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hscscience Ext 1 · Y12
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Module 10 · L10 of 20 ~40 min ⚡ +90 XP available

Expected Frequencies

A quality controller knows from experience that $2\%$ of microchips fail testing. If $5000$ chips are produced today, how many are expected to fail? This question — about expected frequency over many trials — is the bridge between probability theory and real-world planning. The rule is wonderfully simple: expected count = total trials × event probability. This lesson sharpens the calculation and the interpretation.

Today's hook — A fair die is rolled $300$ times. Before reading on, predict the expected frequency of (a) rolling a six and (b) rolling an even number. Then think: if you actually rolled the die $300$ times, would you expect to see exactly that count? Compare your answers after card 05.
0/5QUESTS
01
Recall — your gut answer first
+5 XP warm-up

You already know that for a single trial with success probability $p$, the expected proportion of successes is $p$. Without using any formula — if you run the same experiment $m$ times, how many successes do you expect on average? Write your reasoning below.

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02
The two moves for frequency questions
+5 XP to read

Every expected-frequency question rewards two habits: identify the single-trial probability $p$ and the number of trials $m$, then multiply: expected frequency $= m \times p$. Treat each scenario as a chain of independent Bernoulli trials.

The identify-then-multiply strategy: (1) state $p$, the probability of the event on one trial, (2) state $m$, the number of independent repetitions, (3) compute $E = m \times p$ and round only at the end if a whole-number answer is required.

For $X \sim B(n, p)$ run $m$ trials: expected count $= m \times p$

Identify p (trial) Count m (trials) Multiply E = m·p Interpret as long-run average, not guarantee
$E = m \times p$
$p$ is per single trial
Always extract the probability of the event on one trial. If $p$ is given as a percentage, convert to a decimal before multiplying.
Don't round mid-calculation
Keep $p$ as a fraction or full decimal until the end. Rounding $p$ early can introduce errors of several units in the final expected count.
Expected ≠ certain
The expected frequency is a long-run average. An actual sample will usually differ from it by an amount comparable to $\sqrt{npq}$ (the standard deviation).
03
What you'll master
Know

Key facts

  • Expected frequency: $E = m \times p$ where $m$ = number of trials and $p$ = single-trial probability
  • For $X \sim B(n, p)$, $E(X) = np$ is itself an expected frequency for a single sample
  • The expected count is a theoretical average — actual counts will vary trial-to-trial
Understand

Concepts

  • Why expected frequencies scale linearly with the number of trials
  • Why expected does not mean "exactly" — it is a long-run mean
  • How to recognise that a problem fits the binomial / expected-frequency setup
Can do

Skills

  • Compute expected frequencies in production, sampling, and survey contexts
  • Interpret deviation from expected as natural variability around the mean
  • Compare expected with observed and judge whether the gap is reasonable
04
Key terms
Expected frequencyThe theoretical average number of times an event occurs in $m$ trials: $E = m \times p$.
Observed frequencyThe actual count from a real sample. Usually differs from the expected value due to natural variability.
Bernoulli trialA single experiment with two outcomes (success/failure), with constant probability $p$ of success.
Long-run averageAs the number of trials grows, the observed proportion approaches $p$ and the observed count approaches $E = m \times p$.
Quality controlIndustrial application of expected frequencies: predicting defect counts in production batches and flagging deviations from $E$.
ME12-5NESA outcome: applies appropriate statistical processes to present, analyse and interpret data, including binomial distribution mean, variance and expected frequencies.
05
The expected frequency formula
core concept

If a single trial has probability $p$ of producing a particular event, and the trial is repeated $m$ times independently, then the expected frequency of the event is

$E = m \times p$.

This is exactly $E(X)$ for $X \sim B(m, p)$. The expected frequency is the long-run average count; in any particular sample of $m$ trials the observed count will fluctuate around this value.

Worked through the hook: A fair die rolled $300$ times.

  • (a) Probability of rolling a six on one roll: $p = \tfrac{1}{6}$. Expected frequency: $E = 300 \times \tfrac{1}{6} = 50$ sixes.
  • (b) Probability of rolling an even number on one roll: $p = \tfrac{1}{2}$. Expected frequency: $E = 300 \times \tfrac{1}{2} = 150$ even rolls.
  • Will you get exactly $50$ sixes? Probably not — but typically you'll be within a few of $50$. For $B(300, \tfrac{1}{6})$ the SD is $\sqrt{300 \times \tfrac{1}{6} \times \tfrac{5}{6}} \approx 6.45$, so $50 \pm 6$ is the typical range.
Quality control context. A factory expects $0.5\%$ of products to fail QC. In a batch of $4000$, the expected number of failures is $4000 \times 0.005 = 20$. If a batch shows $35$ failures, that's well above expected — the controller may flag the batch for investigation.

