Mathematics 9709 · Probability & Statistics 2

Combining normal distributions

Add or scale independent normals and you are still on a bell curve, just relocated. So the only work is: find the new mean and variance with the two rules, write $N(\text{mean},\text{variance})$ , then it is an ordinary normal probability. No continuity correction, these are already continuous.

Combine into one bell, then read a tail

The reading-a-tail step is identical to ordinary normal work, so start with the only part that is ever new: combining into one bell's mean and variance.

Worked example

Easiest first: just the mean and variance

A worktop joins chipboard $C\sim N(37,0.3^2)$ and melamine $M\sim N(1,0.01^2)$ , independent. Find the thickness distribution with melamine on (a) one face, (b) both faces.

(a) $C+M$ : mean $37+1=38$ , variance $0.3^2+0.01^2=0.0901$ .
(b) $C+M_1+M_2$ : mean $37+1+1=39$ , variance $0.3^2+0.01^2+0.01^2=0.0902$ .

Answer

N(38,0.0901)\ \text{and}\ N(39,0.0902)

Two melamine layers are two independent variables, so the small variance is added twice.

Worked example

Is one thing more than k times another?

Small bags $X\sim N(2.1,0.2^2)$ , large $Y\sim N(6.6,0.4^2)$ , independent. $P(Y>3X)=P(D>0)$ where $D=Y-3X$ :

$E(D)=6.6-3(2.1)=0.3$ .
$\operatorname{Var}(D)=0.4^2+3^2(0.2^2)=0.16+0.36=0.52$ (the 3 squares).
$P(D>0)=1-\Phi\!\big(\tfrac{0-0.3}{\sqrt{0.52}}\big)=1-\Phi(-0.416)=\Phi(0.416)$ .

Answer

P(D>0)\approx0.661

Combine the two normals into one bell D ~ N(0.3, 0.52); the question is then just a normal tail: P(D > 0) ≈ 0.661.

Build the ONE combined bell N(mean, variance), then z = (value − mean) / √variance and read a plain normal tail.

That last step is always the same move: standardise the value into a $z$ , then read the shaded area off the standard normal. Drag the boundary below to see how $z$ and the tail trade off.

z = 1.00·Φ(z) = P(Z < z) = 0.8413·right tail 0.1587

Here $z=\tfrac{0-0.3}{\sqrt{0.52}}=-0.416$ , so the left area is $\Phi(-0.416)=0.339$ and the answer is the right tail $1-0.339=0.661$ .

Worked example

Coefficient plus an extra variable, upper limit

Three soap bars $S\sim N(50,4^2)$ plus packaging $T\sim N(90,5^2)$ , independent. A package posts cheap if its mass is under 250 g, i.e. $P(3S+T<250)$ .

$E(3S+T)=3(50)+90=240$ .
$\operatorname{Var}(3S+T)=3^2(16)+25=144+25=169$ , so $3S+T\sim N(240,169)$ .
$P(3S+T<250)=\Phi\!\big(\tfrac{250-240}{13}\big)=\Phi(0.769)$ .

Answer

P(3S+T<250)\approx0.779

Worked example

Total of n is a SUM, not a multiple

Battery life $T\sim N(7,2.3^2)$ ; total of four, $S=T_1+\cdots+T_4$ :

$E(S)=4(7)=28$ .
$\operatorname{Var}(S)=4(2.3^2)=21.16$ (a sum, so $\times4$ , NOT $\times16$ ).
$P(S>30)=1-\Phi\!\big(\tfrac{30-28}{\sqrt{21.16}}\big)=1-\Phi(0.435)$ .

Answer

P(S>30)\approx0.332

Worked example

C ≥ 4T: a multiple with a big variance

Conti batteries $C\sim N(25,1.1^2)$ , Thrift $T\sim N(7,2.3^2)$ . Is a Conti at least four times a Thrift, $P(C\ge4T)=P(C-4T\ge0)$ ?

