Your weak spots

Insights load after your first practice round.

Module 10 · L11 of 20 ~40 min ⚡ +90 XP available

Properties of Binomial Distributions

When a quality inspector tests 20 components from a production line that historically rejects 50% of items, the histogram of expected rejections is perfectly symmetric around 10. If the line improves to a 20% reject rate, that same histogram suddenly leans left — most outcomes cluster near 4 with a long right tail. The shape of a binomial distribution carries hidden information about $p$, and recognising it on sight is a Band 5 skill.

Today's hook — Without computing any probability, sketch what you expect the histogram of $X \sim B(10, 0.2)$ to look like. Where is the peak? Which tail is longer? Now repeat for $X \sim B(10, 0.8)$ and $X \sim B(10, 0.5)$. Predict before you read on — the answers reveal themselves in card 05.

0/5QUESTS

Recall — your gut answer first

+5 XP warm-up

A fair coin is tossed 6 times. Let $X$ count the heads. Without computing any probabilities — would you expect $P(X=2)$ and $P(X=4)$ to be equal, or different? Write down your reasoning. What about $P(X=1)$ and $P(X=5)$?

auto-saved

The two moves for shape analysis

+5 XP to read

When you see a binomial $B(n, p)$, two habits diagnose its shape instantly: check $p$ against $0.5$ (symmetric or skewed?), then locate the mean $np$ (that's where the peak lives). Without these two checks students often mis-sketch the histogram and lose method marks on Band 4 shape questions.

The shape diagnostic: (1) is $p = 0.5$? If yes, perfect symmetry around $n/2$. (2) Is $p < 0.5$? Right-skew — long right tail, peak left of centre. (3) Is $p > 0.5$? Left-skew — long left tail, peak right of centre.

Mean: $\mu = np$ · Variance: $\sigma^2 = np(1-p)$ · Std dev: $\sigma = \sqrt{np(1-p)}$

$\mu = np \,,\;\; \sigma^2 = np(1-p)$

$p = 0.5$ means symmetric

The pairing $P(X=k) = P(X=n-k)$ holds only when $p = 0.5$. The peak sits exactly at $n/2$ (or split between $(n-1)/2$ and $(n+1)/2$ for odd $n$).

Skew points to the tail

"Right-skewed" means the long tail is on the right (high values). For $p < 0.5$ failures are rare, so the long tail of high counts is on the right — hence right-skew.

Larger $n$ smooths shape

As $n$ grows, even a skewed binomial approaches a bell curve. By $n = 30$ with moderate $p$, the histogram is visually almost normal. This is the foundation of the normal approximation in later lessons.

What you'll master

Know

Key facts

$B(n, 0.5)$ is symmetric: $P(X = k) = P(X = n - k)$ for all $k$
$B(n, p)$ with $p < 0.5$ is right-skewed; with $p > 0.5$ is left-skewed
Mean $\mu = np$; variance $\sigma^2 = np(1-p)$

Understand

Concepts

Why $p = 0.5$ produces symmetry from the structure of $\binom{n}{k}p^k(1-p)^{n-k}$
Why the skew direction is opposite to where you'd naively expect
How increasing $n$ smooths a skewed distribution toward a bell shape

Can do

Skills

Sketch the shape of $B(n, p)$ at a glance given $n$ and $p$
Identify symmetry or skew from a histogram and deduce $p$
Compute mean and variance and use them to mark expected peak location

Key terms

Binomial distribution$X \sim B(n, p)$ counts the number of successes in $n$ independent trials, each with success probability $p$. PMF: $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$.

Symmetric distributionA distribution where $P(X = k) = P(X = n - k)$ for all $k$. Only occurs in binomial when $p = 0.5$. Visually, the histogram is a mirror image about the centre.

Right-skewed (positive skew)Long tail extends to the right (high values). Occurs in $B(n, p)$ when $p < 0.5$ — successes are rare, so most counts cluster low with a tail of unusually high counts.

