6.2 - Normal and Poisson combinations | Cambridge A-Level Mathematics

Some distributions retain their form when random variables are combined. Learn to find probabilities for transformed normal measurements, independent normal sums and differences, and independent Poisson totals, while keeping the conditions for each result clear.

A transformed normal variable

If a normally distributed reading is doubled, is the new standard deviation doubled or quadrupled? The standard deviation doubles; the variance quadruples. More is true: the entire distribution remains normal.

A shift moves the bell curve, a positive scale stretches its horizontal distances, and a negative scale reflects it as well. These operations preserve the normal shape.

Transforming a normal variable

If $X\sim\mathrm N(\mu,\sigma^2)$ , $\sigma>0$ , and $a\ne0$ , then

aX+b\sim\mathrm N(a\mu+b,a^2\sigma^2).

The second parameter is variance. Standard deviation is $|a|\sigma$ .
Not in the formula booklet: learn this distribution rule.

If $a=0$ , the result is the constant $b$ , with zero variance rather than a normal curve of positive spread. A normal calculation here is exact under the stated model; no continuity correction is involved.

Worked example 1: a scaled normal probability

Question. Let $X\sim\mathrm N(50,4)$ and $Y=2X+10$ . State the distribution of $Y$ and find $\mathrm P(Y<115)$ .

Plan. Transform the mean and variance, then standardise the requested boundary.

\begin{aligned} \mathrm E(Y)&=2(50)+10=110\\ \operatorname{Var}(Y)&=2^2(4)=16\\ Y&\sim\mathrm N(110,16)\\ \mathrm P(Y<115)&=\Phi\left(\frac{115-110}{4}\right) =\Phi(1.25)=0.8944 \end{aligned}

Answer. $Y\sim\mathrm N(110,16)$ , with probability $0.894$ to $3$ significant figures.

Check. Directly, $2X+10<115$ means $X<52.5$ . Standardising with the original standard deviation $2$ gives $(52.5-50)/2=1.25$ , the same boundary.

Reflect the scale while choosing a shift that keeps the new centre positive. The inequality changes direction when it is solved back in terms of $X$ .

Worked example 2: a reflected normal probability

Question. Let $X\sim\mathrm N(50,4)$ and $W=-2X+130$ . State the distribution of $W$ and find $\mathrm P(W>25)$ .

Plan. Square the negative scale for variance, then find the right-tail area for the transformed variable.

\begin{aligned} \mathrm E(W)&=-2(50)+130=30\\ \operatorname{Var}(W)&=(-2)^2(4)=16\\ W&\sim\mathrm N(30,16)\\ \mathrm P(W>25)&=1-\Phi\left(\frac{25-30}{4}\right)\\ &=1-\Phi(-1.25)=0.8944 \end{aligned}

Answer. $W\sim\mathrm N(30,16)$ , with probability $0.894$ .

Check. Solving $-2X+130>25$ gives $X<52.5$ , reversing the inequality when dividing by $-2$ . This is exactly the original-variable event in the preceding check.

The new mean alone does not locate a requested tail. Mark the mean and boundary before choosing the left or right area.

Watch out

A negative multiplier does not give a negative variance or standard deviation. It reverses order, which matters if you solve the event back in terms of the original variable.

Use both the distribution rule and standardisation for a fresh transformation.

Independent normal combinations

Knowing a mean and variance usually does not identify a distribution. Independent normal variables are a useful exception: their linear combination is also normal, so the moment rules give enough information for a new normal probability calculation.

Combining independent normal variables

If $X\sim\mathrm N(\mu_X,\sigma_X^2)$ and $Y\sim\mathrm N(\mu_Y,\sigma_Y^2)$ are independent, then, for constants $a,b$ not both zero,

aX+bY\sim\mathrm N\left( a\mu_X+b\mu_Y,\, a^2\sigma_X^2+b^2\sigma_Y^2 \right).

The variances add with squared coefficients, including for differences. A fixed shift can then be applied using the preceding part.
Not in the formula booklet: learn this distribution rule.

Translate a comparison between measurements into one random variable compared with a fixed number. For example, $X>2Y$ becomes $X-2Y>0$ . The resulting difference can have a small mean even though both original means are large.

Worked example 3: a weighted total

Question. Independent variables have distributions $X\sim\mathrm N(50,9)$ and $Y\sim\mathrm N(24,4)$ . Find $\mathrm P(X+2Y>104)$ .

Plan. Form the weighted total, find its normal distribution and standardise $104$ .

For $T=X+2Y$ ,

\begin{aligned} \mathrm E(T)&=50+2(24)=98\\ \operatorname{Var}(T)&=9+2^2(4)=25\\ T&\sim\mathrm N(98,25)\\ \mathrm P(T>104)&=1-\Phi\left(\frac{104-98}{5}\right)\\ &=1-\Phi(1.2)=0.1151 \end{aligned}

Answer. $0.115$ .

Check. The threshold $104$ is $6$ above the mean, or $1.2$ standard deviations. Its right-tail area must be below $0.5$ . Reversing the standardisation gives $98+1.2(5)=104$ .

Change the sign of the $Y$ contribution to make a comparison. The variance stays $25$ , but the mean changes from $98$ to $2$ .

Worked example 4: comparing two measurements

Question. For the same independent $X,Y$ , find $\mathrm P(X>2Y)$ .

Plan. Define $D=X-2Y$ , then find the normal area above $0$ .

\begin{aligned} \mathrm E(D)&=50-2(24)=2\\ \operatorname{Var}(D)&=9+(-2)^2(4)=25\\ D&\sim\mathrm N(2,25)\\ \mathrm P(X>2Y)&=\mathrm P(D>0)\\ &=1-\Phi\left(\frac{0-2}{5}\right)\\ &=1-\Phi(-0.4)=\Phi(0.4)=0.6554 \end{aligned}

Answer. $0.655$ .

Check. The threshold $0$ lies below the mean $2$ , so the right area exceeds $0.5$ . Equivalently $2Y-X\sim\mathrm N(-2,25)$ , and its left area below $0$ is $\Phi((0+2)/5)=\Phi(0.4)$ .

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6.2 - Normal and Poisson combinations

A transformed normal variable

Independent normal combinations

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