Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 6: Probability & Statistics 2 · official syllabus
LINEAR COMBINATIONS LAB
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Four short missions on transformed means and variances, independent normal differences and Poisson sums. Keep signed mean coefficients separate from squared variance coefficients.
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Mission 1 of 4
Shift and scale the mean
X has expectation 5 and variance 4. Define W=2X-3. Set E(W). An expectation follows the linear transformation, including the constant shift.
Does adding a constant 3 change the expectation?
Both number lines use the same units. Dots mark means, and teal segments span one standard deviation either side, not all possible values.
Your expectation is 5. Double 5, then subtract 3.
Study notes
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Transform a mean 6.2
For constants a and b, E(aX+b)=aE(X)+b. Scaling and shifting the values scale and shift their expectation in the same way. With E(X)=5, E(2X-3)=7. This rule does not assume that X is normal or Poisson.
Linearity also gives E(aX+bY)=aE(X)+bE(Y), without an independence requirement. If E(X)=5 and E(Y)=9, E(X-Y)=-4. A difference can have negative expectation even if both original expectations are positive.
Define the random variable whose expectation is required. A new independent observation X2 with the same distribution as X is not the same variable as X. For example X+X=2X uses the same observation twice, whereas X+X2 combines two observations.
Transform variance and standard deviation 6.2
Var(aX+b)=a squared times Var(X). The constant shift b does not change spread about the mean. A negative multiplier reflects the values but its square still gives non-negative variance. Standard deviation is multiplied by the absolute value of a.
With Var(X)=4 and W=2X-3, Var(W)=4 times 4=16 and standard deviation is 4. With V=-2X+10, variance is still 16, while E(V)=0 when E(X)=5. Means and variances obey different transformation rules.
The variance of a constant is zero. Do not add or subtract b, or b squared, to a variance merely because the variable was shifted by b. Keep squared units for variance and original units for standard deviation.
Combine independent variances 6.2
For independent X and Y, Var(aX+bY)=a squared Var(X)+b squared Var(Y). State the independence assumption when using this result. A negative coefficient is squared, so independent differences add variance contributions rather than subtracting them.
For independent X and Y with variances 4 and 9, Var(X-Y)=4+9=13 and Var(2X-3Y)=4(4)+9(9)=97. Expectations use the signed coefficients: with means 5 and 9, E(2X-3Y)=10-27=-17.
Do not use independent variance addition for two appearances of the same variable. Var(X+X)=Var(2X)=4Var(X), but Var(X+X2)=2Var(X) when X2 is an independent copy. Independence is sufficient for the stated rule; the rule is not a way to infer independence from a single numerical equality.
Transform and combine normal variables 6.2
If X is normal, aX+b is normal for nonzero a, with the transformed mean and variance. With X distributed as N(5,4), 2X-3 has distribution N(7,16). When a=0 the result is a constant, a degenerate case rather than an ordinary positive-variance normal density.
A linear combination of independent normal variables is normal, with the linear-combination expectation and independent variance formula. For independent X distributed as N(5,4) and Y distributed as N(9,9), D=X-Y has distribution N(-4,13). Use sqrt(13), not 13, to standardise D.
For this difference, P(X>Y)=P(D>0)=P(Z>4/sqrt(13)), approximately 0.134 to 3 significant figures. The event concerns the difference from the same pair of observations, not two separate tail probabilities multiplied together.
Add independent Poisson counts 6.2
If independent X and Y have distributions Po(lambda1) and Po(lambda2), their sum has distribution Po(lambda1+lambda2). Both mean and variance of the sum equal that combined parameter. Check that the counts and exposure intervals have been defined consistently.
For independent Po(2) and Po(5) counts, X+Y has distribution Po(7). Its zero-count probability is exp(-7), also equal to P(X=0)P(Y=0)=exp(-2)exp(-5) under independence.
This rule does not say every linear combination of Poisson variables is Poisson. A difference can be negative, while 2X takes only even values. For X distributed as Po(2), E(2X)=4 and Var(2X)=8, so 2X cannot have the Po(4) distribution, which would have variance 4.
Choose the correct identity before calculating 6.2
Write the required variable as an explicit expression, keeping signed coefficients. Calculate its expectation and variance separately. Decide whether the given distribution and independence information also establish a distributional form; two moments alone do not identify an arbitrary distribution.
If a normal probability is required, standardise only after finding the resulting mean and variance. Linear combinations of continuous normal variables need no continuity correction. A continuity correction belongs to a separate approximation from a discrete count.
Check that a variance is non-negative and that any claimed Poisson variable is an appropriate integer count. Distinguish multiplying one observation from summing independent observations. A formula without its assumptions can give a plausible number for the wrong experiment.
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