Left-skewed (negative skew)Long tail extends to the left (low values). Occurs in $B(n, p)$ when $p > 0.5$ — successes are common, so most counts cluster high with a tail of unusually low counts.

Mean $\mu = np$The expected number of successes in $n$ trials. The histogram peak sits at or beside $\mu$. For $B(20, 0.3)$, $\mu = 6$.

ME12-5NESA outcome: applies appropriate statistical processes to present, analyse and interpret data, including from binomial distributions.

Symmetry when $p = 0.5$

core concept

The binomial PMF is $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$. When $p = 0.5$, we have $p = 1 - p$, so:

$$P(X = k) = \binom{n}{k}(0.5)^k(0.5)^{n-k} = \binom{n}{k}(0.5)^n.$$

Now use the symmetry of binomial coefficients $\binom{n}{k} = \binom{n}{n-k}$:

$$P(X = k) = \binom{n}{n-k}(0.5)^n = P(X = n - k).$$

So the histogram is a perfect mirror about $n/2$. Worked through the hook: For 6 coin tosses, $P(X = 2) = P(X = 4)$ and $P(X = 1) = P(X = 5)$ — exactly as your gut said.

$B(10, 0.2)$: peak near $k = 2$ (since $np = 2$), short left tail, long right tail — right-skewed.
$B(10, 0.8)$: peak near $k = 8$, long left tail, short right tail — left-skewed.
$B(10, 0.5)$: peak at $k = 5$, perfectly symmetric about 5.

Why $p = 0.5$ is special. Only at $p = 0.5$ does success have the same probability as failure, so swapping "success" and "failure" leaves the distribution unchanged. This swap operation maps $k \to n-k$, which is exactly the symmetry $P(X = k) = P(X = n - k)$.

The binomial PMF is $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$. When $p = 0.5$, we have $p = 1 - p$, so:

Pause — copy the symmetry proof: $P(X=k)=\binom{n}{k}(0.5)^n=\binom{n}{n-k}(0.5)^n=P(X=n-k)$ for $p=0.5$ into your book.

Quick check: $X \sim B(8, 0.5)$. Which of the following equals $P(X = 3)$?

Skewness when $p \neq 0.5$

core concept

We just saw that when $p=0.5$, $P(X=k)=\binom{n}{k}(0.5)^n$ and symmetry of binomial coefficients $\binom{n}{k}=\binom{n}{n-k}$ gives $P(X=k)=P(X=n-k)$, producing a perfectly symmetric histogram. That raises a question: how does the symmetry break when $p\neq0.5$, and which direction does the skew go? This card answers it → if $p<0.5$, most mass sits near small $k$ (right-skewed); if $p>0.5$, mass sits near large $k$ (left-skewed).

When $p \neq 0.5$, the probabilities $p^k$ and $(1-p)^{n-k}$ change at different rates as $k$ grows, breaking the symmetry. The result is a skewed distribution.

Rule of thumb:

If $p < 0.5$, the distribution is right-skewed (positive skew). Most probability mass sits near small $k$, with a long tail of small probabilities at large $k$.
If $p > 0.5$, the distribution is left-skewed (negative skew). Most mass sits near large $k$, with a long tail at small $k$.

Worked example: $X \sim B(10, 0.2)$. Mean $\mu = 10 \times 0.2 = 2$. Computed values:

$P(X = 0) \approx 0.107$
$P(X = 1) \approx 0.268$
$P(X = 2) \approx 0.302$ (peak)
$P(X = 3) \approx 0.201$
$P(X = 4) \approx 0.088$
$P(X = 5) \approx 0.026$
$P(X = 6) \approx 0.0055$

The peak is at $k = 2$, and the tail extends much further to the right than to the left (which is cut off at 0). Classic right-skew.

$$\mu = np \,,\quad \sigma^2 = np(1-p) \,,\quad \sigma = \sqrt{np(1-p)}$$

Common mistake. Students often confuse the direction of skew. Remember: "skewed" describes the tail, not the peak. A right-skewed distribution has its peak on the left and its tail on the right. The naming follows the tail.

