Mathematics 9709 · Probability & Statistics 1

Binomial mean & variance

Over $n$ trials, the expected number of successes is $np$ .
Example: 100 fair coin flips gives an expected 50 heads.
So for a binomial, $\mu=np$ .
The spread result to memorise is $\sigma^2=npq$ .
The standard deviation is the square root: $\sqrt{npq}$ .

μ = np, σ² = npq, and the reverse trick

$\mu=np,\qquad \sigma^2=npq,\qquad \text{SD}=\sqrt{npq}$

Worked example

Forward

$X\sim B(12,0.3)$ : $\mu=3.6$ , $\sigma^2=12(0.3)(0.7)=2.52$ , $\text{SD}=\sqrt{2.52}$ .

Answer

\mu=3.6,\ \sigma^2=2.52,\ \text{SD}\approx1.59

B(20, 0.40)·mean np = 8.0·variance npq = 4.80

Reverse: $q=\operatorname{Var}\div E$ ( $npq\div np$ ).
$p=1-q$ , then $n=\mu/p$ .
“Least n” from $q^n<k$ : trial or logs, round up.

Reverse problem: q = Var ÷ E (npq ÷ np), then p = 1 − q, then n = μ/p.

Worked example

Reverse: find n and p

$E(X)=12,\ \operatorname{Var}(X)=7.5$ : $q=\tfrac{7.5}{12}=0.625$ , $p=0.375$ , $n=\tfrac{12}{0.375}=32$ . Then $P(X=11)={}^{32}C_{11}(0.375)^{11}(0.625)^{21}$ .

Answer

X\sim B(32,0.375),\ P(X{=}11)\approx0.138

Worked example

Least n (log method)

$X\sim B(n,0.4)$ , least $n$ with $P(X=0)<0.1$ : $0.6^n<0.1$ . Since $0.6^4=0.1296,\ 0.6^5=0.0778$ :

Answer

n=5

Why npq for the variance

Each trial is a Bernoulli with variance $p(1-p)$ ; $n$ independent trials add variances, giving $np(1-p)=npq$ — that is the whole story behind $\sigma^2=npq$ .

Common mistake

Quoting

npq

as the SD (it is the variance — take the root). In the reverse problem divide

\operatorname{Var}\div E

(not

E\div\operatorname{Var}

), and for “least

\ n

” always round up.

Big n: approximate with a Normal

When $n$ is large enough, binomial bars start to look bell-shaped.
Use this only when $np>5$ and $nq>5$ .
Approximate $X\sim B(n,p)$ by $N(np,\,npq)$ .
The mean stays $np$ and the variance stays $npq$ .
Because integers become a smooth curve, use a continuity correction: shift the boundary by 0.5.
The Normal chapter gives the full method (Ch8).

State $X\sim B(n,p)$ , then approx $X\sim N(np,npq)$ (name both).
Compute mean $np$ and variance $npq$ (not the SD yet).
Continuity-correct the boundary by 0.5 (table below).
Standardise $Z=\dfrac{\text{c.c. boundary}-np}{\sqrt{npq}}$ — use $\sqrt{npq}$ , never $npq$ .
Read Φ; finish (1−Φ for an upper tail, Φ(b)+Φ(a)−1 for a band).

Worked example

One-sided normal approximation

$X\sim B(110,0.25)$ , P(fewer than 22). Mean $np=27.5$ , variance $npq=20.625$ . C.C. $P(X\le21)\to21.5$ ; $Z=\dfrac{21.5-27.5}{\sqrt{20.625}}=-1.321$ ; $\Phi(-1.321)=1-0.9066$ . (9709/52/M/J/24 Q6)

Answer

\approx0.0934

Want P(…)	Corrected boundary	Then
$X<k$	$k-0.5$	$\Phi$
$X\le k$	$k+0.5$	$\Phi$
$X>k$	$k+0.5$	$1-\Phi$
$X\ge k$	$k-0.5$	$1-\Phi$
$a<X<b$	$a+0.5,\ b-0.5$	$\Phi(b)+\Phi(a)-1$

Name N(np, npq), shift the boundary by 0.5, standardise with √(npq) — never npq, never √σ.

Examiner note

Write np and npq as numbers before standardising (a separate B1), and the 0.5 continuity correction must be visibly present (its own M1). Standardise with

\sqrt{npq}

, not

npq

\sqrt\sigma

. (9709/52/O/N/24 Q7)

Common mistake

Standardising with the variance

npq

(or

\sqrt\sigma

) instead of

\sqrt{npq}

, and the wrong continuity-correction direction on a strict band (use

28.5

and

34.5

for “more than 28 but fewer than 35”).