$E(C-4T)=25-4(7)=-3$ .
$\operatorname{Var}(C-4T)=1.1^2+4^2(2.3^2)=1.21+84.64=85.85$ (the 4 squares).
$P(C-4T>0)=1-\Phi\!\big(\tfrac{0-(-3)}{\sqrt{85.85}}\big)=1-\Phi(0.324)$ .

Answer

P(C\ge4T)\approx0.373

Worked example

Differ by more than k: two tails

Independent $X\sim N(24,2^2)$ , $Y\sim N(25,3^2)$ . Find the chance they differ by more than 4, i.e. $P(|X-Y|>4)$ .

Form $D=X-Y\sim N(24-25,\ 2^2+3^2)=N(-1,13)$ .
$|D|>4$ means $D>4$ OR $D<-4$ , so add BOTH tails.
$P(D>4)=1-\Phi\!\big(\tfrac{4-(-1)}{\sqrt{13}}\big)=1-\Phi(1.387)=0.083$ .
$P(D<-4)=\Phi\!\big(\tfrac{-4-(-1)}{\sqrt{13}}\big)=\Phi(-0.832)=0.203$ .

Answer

P(|X-Y|>4)\approx0.285

Because the means are not equal, the two tails differ in size; you must compute both, never just double one.

Examiner note

“Differ by more than k” (

|X-Y|>k

) is two-sided:

P(D>k)+P(D<-k)

D=X-Y

. A cost or score coefficient scales the variance SQUARED. (9709/62/M/J/23 Q5) (9709/62/O/N/23 Q6)

Why 'differ by more than k' needs two tails

$|X-Y|>k$ means $X-Y>k$ OR $X-Y<-k$ . Form $D=X-Y\sim N(\mu_X-\mu_Y,\ \sigma_X^2+\sigma_Y^2)$ and add both tails. They are equal only when $\mu_X=\mu_Y$ ; otherwise a single tail under- or over-counts.

Common mistake

“Total of

n

” is a sum (

\operatorname{Var}\times n

), not

nX

(

\times n^2

). Keep variance in the

N(\cdot,\cdot)

slot, standardise by

\sqrt{\text{variance}}

, and apply no continuity correction (these are already continuous normals).

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Combining normal distributions

Combine into one bell, then read a tail

The reading-a-tail step is identical to ordinary normal work, so start with the only part that is ever new: combining into one bell's mean and variance.

Worked example

Easiest first: just the mean and variance

A worktop joins chipboard $C\sim N(37,0.3^2)$ and melamine $M\sim N(1,0.01^2)$ , independent. Find the thickness distribution with melamine on (a) one face, (b) both faces.

(a) $C+M$ : mean $37+1=38$ , variance $0.3^2+0.01^2=0.0901$ .
(b) $C+M_1+M_2$ : mean $37+1+1=39$ , variance $0.3^2+0.01^2+0.01^2=0.0902$ .

Answer

N(38,0.0901)\ \text{and}\ N(39,0.0902)

Two melamine layers are two independent variables, so the small variance is added twice.

Worked example

Is one thing more than k times another?

Small bags $X\sim N(2.1,0.2^2)$ , large $Y\sim N(6.6,0.4^2)$ , independent. $P(Y>3X)=P(D>0)$ where $D=Y-3X$ :

$E(D)=6.6-3(2.1)=0.3$ .
$\operatorname{Var}(D)=0.4^2+3^2(0.2^2)=0.16+0.36=0.52$ (the 3 squares).
$P(D>0)=1-\Phi\!\big(\tfrac{0-0.3}{\sqrt{0.52}}\big)=1-\Phi(-0.416)=\Phi(0.416)$ .

Answer

P(D>0)\approx0.661

Combine the two normals into one bell D ~ N(0.3, 0.52); the question is then just a normal tail: P(D > 0) ≈ 0.661.

