Normal Approximation to the Binomial

What’s the probability of getting at least $80$ “yes” answers in a survey of $200$ people? With the binomial distribution, you’d have to add up $121$ separate probabilities. But when the number of trials is large, the binomial distribution’s bars line up almost perfectly under a normal curve. That lets you replace a huge sum with one quick z-score calculation.

Key ideas

Binomial distributions become bell-shaped

Recall that a binomial random variable $X$ counts the successes in $n$ independent trials, each with probability of success $p$. It has

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

For a small $n$, the distribution can be lopsided, especially when $p$ is close to $0$ or $1$. As $n$ grows, the histogram becomes more and more symmetric and bell-shaped, centred at $np$.

The binomial distribution with $n = 40$, $p = 0.3$, and the normal curve $N(12, 8.4)$. The orange bars are $P(X \le 10)$.

The approximation

If $X$ is binomial with $n$ trials and probability $p$, and $n$ is large enough, then $X$ is approximately normal:

$$X \approx N\big(np,\ np(1 - p)\big)$$

A common rule of thumb for “large enough” is

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

That is, you should expect at least $5$ successes and at least $5$ failures. (Some textbooks use $10$ instead of $5$ for a more accurate approximation.)

The continuity correction

A binomial variable only takes whole numbers, but the normal curve is continuous. Each bar for the value $k$ stretches from $k - 0.5$ to $k + 0.5$. So to match the bars, adjust each boundary by $0.5$, making sure you include the bars you want:

Binomial	Normal, with continuity correction
$P(X = k)$	$P(k - 0.5 \lt Y \lt k + 0.5)$
$P(X \le k)$	$P(Y \lt k + 0.5)$
$P(X \lt k)$	$P(Y \lt k - 0.5)$
$P(X \ge k)$	$P(Y \gt k - 0.5)$
$P(X \gt k)$	$P(Y \gt k + 0.5)$
$P(a \le X \le b)$	$P(a - 0.5 \lt Y \lt b + 0.5)$

Here $Y$ is the approximating normal variable. Notice that $P(X \lt k)$ and $P(X \le k)$ are different for a binomial variable (the bar at $k$ is in one but not the other), even though they’re the same for a normal variable.

Hypergeometric distributions too

The hypergeometric distribution counts successes when you sample without replacement. When the sample is a small part of the population (a common guideline is at most $5\%$ to $10\%$), removing a few items barely changes the probabilities, so it behaves like a binomial with $p = \dfrac{\text{successes in population}}{\text{population size}}$. With a large enough sample, it’s approximately normal too, and you can use the same method.

Why bother, with technology?

A calculator can find exact binomial probabilities instantly with binomcdf. But the normal approximation explains why so many counts in real life look bell-shaped, and it’s the foundation for the margin of error in polls.

Worked examples

Example 1: Checking the conditions

Can the normal approximation be used?

• (a) $n = 40$, $p = 0.3$
• (b) $n = 50$, $p = 0.04$

Solution.

(a) $np = 40(0.3) = 12$ and $n(1 - p) = 40(0.7) = 28$. Both are at least $5$, so yes.

(b) $np = 50(0.04) = 2$, which is less than $5$. No: you’d expect only about $2$ successes, so the distribution is squashed against $0$ and skewed, not bell-shaped.

Example 2: Continuity corrections

$X$ is binomial and approximated by a normal variable $Y$. Write each probability with a continuity correction.

• (a) $P(X \ge 30)$
• (b) $P(X \lt 30)$
• (c) $P(25 \le X \le 32)$

Solution. Think about which bars to include.

(a) Include the bar at $30$, which starts at $29.5$: $P(Y \gt 29.5)$.

(b) Exclude the bar at $30$, so stop at the end of the bar at $29$: $P(Y \lt 29.5)$.

(c) Include both end bars: $P(24.5 \lt Y \lt 32.5)$.

Example 3: Approximation vs exact

For $n = 40$ and $p = 0.3$, use the normal approximation to estimate $P(X \le 10)$. Compare with the exact binomial probability.

