for X ~ B(10, p)?

How is B(15, 0.1) shaped?

A package of 4 bulbs, each defective with probability 0.005. What is P(no defective bulbs)?

In a spreadsheet, what does =BINOM.DIST(3, 12, 0.4, TRUE) give?

For parents and teachers

Big idea (B1.4, B1.6): the binomial distribution lists P(X = k) for every k from 0 to n. Its expected value is np and its standard deviation is √(npq). Its shape depends on p (symmetric at 0.5, skewed otherwise) and becomes more bell-shaped as n grows.

How to help: ask quick estimation questions: "If 30% of buses are late and you ride 20 times, about how many late buses?" (6.) Ask your teen to translate phrases like "at least" and "fewer than" into inequalities before calculating.

Common difficulties:

Off-by-one errors with "more than" and "at least".
Adding many terms instead of using the complement.
Rounding each term too early, so the table doesn't add to 1.
Mixing up FALSE (single value) and TRUE (cumulative) in BINOM.DIST.