Build the ONE combined bell N(mean, variance), then z = (value − mean) / √variance and read a plain normal tail.

That last step is always the same move: standardise the value into a $z$ , then read the shaded area off the standard normal. Drag the boundary below to see how $z$ and the tail trade off.

z = 1.00·Φ(z) = P(Z < z) = 0.8413·right tail 0.1587

Here $z=\tfrac{0-0.3}{\sqrt{0.52}}=-0.416$ , so the left area is $\Phi(-0.416)=0.339$ and the answer is the right tail $1-0.339=0.661$ .

Worked example

Coefficient plus an extra variable, upper limit

Three soap bars $S\sim N(50,4^2)$ plus packaging $T\sim N(90,5^2)$ , independent. A package posts cheap if its mass is under 250 g, i.e. $P(3S+T<250)$ .

$E(3S+T)=3(50)+90=240$ .
$\operatorname{Var}(3S+T)=3^2(16)+25=144+25=169$ , so $3S+T\sim N(240,169)$ .
$P(3S+T<250)=\Phi\!\big(\tfrac{250-240}{13}\big)=\Phi(0.769)$ .

Answer

P(3S+T<250)\approx0.779

Worked example

Total of n is a SUM, not a multiple

Battery life $T\sim N(7,2.3^2)$ ; total of four, $S=T_1+\cdots+T_4$ :

$E(S)=4(7)=28$ .
$\operatorname{Var}(S)=4(2.3^2)=21.16$ (a sum, so $\times4$ , NOT $\times16$ ).
$P(S>30)=1-\Phi\!\big(\tfrac{30-28}{\sqrt{21.16}}\big)=1-\Phi(0.435)$ .

Answer

P(S>30)\approx0.332

Worked example

C ≥ 4T: a multiple with a big variance

Conti batteries $C\sim N(25,1.1^2)$ , Thrift $T\sim N(7,2.3^2)$ . Is a Conti at least four times a Thrift, $P(C\ge4T)=P(C-4T\ge0)$ ?

$E(C-4T)=25-4(7)=-3$ .
$\operatorname{Var}(C-4T)=1.1^2+4^2(2.3^2)=1.21+84.64=85.85$ (the 4 squares).
$P(C-4T>0)=1-\Phi\!\big(\tfrac{0-(-3)}{\sqrt{85.85}}\big)=1-\Phi(0.324)$ .

Answer

P(C\ge4T)\approx0.373

Worked example

Differ by more than k: two tails

Independent $X\sim N(24,2^2)$ , $Y\sim N(25,3^2)$ . Find the chance they differ by more than 4, i.e. $P(|X-Y|>4)$ .

Form $D=X-Y\sim N(24-25,\ 2^2+3^2)=N(-1,13)$ .
$|D|>4$ means $D>4$ OR $D<-4$ , so add BOTH tails.
$P(D>4)=1-\Phi\!\big(\tfrac{4-(-1)}{\sqrt{13}}\big)=1-\Phi(1.387)=0.083$ .
$P(D<-4)=\Phi\!\big(\tfrac{-4-(-1)}{\sqrt{13}}\big)=\Phi(-0.832)=0.203$ .

Answer

P(|X-Y|>4)\approx0.285

Because the means are not equal, the two tails differ in size; you must compute both, never just double one.

Examiner note

“Differ by more than k” (

|X-Y|>k

) is two-sided:

P(D>k)+P(D<-k)

D=X-Y

. A cost or score coefficient scales the variance SQUARED. (9709/62/M/J/23 Q5) (9709/62/O/N/23 Q6)

Why 'differ by more than k' needs two tails

Common mistake

“Total of

n

” is a sum (

\operatorname{Var}\times n

), not

nX

(

\times n^2

). Keep variance in the

N(\cdot,\cdot)

slot, standardise by

\sqrt{\text{variance}}

, and apply no continuity correction (these are already continuous normals).