Solution. The conditions hold (Example 1). The approximating normal has

$$\mu = 12, \qquad \sigma = \sqrt{40(0.3)(0.7)} = \sqrt{8.4} \approx 2.8983$$

With the continuity correction, $P(X \le 10) \approx P(Y \lt 10.5)$:

$$z = \frac{10.5 - 12}{2.8983} \approx -0.5175$$ $$P(Y \lt 10.5) = P(Z \lt -0.5175) \approx 0.3024$$

(Rounding to $z = -0.52$ and using a table gives $0.3015$.)

The exact binomial value is

$$P(X \le 10) = \sum_{k=0}^{10} \binom{40}{k}(0.3)^k(0.7)^{40-k} \approx 0.3087$$

(On a TI-84: binomcdf(40, 0.3, 10).) The approximation is off by less than $0.01$. That’s the orange area in the diagram compared with the area under the curve to the left of $10.5$.

Example 4: A survey

In a large town, $35\%$ of residents support building a new skate park. A random sample of $200$ residents is surveyed. Estimate the probability that at least $80$ of them support it.

Solution. $X$ is binomial with $n = 200$, $p = 0.35$. Check: $np = 70 \ge 5$ and $n(1 - p) = 130 \ge 5$. ✓

$$\mu = 70, \qquad \sigma = \sqrt{200(0.35)(0.65)} = \sqrt{45.5} \approx 6.7454$$

“At least $80$” includes the bar at $80$, so use $P(Y \gt 79.5)$:

$$z = \frac{79.5 - 70}{6.7454} \approx 1.4084$$ $$P(Y \gt 79.5) = 1 - P(Z \lt 1.4084) \approx 1 - 0.9205 = 0.0795$$

The exact binomial probability is $0.0805$, so the approximation is very close. There’s about an $8\%$ chance that $80$ or more of the $200$ support the skate park.

Common mistakes

Skipping the condition check. If $np$ or $n(1 - p)$ is less than $5$, the binomial is too lopsided for a normal curve. Check both before you start.

Using the variance as the standard deviation. The standard deviation is $\sqrt{np(1 - p)}$, not $np(1 - p)$.

Forgetting the continuity correction. In Example 3, using $10$ instead of $10.5$ gives $0.2451$, much further from the exact $0.3087$ than the corrected $0.3024$.

Correcting in the wrong direction. Picture the bars. “At least $80$” must include the whole bar at $80$, which starts at $79.5$, so you use $79.5$, not $80.5$. “More than $80$” excludes that bar, so you use $80.5$.

Treating “fewer than” and “at most” the same. For a count, $X \lt 30$ means $X \le 29$. The corrected boundary is $29.5$, not $30.5$.

Practice

1. (Warm-up) Can the normal approximation be used? Show the check.

• (a) $n = 30$, $p = 0.5$
• (b) $n = 25$, $p = 0.2$
• (c) $n = 100$, $p = 0.97$

Solution

(a) $np = 15$, $n(1 - p) = 15$. Yes.

(b) $np = 5$, $n(1 - p) = 20$. Yes, just barely.

(c) $np = 97$, but $n(1 - p) = 100(0.03) = 3 \lt 5$. No: too few expected failures.

2. (Warm-up) $X$ is binomial with $n = 80$ and $p = 0.25$. State the normal distribution that approximates it, including the standard deviation.

Solution

$\mu = 80(0.25) = 20$ and $\sigma^2 = 80(0.25)(0.75) = 15$, so $X \approx N(20, 15)$, with $\sigma = \sqrt{15} \approx 3.873$.

3. (Warm-up) Write each binomial probability as a normal probability with a continuity correction.

• (a) $P(X = 20)$
• (b) $P(X \gt 20)$
• (c) $P(X \le 20)$
• (d) $P(15 \le X \lt 25)$

Solution

(a) $P(19.5 \lt Y \lt 20.5)$

(b) $P(Y \gt 20.5)$

(c) $P(Y \lt 20.5)$

(d) $15 \le X \lt 25$ means $X$ from $15$ to $24$: $P(14.5 \lt Y \lt 24.5)$.

4. (Core) A fair coin is flipped $100$ times. Use the normal approximation to estimate the probability of at least $60$ heads. Compare with the exact value of $0.0284$.

Solution

$\mu = 50$, $\sigma = \sqrt{100(0.5)(0.5)} = 5$. Use $P(Y \gt 59.5)$:

$$z = \frac{59.5 - 50}{5} = 1.9, \qquad P(Z \gt 1.9) = 1 - 0.9713 = 0.0287$$

The approximation $0.0287$ is very close to the exact $0.0284$.

