How do you find a binomial probability using the normal approximation?

Write X as approximately N(np, np(1-p)), then use technology to find the required normal probability such as P(X less than or equal to k), P(X greater than or equal to k) or P(a less than or equal to X less than or equal to b). Standardising with z equals (value minus mu) divided by sigma also gives the answer.

How do you find a binomial probability using the normal approximation?

Write X as approximately N(np, np(1-p)), then use technology to find the required normal probability such as P(X less than or equal to k), P(X greater than or equal to k) or P(a less than or equal to X less than or equal to b). Standardising with z equals (value minus mu) divided by sigma also gives the answer.

Year 12 Methods (Unit 3 & 4) The normal distribution

The normal approximation to the binomial distribution

Q: When can you use a normal distribution to approximate a binomial?

When n is large and p is not too extreme. The usual rule of thumb is that both np and n times (1 minus p) are at least 5. Then the binomial X with parameters n and p is approximately normal with mean np and variance np times (1 minus p).

Q: What mean and variance does the approximating normal have?

The mean is mu equals np and the variance is sigma squared equals np times (1 minus p), so the standard deviation is sigma equals the square root of np times (1 minus p). A common error is to use np as the standard deviation, forgetting both the factor (1 minus p) and the square root.

Q: Does QCAA Mathematical Methods use a continuity correction?

No. The QCAA Mathematical Methods syllabus does not include a continuity correction; probabilities are found directly from the approximating normal using technology. The plain normal approximation X approximately N(np, np(1-p)) is used throughout.

Q: When is the normal approximation poor?

When n is small or p is close to 0 or 1, so that np or n times (1 minus p) is less than 5. Then the binomial is too skewed for a symmetric normal curve to fit, and the exact binomial probability should be used instead.

Q: Why is the binomial approximately normal for large n?

A binomial count is a sum of many independent Bernoulli trials. For large n the probability histogram becomes bell shaped and symmetric about the mean np, so a normal curve with the same mean and variance closely matches the bars.

20 practice questions 0 video lessons Theory + worked examples

In Year 12 Mathematical Methods (Queensland, QCAA), the normal approximation to the binomial replaces a binomial variable \(X\sim B(n,p)\) with a normal distribution \(N\big(np,\,np(1-p)\big)\) when \(n\) is large and \(p\) is not too extreme (rule of thumb \(np\ge 5\) and \(n(1-p)\ge 5\)). You then use technology to estimate \(P(X\le k)\), \(P(X\ge k)\) or \(P(a\le X\le b)\). No continuity correction is used.

A binomial random variable \(X\sim B(n,p)\) counts the successes in \(n\) independent trials, each with success probability \(p\). Its probability histogram is a set of bars at \(x=0,1,\dots,n\). As \(n\) grows and \(p\) stays away from \(0\) and \(1\), that histogram becomes bell shaped and symmetric about its mean, so a normal curve fits it closely.

The normal approximation to the binomial uses the normal distribution with the same mean and variance as the binomial: \(\mu=np\) and \(\sigma^2=np(1-p)\). We write \(X\approx N\big(np,\,np(1-p)\big)\). Because a normal distribution is continuous and easy to evaluate with technology, it gives quick estimates of binomial probabilities for large \(n\).

The approximation is reliable when \(np\ge 5\) and \(n(1-p)\ge 5\). If \(n\) is small or \(p\) is close to \(0\) or \(1\), the binomial is too skewed and the approximation is poor — then the exact binomial should be used. In QCAA Methods the plain normal approximation is used, with no continuity correction.

Key idea. If \(np\ge 5\) and \(n(1-p)\ge 5\), then \(X\sim B(n,p)\) is approximately \(N\big(np,\,np(1-p)\big)\): mean \(\mu=np\), standard deviation \(\sigma=\sqrt{np(1-p)}\).

The normal curve \(N\big(np,\,np(1-p)\big)\) sits over the binomial bars, centred at \(\mu=np\)

A probability such as \(P(X\ge k)\) is the shaded area under the approximating normal curve

The approximating normal (valid when \(np\ge 5\) and \(n(1-p)\ge 5\)):

\[X\sim B(n,p)\ \Rightarrow\ X\approx N\big(np,\;np(1-p)\big)\]

X \approx N (n p, n p (1 - p))

Mean, variance and standard deviation:

\[\mu=np \qquad \sigma^{2}=np(1-p) \qquad \sigma=\sqrt{np(1-p)}\]

σ = \sqrt{n p (1 - p)}

Standardising to find a probability (technology, no continuity correction):

\[P(X\le k)\approx P\!\left(Z\le \dfrac{k-np}{\sqrt{np(1-p)}}\right)\]

Z = \frac{k - n p}{\sqrt{n p (1 - p)}}

Validity. Use the approximation only when \(np\ge 5\) and \(n(1-p)\ge 5\). Otherwise the binomial is too skewed — use the exact binomial instead.

