N3StatisticsStretch

Normal Approximation to the Binomial

A binomial with a large number of trials becomes tedious to compute directly and, when $p$ is not too extreme, its histogram looks strikingly like a bell curve. The $\textbf{normal approximation}$ exploits that resemblance — provided you account for the fact that one distribution is discrete and the other is not.

35 min Statistical Distributions

The big picture

The mismatch this lesson resolves is a real one, not a technicality. A binomial puts probability on whole numbers as separate bars; a normal spreads it continuously, so any single value has zero probability. The $\textbf{continuity correction}$ bridges them by treating the bar at $r$ as covering the interval from $r-0.5$ to $r+0.5$ — which is exactly what the histogram shows. Understanding it that way means you can work the adjustment out from a sketch every time instead of recalling four separate rules, and it explains why the correction matters most when $n$ is small and the bars are wide.

What you'll be able to do

State the conditions under which the approximation is reasonable
Find the matching normal parameters from a binomial
Apply the continuity correction correctly to each type of inequality
Judge when the approximation is appropriate rather than exact calculation

When the approximation is valid

The binomial histogram is roughly symmetric and bell-shaped when $n$ is large and $p$ is not close to 0 or 1. The usual working conditions are $np > 5$ and $n(1-p) > 5$ , which together ensure the distribution is not squashed against either end.

If $p$ is very small, the distribution is strongly skewed and the approximation is poor — the normal would assign probability to negative counts that cannot occur.

The approximation matters less than it once did now that calculators handle cumulative binomials directly, but it remains examinable and it is the same reasoning that underlies testing a sample mean later.

np > 5 \quad \text{and} \quad n(1-p) > 5

Working conditions for the approximation.

Why this works

The two conditions are symmetric for a reason: $np$ is the expected number of successes and $n(1-p)$ the expected number of failures. Both need to be comfortably above zero or the distribution is pressed against a boundary the normal does not have.

Matching the parameters

The approximating normal is chosen to have the same mean and variance as the binomial it replaces. Since the binomial has mean $np$ and variance $np(1-p)$ , those become $\mu$ and $\sigma^{2}$ .

Matching the first two moments is what makes the curves line up; the shapes then agree well provided the conditions above hold.

X \sim B(n,p) \;\Rightarrow\; Y \sim N\left(np,\; np(1-p)\right)

Same mean, same variance.

Example 1For

X \sim B(100, 0.4)

, state the approximating normal distribution.

1Check conditions:

np = 40 > 5

and

n(1-p) = 60 > 5

✓.

2Mean:

np = 40

3Variance:

np(1-p) = 100 \times 0.4 \times 0.6 = 24

Answer

Y \sim N(40, 24)

, so

\sigma = \sqrt{24} = 4.90

Tip — State the condition check explicitly. "Since $np = 40 > 5$ and $n(1-p) = 60 > 5$ , a normal approximation is reasonable" is usually a mark in itself.

The continuity correction

On a binomial histogram, the bar for $X = r$ runs from $r - 0.5$ to $r + 0.5$ and its area is $P(X = r)$ . The normal curve has no bars, so to capture the same area you must integrate across that whole interval.

So every discrete statement is widened to the matching continuous interval. $P(X = 8)$ becomes $P(7.5 < Y < 8.5)$ . $P(X \leq 8)$ includes the whole bar at 8, so it becomes $P(Y < 8.5)$ . $P(X < 8)$ excludes that bar, so it becomes $P(Y < 7.5)$ .

The same logic in the other direction: $P(X \geq 8)$ includes the bar at 8 and becomes $P(Y > 7.5)$ , while $P(X > 8)$ excludes it and becomes $P(Y > 8.5)$ .

Rather than memorising these, ask each time whether the bar at the boundary is included. If it is, extend the interval outwards by 0.5; if not, pull it in by 0.5.

P(X \leq r) \to P(Y < r+0.5)

Bar included: extend outwards.

P(X < r) \to P(Y < r-0.5)

Bar excluded: pull inwards.

P(X \geq r) \to P(Y > r-0.5)

Bar included, other tail.

P(X = r) \to P(r-0.5 < Y < r+0.5)

A single bar becomes an interval.

Example 1For

X \sim B(100, 0.4)

, use a normal approximation to estimate

P(X \geq 45)

1Approximating distribution:

Y \sim N(40, 24)

\sigma = 4.899

2"At least 45" includes the bar at 45, so the correction gives

P(Y > 44.5)

3Standardise:

z = \dfrac{44.5 - 40}{4.899} = 0.9186

P(Z > 0.9186) = 1 - 0.8208

Answer

\approx 0.179

Why this works

Using 45 instead of 44.5 would give $z = 1.021$ and a probability of about $0.154$ — a difference of roughly 14%. The correction is not a rounding nicety; it materially changes the answer.