5. (Core) A basketball player makes $75\%$ of her free throws. She takes $48$ free throws in practice. Estimate the probability that she makes at most $32$.

Solution

Check: $np = 36$ and $n(1 - p) = 12$, both at least $5$. ✓

$\mu = 36$, $\sigma = \sqrt{48(0.75)(0.25)} = \sqrt{9} = 3$. Use $P(Y \lt 32.5)$:

$$z = \frac{32.5 - 36}{3} \approx -1.1667, \qquad P(Z \lt -1.1667) \approx 0.1217$$

(The exact binomial value is $0.1232$.)

6. (Core) $X$ is binomial with $n = 50$ and $p = 0.3$. Use the normal approximation to estimate $P(X = 15)$, and compare with the exact value $\binom{50}{15}(0.3)^{15}(0.7)^{35}$.

Solution

$\mu = 15$, $\sigma = \sqrt{50(0.3)(0.7)} = \sqrt{10.5} \approx 3.2404$. Use $P(14.5 \lt Y \lt 15.5)$:

$$z = \frac{14.5 - 15}{3.2404} \approx -0.1543, \qquad z = \frac{15.5 - 15}{3.2404} \approx 0.1543$$$$P(-0.1543 \lt Z \lt 0.1543) \approx 0.5613 - 0.4387 = 0.1226$$

The exact value is $\binom{50}{15}(0.3)^{15}(0.7)^{35} \approx 0.1223$. Very close.

(Without the continuity correction, you’d be finding $P(Y = 15)$, which is $0$ for a continuous variable.)

7. (Core) A warehouse has $2000$ light bulbs, and $160$ of them are defective. An inspector randomly selects $100$ bulbs (without replacement). Use a normal approximation to estimate the probability that at most $5$ are defective.

Solution

The count of defective bulbs is hypergeometric. The sample is $\tfrac{100}{2000} = 5\%$ of the population, so treat it as approximately binomial with $p = \tfrac{160}{2000} = 0.08$.

Check: $np = 8 \ge 5$ and $n(1 - p) = 92 \ge 5$. ✓

$\mu = 8$, $\sigma = \sqrt{100(0.08)(0.92)} = \sqrt{7.36} \approx 2.7129$. Use $P(Y \lt 5.5)$:

$$z = \frac{5.5 - 8}{2.7129} \approx -0.9215, \qquad P(Z \lt -0.9215) \approx 0.1784$$

(The exact hypergeometric probability is $0.1730$, so the estimate is reasonable.)

8. (Challenge) Return to Example 3 ($n = 40$, $p = 0.3$, $P(X \le 10)$).

• (a) Estimate $P(X \le 10)$ without the continuity correction (use $P(Y \lt 10)$).
• (b) Using the diagram, explain why the continuity correction gives a better answer.

Solution

(a)

$$z = \frac{10 - 12}{2.8983} \approx -0.6901, \qquad P(Z \lt -0.6901) \approx 0.2451$$

That’s much further from the exact $0.3087$ than the corrected estimate $0.3024$.

(b) The bar for $X = 10$ stretches from $9.5$ to $10.5$. Stopping the normal area at $10$ leaves out the right half of that bar, which is a noticeable chunk of probability (the bar has height about $0.11$). Stopping at $10.5$ includes the whole bar.

9. (Challenge) An airline has $200$ seats on a flight and sells $210$ tickets, because on average $6\%$ of ticketed passengers don’t show up. Assume passengers show up independently. Estimate the probability that more than $200$ passengers show up, so the flight is overbooked.

Solution

Let $X$ be the number who show up: binomial with $n = 210$ and $p = 0.94$.

Check: $np = 197.4 \ge 5$ and $n(1 - p) = 12.6 \ge 5$. ✓

$\mu = 197.4$, $\sigma = \sqrt{210(0.94)(0.06)} = \sqrt{11.844} \approx 3.4415$.

“More than $200$” excludes the bar at $200$, so use $P(Y \gt 200.5)$:

$$z = \frac{200.5 - 197.4}{3.4415} \approx 0.9008, \qquad P(Z \gt 0.9008) \approx 1 - 0.8161 = 0.1839$$

There’s about an $18\%$ chance the flight is overbooked. (The exact binomial probability is $0.1856$.)