When $p \neq 0.5$, the probabilities $p^k$ and $(1-p)^{n-k}$ change at different rates as $k$ grows, breaking the symmetry. The result is a skewed distribution.

Pause — copy the skewness rule: $p<0.5\Rightarrow$ right-skewed (positive skew); $p>0.5\Rightarrow$ left-skewed (negative skew); $p=0.5\Rightarrow$ symmetric into your book.

Did you get this? True or false: for $X \sim B(15, 0.7)$, the histogram is left-skewed with the peak near $k = 10$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · IDENTIFY SHAPE FROM (n, p)

Describe the shape of each distribution, locating the approximate peak and tail direction: (a) $X \sim B(12, 0.5)$, (b) $Y \sim B(20, 0.15)$, (c) $W \sim B(25, 0.9)$.

(a) $X \sim B(12, 0.5)$: $p = 0.5$ so the distribution is symmetric. Mean $\mu = 12 \times 0.5 = 6$. Peak at $k = 6$, mirror symmetry about 6.

When $p = 0.5$, no further work is needed — the shape is determined. Pair $P(X=k)$ with $P(X=12-k)$.

PROBLEM 2 · USE SYMMETRY TO FIND A PROBABILITY

$X \sim B(10, 0.5)$ and $P(X = 3) = 0.1172$. Without further computation, state $P(X = 7)$ and explain the principle used.

Since $p = 0.5$, the distribution is symmetric: $P(X = k) = P(X = n - k) = P(X = 10 - k)$.

Symmetry is the key. State it first, then apply.

PROBLEM 3 · SHAPE CHANGE AS n GROWS

A factory's defect rate is $p = 0.25$. Compare the shape of $B(8, 0.25)$ and $B(40, 0.25)$. Describe how the shape evolves as $n$ increases, computing the mean and standard deviation in each case.

$B(8, 0.25)$: $\mu = 8 \times 0.25 = 2$, $\sigma = \sqrt{8 \times 0.25 \times 0.75} = \sqrt{1.5} \approx 1.22$. With $p < 0.5$, the distribution is right-skewed.

Small $n$ makes the right-skew very visible — peak at $k = 2$, with the left tail cut off at 0.

Fill the gap: If $X \sim B(20, 0.3)$, then $\mu = np = $ and $\sigma^2 = np(1-p) = 4.2$. The distribution is right-skewed because $p < 0.5$.

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Reversing skew direction

Many students say "$p < 0.5$ gives left-skew" because they associate the peak's location (on the left) with the skew name. The skew name is defined by where the long tail sits, not the peak. $p < 0.5$ has peak left, tail right — right-skewed (positive skew).

Trap 02

Assuming symmetry for any $p$

It is tempting to write $P(X = k) = P(X = n - k)$ for any $X \sim B(n, p)$. This is true only when $p = 0.5$. For $B(10, 0.3)$, $P(X = 2) \approx 0.234$ but $P(X = 8) \approx 0.0014$ — not even close. Always check $p$ before invoking symmetry.

Trap 03

Confusing $\sigma^2 = np(1-p)$ with $\mu(1-p)$

The variance formula has the full product $np(1-p)$, often written as $\mu(1-p)$. Be careful to not write $\sigma^2 = np$ (that's the mean) or $\sigma = np(1-p)$ (that's the variance, not the standard deviation). Always check units: $\sigma$ has the same units as the count, $\sigma^2$ is in squared count.

Did you get this? True or false: a binomial distribution with $n = 30$ and $p = 0.4$ is symmetric because $n$ is large.

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Describe the shape of $X \sim B(16, 0.5)$. State whether it is symmetric or skewed, give the mean, and identify where the peak sits.

For $X \sim B(20, 0.1)$, compute $\mu$ and $\sigma$. Describe the shape and identify the approximate peak.

$X \sim B(12, 0.5)$ and $P(X = 4) = 0.1208$. Without further computation, find $P(X = 8)$ and justify.