Choosing whether to approximate

Exact binomial calculation is available on a calculator and is always preferable when the question permits it. Use the approximation when a question explicitly asks for it, or when $n$ is too large for the exact function to be practical.

When you do approximate, say so and state the conditions you checked. An answer that silently uses a normal where an exact binomial was expected can lose marks even when the arithmetic is right.

Remember the approximation is only that: the exact and approximate values differ slightly, and the gap grows as $n$ falls or $p$ moves towards an extreme.

Example 1A fair coin is tossed 400 times. Estimate the probability of at most 190 heads.

X \sim B(400, 0.5)

;

np = 200 > 5

and

n(1-p) = 200 > 5

✓.

Y \sim N(200, 100)

, so

\sigma = 10

3"At most 190" includes the bar at 190, so use

P(Y < 190.5)

z = \dfrac{190.5 - 200}{10} = -0.95

Answer

P(Z < -0.95) \approx 0.171

Tip — Sketch the binomial bars near the boundary and the normal curve over them. It makes the direction of the 0.5 shift obvious and takes seconds.

Think like an examiner

Check and state both conditions before approximating.
Write out the approximating distribution with its variance, then take the square root for standardising.
Show the continuity correction as a separate line so the marker sees the adjusted boundary.
Say explicitly that the answer is an approximation.

Common misconceptions

Omitting the continuity correction.

The discrete bars must be widened by

\pm0.5

; the error is often several percent.

Adjusting in the wrong direction for a strict inequality.

Ask whether the boundary bar is included. Included means extend outwards, excluded means pull in.

Using

np(1-p)

as the standard deviation.

That is the variance. Take its square root before standardising.

Approximating when

p

is very small and

np < 5

The distribution is too skewed. State that the conditions fail and use the exact binomial.

Normal approximation

np > 5,\; n(1-p) > 5

Validity conditions.

Y \sim N(np,\, np(1-p))

Matching parameters.

P(X\leq r) \to P(Y<r+0.5)

Continuity correction.

z = \tfrac{y-\mu}{\sigma}

Then standardise as usual.

Stretch yourself

A machine produces items with a 20% defect rate. In a batch of 150, use a normal approximation to estimate the probability that between 25 and 35 items inclusive are defective.

Hint — Both endpoints are included, so both bars must be captured. Widen the interval at both ends.

Reveal solution

Solution

X \sim B(150, 0.2)

. Conditions:

np = 30 > 5

and

n(1-p) = 120 > 5

✓. The approximating normal has mean

30

and variance

150 \times 0.2 \times 0.8 = 24

, so

\sigma = \sqrt{24} = 4.899

. "Between 25 and 35 inclusive" includes both boundary bars, so the corrected interval runs from

24.5

35.5

. Standardising:

z_{1} = \dfrac{24.5-30}{4.899} = -1.1227

and

z_{2} = \dfrac{35.5-30}{4.899} = 1.1227

. By symmetry the answer is

P(-1.1227 < Z < 1.1227) = 2 \times 0.8692 - 1 \approx 0.738

. The symmetry here is not a coincidence — 25 and 35 are equally spaced either side of the mean of 30, so the corrected bounds are too. Spotting that halves the calculator work.

\blacksquare

Questions students ask

Why is a continuity correction needed at all?

The binomial is discrete and the normal continuous. Each binomial bar spans half a unit either side of its value, so the equivalent normal probability must be taken across that full interval.

How do I decide whether to add or subtract the 0.5?

Ask whether the bar at the boundary is part of the event. If it is, extend the interval outwards to include it; if not, move inwards to exclude it.

Should I approximate when the exact binomial is available?

No, unless the question asks for an approximation. Exact calculation is more accurate; the approximation exists for cases where exact computation is impractical.

What happens if the conditions are not met?

The approximation becomes unreliable because the binomial is too skewed to resemble a bell curve. State that the conditions fail and use the exact distribution instead.

Key takeaways

A binomial can be approximated by a normal when $np > 5$ and $n(1-p) > 5$ .
The approximating normal matches the binomial mean $np$ and variance $np(1-p)$ .
Discrete bars span $\pm 0.5$ , so every boundary needs a continuity correction.
Extend outwards when the boundary bar is included, inwards when it is not.
Exact binomial calculation is preferable whenever it is practical.

How this fits the course

Build on

The Language of Hypothesis Testing

Leads to

The Language of Hypothesis Testing

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

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