The normal approximation to the binomial

◈ 4 cards

When np ≥ 5 and n(1 − p) ≥ 5, X ~ Bin(n, p) is close to N(np, np(1 − p)); apply the ±0.5 continuity correction, then standardise and read the table.

A binomial with a hundred terms

Northfield Credit sends 100 payment reminders a month, and historically 30 % of reminders are answered within a week. $X$ = the number answered within a week is $Bin (100, 0.30)$ . The credit manager wants $P (X \leq 25)$ . Module 6's route — sum 26 pmf terms with $(x 100)$ in each — is not a hand calculation. But a binomial with many trials is bell-shaped, and the normal can stand in for it.

The conditions and the parameters

The approximation is acceptable when

$n p \geq 5 and n (1 - p) \geq 5$

— enough expected successes and enough expected failures for the bell not to be cut off at 0 or at $n$ . Here $n p = 30$ and $n (1 - p) = 70$ ; both pass. The approximating normal takes the binomial's own mean and variance (L6.4):

$μ = n p = 30, σ^{2} = n p (1 - p) = 21, σ = 21 = 4.583$

so $X \approx N (30, 21)$ .

The continuity correction

$X$ is a count; the normal is continuous. On the binomial, " $X \leq 25$ " is the bar at 25 and everything below it. On the normal, the bar at 25 is the strip from 24.5 to 25.5 — so " $X \leq 25$ " must become " $X < 25.5$ " to keep the whole bar. This ±0.5 continuity correction moves the boundary half a unit away from the region you are keeping:

Binomial	Normal
$P (X \leq k)$	$P (X < k + 0.5)$
$P (X < k)$	$P (X < k - 0.5)$
$P (X \geq k)$	$P (X > k - 0.5)$
$P (X > k)$	$P (X > k + 0.5)$
$P (X = k)$	$P (k - 0.5 < X < k + 0.5)$

The boundary discipline of Module 6 returns: $\leq 25$ and $< 26$ are the same binomial event and both become 25.5.

Worked example

$P (X \leq 25) \approx P (Z < \frac{25.5 - 30}{4.583}) = P (Z < - 0.98) = Φ (- 0.98) = 0.1635$

The exact binomial sum is 0.1631 — the approximation is off by four in the fourth decimal. Without the correction, $z = (25 - 30) /4.583 = - 1.09$ and $Φ (- 1.09) = 0.1379$ : off by 0.025, a visibly wrong answer. The half-unit is what makes the approximation usable at this $n$ .

For the upper tail, "at least 36 answered": $P (X \geq 36) \approx P (X > 35.5)$ ,

$z = \frac{35.5 - 30}{4.583} = 1.20, P = 1 - Φ (1.20) = 1 - 0.8849 = 0.1151$

(exact 0.1161).

Which version to state

IBS1 presents the approximation without the correction; IS1 and the course use it. On the paper, write the corrected boundary explicitly — " $P (X \leq 25) \approx P (X < 25.5)$ " — so the marker sees which convention you applied. Module 12 approximates the sample proportion the same way, with the same $n p \geq 5$ conditions, but there the correction is dropped, because $\overset{p}{^}$ is treated as a mean rather than a count.