Expected frequency formula: $E = m\times p$ where $m$ is the number of trials and $p$ is the probability of the event on each trial.

Pause — copy the expected frequency formula $E=mp$ and work through the hook example (die rolled 300 times, expected frequency of each face $= 300\times\frac{1}{6}=50$) into your book.

Quick check: A factory produces components, $3\%$ of which are defective. In a batch of $1500$ components, what is the expected number of defectives?

06
Observed vs expected — judging the gap
core concept

We just saw that if a single trial has probability $p$ and is repeated $m$ times, the expected frequency is $E=mp$. That raises a question: when actual observed counts differ from expected, how do you decide if the gap is "within normal variability" or genuinely suspicious? This card answers it → gaps up to about $1$–$2$ standard deviations from the expected value are typical; gaps beyond $2$ standard deviations warrant further investigation.

The expected frequency tells you what to predict; the observed frequency is what actually happens. The gap between them is natural variability. A useful rule of thumb: differences of up to about $1$–$2$ standard deviations from the expected value are common, while gaps larger than $3$ SDs are unusual enough to investigate.

Example: A coin is tossed $400$ times. Expected number of heads $= 400 \times 0.5 = 200$.

  • Variance: $npq = 400 \times 0.5 \times 0.5 = 100$, so SD $= 10$.
  • Observing $195$ heads (deviation $5$, that is $0.5$ SDs) — completely typical.
  • Observing $230$ heads (deviation $30$, that is $3$ SDs) — unusual; the coin may not be fair.

So expected frequencies are not just calculations — they give you a baseline to compare real-world data against.

$$\text{Expected frequency} = m \times p, \quad \text{Observed} \approx E \pm \sqrt{mpq}$$
Common mistake. Treating expected as exact. The expected frequency is a mean; real samples vary around it. Always interpret a single observation in the context of the standard deviation.

The expected frequency tells you what to predict; the observed frequency is what actually happens. The gap between them is natural variability. A useful rule of thumb: differences of up to about $1$–$2$ standard deviations from the expected value are...

Pause — copy the rule of thumb: observed frequency within $\pm 2$ SD of expected is normal variation; a larger gap suggests the model may not hold into your book.

Did you get this? True or false: in $1000$ tosses of a fair coin, you should expect to see exactly $500$ heads every time.

PROBLEM 1 · QUALITY CONTROL

In a manufacturing process, $2\%$ of microchips are defective. A batch of $5000$ microchips is produced. (a) Find the expected number of defective chips. (b) If $130$ defective chips are actually found, comment on whether this is unusual.

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Identify: $p = 0.02$, $m = 5000$. Expected frequency: $E = m \times p = 5000 \times 0.02 = 100$ defective chips.
Convert percentage to decimal: $2\% = 0.02$. Multiply by the number of trials.
PROBLEM 2 · SAMPLING + ROUNDING

A survey finds that $37\%$ of voters in an electorate support Party A. If $250$ voters are polled at random, what is the expected number who support Party A? Why might the actual number be a whole number near $93$?

1
$p = 0.37$, $m = 250$. Expected: $E = 250 \times 0.37 = 92.5$.
Don't round $p$ early. Keep $0.37$ exactly.
PROBLEM 3 · REVERSE PROBLEM

A pharmacist knows that $15\%$ of customers ask for a generic substitute. How many customers must be served, on average, so that the expected number of generic-substitute requests reaches $30$?

1
Set up: $E = m \times p$. We're given $E = 30$, $p = 0.15$, and we need $m$. So $30 = m \times 0.15$.
This is a reverse expected-frequency problem — solve for the number of trials.

Fill the gap: A spinner lands on red with probability $0.4$. If the spinner is spun $250$ times, the expected number of reds is $250 \times 0.4 = $.

Trap 01
Treating expected frequency as exact
Students often expect the actual count to equal $E = m \times p$. It usually won't — natural variability scatters observed counts around $E$ with SD $\sqrt{mpq}$. Always describe expected as a mean or long-run average.
Trap 02
Forgetting to convert percentages
If $p$ is given as $3\%$, you must use $p = 0.03$, not $p = 3$. Multiplying by $3$ instead of $0.03$ inflates the expected count by a factor of $100$ — a classic costly slip.
Trap 03
Rounding $E$ when it doesn't need to be a whole number
The expected frequency is a theoretical mean. A value like $E = 92.5$ is mathematically correct — do not round to "$93$" unless the question specifically asks for a whole-number prediction. Even then, state the exact mean first.