Compare the shape of $B(10, 0.3)$ and $B(50, 0.3)$. Compute $\mu$ and $\sigma$ in each case, and describe how skew changes as $n$ grows.

A histogram for $X \sim B(n, p)$ has its peak at $k = 15$ and is left-skewed. Given $n = 18$, estimate $p$ and justify your choice.

Odd one out: Three of these statements about $X \sim B(20, 0.5)$ are correct. Which one is NOT?

Revisit your thinking

Earlier you sketched $B(10, 0.2)$, $B(10, 0.8)$ and $B(10, 0.5)$ from intuition.

$B(10, 0.5)$ is symmetric with peak at 5. $B(10, 0.2)$ has peak at 2 with a long right tail (right-skewed). $B(10, 0.8)$ is its mirror image: peak at 8 with a long left tail (left-skewed). Notice that $B(10, 0.2)$ and $B(10, 0.8)$ are exactly mirror images of each other — a consequence of swapping "success" and "failure".

auto-saved

Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 32 marks

Q1. $X \sim B(14, 0.5)$ and $P(X = 5) = 0.1222$. Use symmetry to state $P(X = 9)$ and explain your reasoning. (2 marks)

auto-saved

ApplyBand 43 marks

Q2. $Y \sim B(25, 0.4)$. Compute $\mu$ and $\sigma$. State the shape (symmetric or skewed direction) and identify the approximate location of the peak. (3 marks)

auto-saved

AnalyseBand 53 marks

Q3. A histogram for $X \sim B(n, p)$ shows perfect symmetry about $k = 12$, with peak value $P(X = 12) \approx 0.176$. Use these properties to deduce $n$ and $p$, and verify by computing the variance. (3 marks)

auto-saved

Comprehensive answers (click to reveal)

Activity answers:

1. $p = 0.5$ so symmetric. $\mu = 16 \times 0.5 = 8$. Peak at $k = 8$. Histogram is a mirror image about 8.

2. $\mu = 20 \times 0.1 = 2$, $\sigma = \sqrt{20 \times 0.1 \times 0.9} = \sqrt{1.8} \approx 1.34$. Right-skewed (since $p < 0.5$), peak at $k = 2$.

3. By symmetry ($p = 0.5$): $P(X = 8) = P(X = 12 - 8) = P(X = 4) = 0.1208$.

4. $B(10, 0.3)$: $\mu = 3$, $\sigma = \sqrt{2.1} \approx 1.45$. $B(50, 0.3)$: $\mu = 15$, $\sigma = \sqrt{10.5} \approx 3.24$. Both right-skewed (same $p < 0.5$) but $B(50, 0.3)$ is visibly more bell-shaped because skew shrinks like $1/\sqrt{n}$.

5. Left-skewed implies $p > 0.5$. Peak at $np = 15$ with $n = 18$ gives $p = 15/18 \approx 0.833$.

Q1 (2 marks): $p = 0.5 \Rightarrow P(X = k) = P(X = n - k)$ [1]. So $P(X = 9) = P(X = 14 - 9) = P(X = 5) = 0.1222$ [1].

Q2 (3 marks): $\mu = 25 \times 0.4 = 10$ [1]. $\sigma^2 = 25 \times 0.4 \times 0.6 = 6$, $\sigma = \sqrt 6 \approx 2.45$ [1]. $p < 0.5$ so right-skewed, peak near $k = 10$ [1].

Q3 (3 marks): Symmetry $\Rightarrow p = 0.5$ [1]. Peak at $np = 12 \Rightarrow n = 24$ [1]. Verify: $\sigma^2 = 24 \times 0.5 \times 0.5 = 6$; the peak height $\binom{24}{12}(0.5)^{24} \approx 0.176$ confirms [1].

Boss battle · The Shape Sleuth

earn bronze · silver · gold

Five timed questions on binomial shape, symmetry, skew and mean/variance. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering binomial shape questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 10 · Binomial Distribution Lesson 12 · Mode of a Binomial Distribution →

Module overview · Extension 1 · Checkpoint 3