How to use the normal approximation to the binomial

Identify \(n\) and \(p\). Recognise the count as \(X\sim B(n,p)\).
Check the approximation is appropriate. Confirm \(np\ge 5\) and \(n(1-p)\ge 5\). If either fails, use the exact binomial.
Find the parameters. Compute \(\mu=np\) and \(\sigma=\sqrt{np(1-p)}\), and write \(X\approx N\big(np,\,np(1-p)\big)\).
Evaluate the probability with technology. Find \(P(X\le k)\), \(P(X\ge k)\) or \(P(a\le X\le b)\) directly, or standardise with \(z=\dfrac{x-\mu}{\sigma}\). No continuity correction is applied.

Set-up clue. On a calculator, enter the normal distribution with mean \(np\) and standard deviation \(\sqrt{np(1-p)}\) (the standard deviation, not the variance) before finding the tail or interval probability.

Example 1 — Mean and standard deviation

\(X\sim B(80,0.25)\). Find the mean and standard deviation of the approximating normal.

Solution

Use \(\mu=np\) and \(\sigma=\sqrt{np(1-p)}\).

\(\mu\)	\(=\)	\(80\times 0.25=20\)
\(\sigma^2\)	\(=\)	\(80\times 0.25\times 0.75=15\)
\(\sigma\)	\(=\)	\(\sqrt{15}\approx 3.87\)

So \(X\approx N(20,\,15)\).

μ = 20

Example 2 — Is the approximation appropriate?

Decide whether the normal approximation is appropriate for \(X\sim B(20,0.1)\).

Solution

Check both \(np\) and \(n(1-p)\).

\(np\)	\(=\)	\(20\times 0.1=2\)
\(n(1-p)\)	\(=\)	\(20\times 0.9=18\)
\(np=2\)	\(<\)	\(5\)

Not appropriate: \(np<5\), so use the exact binomial.

n p = 2

Example 3 — An upper-tail probability

A fair coin is tossed \(100\) times; \(X\) is the number of heads. Estimate \(P(X\ge 60)\).

Solution

\(np=50\ge 5\) and \(n(1-p)=50\ge 5\), so use \(N(50,25)\).

\(\mu\)	\(=\)	\(50,\ \sigma=\sqrt{25}=5\)
\(z\)	\(=\)	\(\dfrac{60-50}{5}=2\)
\(P(X\ge 60)\)	\(\approx\)	\(P(Z\ge 2)=0.0228\)

P (X \geq 60) \approx 0.0228

Example 4 — A modelling problem

Items are defective independently with probability \(0.08\). In \(150\) items, \(X\) is the number defective. Estimate \(P(10\le X\le 15)\).

Solution

\(np=12\ge 5,\ n(1-p)=138\ge 5\), so use the approximation.

\(\mu\)	\(=\)	\(12\)
\(\sigma\)	\(=\)	\(\sqrt{11.04}\approx 3.32\)
\(P(10\le X\le 15)\)	\(\approx\)	\(0.5431\)

Using \(N(12,\,11.04)\) with technology.

P (10 \leq X \leq 15) \approx 0.5431

Common pitfalls

The standard deviation is \(\sqrt{np(1-p)}\), not \(np\). A frequent error is to use \(\sigma=np\), forgetting both the factor \((1-p)\) and the square root. The variance is \(np(1-p)\); the standard deviation is its square root.

Do not use \((n,p)\) as the normal parameters. The approximating normal is \(N\big(np,\,np(1-p)\big)\), not \(N(n,p)\). Convert to \(\mu=np\) and \(\sigma^2=np(1-p)\) first.

Check the approximation is valid first. If \(np<5\) or \(n(1-p)<5\) (small \(n\) or extreme \(p\)), the binomial is skewed and the normal fit is poor — use the exact binomial instead.

Frequently asked questions

When can you use a normal distribution to approximate a binomial?

When \(n\) is large and \(p\) is not too extreme — the rule of thumb is \(np\ge 5\) and \(n(1-p)\ge 5\). Then \(X\sim B(n,p)\) is approximately \(N\big(np,\,np(1-p)\big)\).

What mean and variance does the approximating normal have?

The same as the binomial: \(\mu=np\) and \(\sigma^2=np(1-p)\), so \(\sigma=\sqrt{np(1-p)}\).

How do you find a binomial probability using the approximation?

Write \(X\approx N\big(np,\,np(1-p)\big)\) and use technology for \(P(X\le k)\), \(P(X\ge k)\) or \(P(a\le X\le b)\), or standardise with \(z=\dfrac{x-\mu}{\sigma}\).

Does QCAA Mathematical Methods use a continuity correction?

No. The QCAA syllabus does not include a continuity correction; probabilities are read directly from the approximating normal using technology.

When is the normal approximation poor?

When \(n\) is small or \(p\) is close to \(0\) or \(1\), so \(np<5\) or \(n(1-p)<5\). The binomial is then too skewed — use the exact binomial.

Why is the binomial approximately normal for large \(n\)?

A binomial count is a sum of many independent trials, so for large \(n\) its histogram is bell shaped and symmetric about \(np\), matching a normal curve of the same mean and variance.

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