Did you get this? True or false: if $4\%$ of items are defective, then in $250$ items the expected number of defectives is $250 \times 4 = 1000$.

Work mode · how are you completing this lesson?
1

A fair die is rolled $180$ times. Find the expected number of (a) ones and (b) numbers greater than $4$.

2

A biased coin lands heads with probability $0.7$. If it is tossed $400$ times, find the expected number of heads and the standard deviation.

3

A factory produces $8000$ light bulbs per day. From experience, $1.25\%$ are defective. How many defective bulbs are expected per day?

4

A survey reports that $22\%$ of households own two cars. If $450$ households are surveyed, how many are expected to own two cars? Briefly justify why the actual count might differ.

5

In a quality test, the expected number of faulty items in a sample of $m$ is $24$. The single-item fault rate is $0.06$. Find $m$.

Odd one out: Three of these statements about expected frequencies are correct. Which one is NOT?

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Revisit your thinking

Earlier you predicted expected counts for $300$ die rolls and asked whether the actual count would land exactly on the expected value.

Expected sixes $= 50$, expected evens $= 150$. But in any single run of $300$ rolls, the observed sixes will scatter around $50$ with SD $\approx 6.45$, so seeing $43$ or $57$ sixes is unremarkable. "Expected" is a theoretical mean, not a guarantee. The strength of the formula $E = m \times p$ is that it gives a baseline against which real data can be compared.

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01
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.

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Short answer
ApplyBand 32 marks

Q1. A die is rolled $600$ times. Find the expected number of times a $5$ or $6$ appears. (2 marks)

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ApplyBand 43 marks

Q2. A factory produces components. The probability that a randomly chosen component is faulty is $0.025$. In a daily batch of $2000$ components, find the expected number of faulty components and the standard deviation. (3 marks)

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AnalyseBand 53 marks

Q3. A quality engineer expects $3\%$ of bottles to be cracked. In a sample of $500$ bottles, $25$ are found to be cracked. (a) Calculate the expected number of cracked bottles. (b) Calculate the standard deviation. (c) Decide, with reasoning, whether the observed count is unusual. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers:

1. (a) $p = 1/6$, $E = 180 \times \tfrac{1}{6} = 30$ ones. (b) Numbers greater than $4$ are $5$ and $6$, so $p = 2/6 = 1/3$; $E = 180 \times \tfrac{1}{3} = 60$.

2. $E = 400 \times 0.7 = 280$ heads. $\text{SD} = \sqrt{400 \times 0.7 \times 0.3} = \sqrt{84} \approx 9.165$.

3. $E = 8000 \times 0.0125 = 100$ defective bulbs per day.

4. $E = 450 \times 0.22 = 99$ households. The actual count varies because the survey is a random sample; SD $= \sqrt{450 \times 0.22 \times 0.78} \approx 8.79$, so typical observed counts fall roughly in $90$–$108$.

5. $24 = m \times 0.06 \Rightarrow m = 24/0.06 = 400$.

Q1 (2 marks): $p = \tfrac{2}{6} = \tfrac{1}{3}$ [1]. $E = 600 \times \tfrac{1}{3} = 200$ times [1].

Q2 (3 marks): $E = 2000 \times 0.025 = 50$ faulty components [1]. $\text{Var} = 2000 \times 0.025 \times 0.975 = 48.75$ [1]. $\text{SD} = \sqrt{48.75} \approx 6.982$ [1].

Q3 (3 marks): (a) $E = 500 \times 0.03 = 15$ cracked bottles [1]. (b) $\text{SD} = \sqrt{500 \times 0.03 \times 0.97} = \sqrt{14.55} \approx 3.814$ [1]. (c) Deviation $= 25 - 15 = 10$, which is $10/3.814 \approx 2.62$ SDs above expected — moderately unusual (between $2$ and $3$ SDs); worth investigating, but not extreme [1].

01
Boss battle · The Frequency Forecaster
earn bronze · silver · gold

Five timed questions on expected frequencies and comparing observed counts to expected. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena
02
Science Jump · platform challenge

Climb platforms by answering expected frequency questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

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