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Binomial mean & variance

Over $n$ trials, the expected number of successes is $np$ .
Example: 100 fair coin flips gives an expected 50 heads.
So for a binomial, $\mu=np$ .
The spread result to memorise is $\sigma^2=npq$ .
The standard deviation is the square root: $\sqrt{npq}$ .

μ = np, σ² = npq, and the reverse trick

$\mu=np,\qquad \sigma^2=npq,\qquad \text{SD}=\sqrt{npq}$

Worked example

Forward

$X\sim B(12,0.3)$ : $\mu=3.6$ , $\sigma^2=12(0.3)(0.7)=2.52$ , $\text{SD}=\sqrt{2.52}$ .

Answer

\mu=3.6,\ \sigma^2=2.52,\ \text{SD}\approx1.59

B(20, 0.40)·mean np = 8.0·variance npq = 4.80

Reverse: $q=\operatorname{Var}\div E$ ( $npq\div np$ ).
$p=1-q$ , then $n=\mu/p$ .
“Least n” from $q^n<k$ : trial or logs, round up.

Reverse problem: q = Var ÷ E (npq ÷ np), then p = 1 − q, then n = μ/p.

Worked example

Reverse: find n and p

$E(X)=12,\ \operatorname{Var}(X)=7.5$ : $q=\tfrac{7.5}{12}=0.625$ , $p=0.375$ , $n=\tfrac{12}{0.375}=32$ . Then $P(X=11)={}^{32}C_{11}(0.375)^{11}(0.625)^{21}$ .

Answer

X\sim B(32,0.375),\ P(X{=}11)\approx0.138

Worked example

Least n (log method)

$X\sim B(n,0.4)$ , least $n$ with $P(X=0)<0.1$ : $0.6^n<0.1$ . Since $0.6^4=0.1296,\ 0.6^5=0.0778$ :

Answer

n=5

Why npq for the variance

Each trial is a Bernoulli with variance $p(1-p)$ ; $n$ independent trials add variances, giving $np(1-p)=npq$ — that is the whole story behind $\sigma^2=npq$ .

Common mistake

Quoting

npq

as the SD (it is the variance — take the root). In the reverse problem divide

\operatorname{Var}\div E

(not

E\div\operatorname{Var}

), and for “least

\ n

” always round up.

Big n: approximate with a Normal

When $n$ is large enough, binomial bars start to look bell-shaped.
Use this only when $np>5$ and $nq>5$ .
Approximate $X\sim B(n,p)$ by $N(np,\,npq)$ .
The mean stays $np$ and the variance stays $npq$ .
Because integers become a smooth curve, use a continuity correction: shift the boundary by 0.5.
The Normal chapter gives the full method (Ch8).

State $X\sim B(n,p)$ , then approx $X\sim N(np,npq)$ (name both).
Compute mean $np$ and variance $npq$ (not the SD yet).
Continuity-correct the boundary by 0.5 (table below).
Standardise $Z=\dfrac{\text{c.c. boundary}-np}{\sqrt{npq}}$ — use $\sqrt{npq}$ , never $npq$ .
Read Φ; finish (1−Φ for an upper tail, Φ(b)+Φ(a)−1 for a band).

Worked example

One-sided normal approximation

$X\sim B(110,0.25)$ , P(fewer than 22). Mean $np=27.5$ , variance $npq=20.625$ . C.C. $P(X\le21)\to21.5$ ; $Z=\dfrac{21.5-27.5}{\sqrt{20.625}}=-1.321$ ; $\Phi(-1.321)=1-0.9066$ . (9709/52/M/J/24 Q6)

Answer

\approx0.0934

Want P(…)	Corrected boundary	Then
$X<k$	$k-0.5$	$\Phi$
$X\le k$	$k+0.5$	$\Phi$
$X>k$	$k+0.5$	$1-\Phi$
$X\ge k$	$k-0.5$	$1-\Phi$
$a<X<b$	$a+0.5,\ b-0.5$	$\Phi(b)+\Phi(a)-1$

Name N(np, npq), shift the boundary by 0.5, standardise with √(npq) — never npq, never √σ.

Examiner note

Write np and npq as numbers before standardising (a separate B1), and the 0.5 continuity correction must be visibly present (its own M1). Standardise with

\sqrt{npq}

, not

npq

\sqrt\sigma

. (9709/52/O/N/24 Q7)

Common mistake

Standardising with the variance

npq

(or

\sqrt\sigma

) instead of

\sqrt{npq}

, and the wrong continuity-correction direction on a strict band (use

28.5

and

34.5

for “more than 28 but fewer than 35”).