Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 5: Probability & Statistics 1 · official syllabus
NORMAL DISTRIBUTION LAB
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Four short missions on standardisation, normal table tails, inverse cutoffs and continuity correction. Sketch the event, predict the rule, then calculate with the stated model.
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Mission 1 of 4
Measure distance in standard deviations
X has normal distribution N(100,20 squared): mean 100 and standard deviation 20. Set the standardised value of x=130. Subtract the mean, then divide by the standard deviation, not the variance.
The denominator in z=(x-mean)/... is the...
The curve uses standardised z units. The dashed line follows your z value; the right-hand shaded region shows the area above that boundary.
Your z is 0.0. Subtract 100 from 130 and divide by 20.
Study notes
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Describe a normal model and sketch its area 5.5
For X distributed as N(mu,sigma squared), mu is the mean and sigma is the positive standard deviation. The second parameter is the variance. The density curve is symmetric about mu, and its total area is 1. A larger sigma spreads the curve out; it does not change its total area.
Normal probabilities are areas under a continuous density curve. The height at a point is not the probability of that exact value. For a continuous normal variable, P(X=a)=0, so strict and non-strict endpoint inequalities give the same probability. This does not hold for a discrete binomial count.
Sketch the curve, mark the mean and requested boundary or boundaries, then shade the event before using a table. The position of the boundary relative to the mean gives a useful check: a right tail above the mean must be less than 1/2.
Standardise using the standard deviation 5.5
Z=(X-mu)/sigma has the standard normal distribution N(0,1). Write the event with its inequality before standardising. Because sigma is positive, standardisation preserves the inequality direction. Show the substitution and working, not only a calculator probability.
Worked example: X has distribution N(100,400), so sigma=sqrt(400)=20. For x=130, z=(130-100)/20=1.5. Dividing by 400 would incorrectly use the variance instead of the standard deviation.
Values below the mean standardise to negative z; those above it to positive z. A value one standard deviation below the mean gives z=-1. Check signs against the sketch before reading a table.
Read cumulative normal tables and use symmetry 5.5
Here Phi(z)=P(Z<=z), the area to the left of z. Check a table's heading: some tables instead give the area between 0 and z or an upper tail. Convert its entry to the event required rather than assuming every table is cumulative.
For a cumulative table, P(Z>z)=1-Phi(z), Phi(-z)=1-Phi(z), and P(a<Z<b)=Phi(b)-Phi(a). As an example, Phi(1.50) is approximately 0.9332, so the upper tail is approximately 0.0668, both to 4 d.p. By symmetry Phi(-1.50) is approximately 0.0668.
Worked interval example: P(-1.50<Z<1.50)=2Phi(1.50)-1, approximately 0.8664 using the rounded table value. The two omitted tails together have probability approximately 0.1336. Keep the calculation as an area difference so a negative or over-1 result is easy to detect.
Work backwards from a probability 5.5
Convert the stated tail to a cumulative area, find the corresponding standard normal quantile z, then use x=mu+sigma z. Keep the sign consistent with the sketch. A small upper tail requires a positive z; a small lower tail a negative z.
Worked example: X has mean 100 and standard deviation 20. For P(X>c)=0.05, the cumulative area is 0.95. Using the approximate table quantile z=1.645 gives c approximately 100+20(1.645)=132.9. The quantile is rounded, so the cutoff is an approximation, not an exact identity with the true 95th percentile.
If mu or sigma is unknown, standardisation gives a relationship such as (c-mu)/sigma=z. Two independent percentile conditions can supply two equations. Solve them and check sigma>0. Do not replace sigma by sigma squared in either equation.
Approximate a binomial only under suitable conditions 5.5
For a binomial count X with parameters n and p, use a normal approximation with mean np and variance np(1-p) when n is sufficiently large for both np>5 and n(1-p)>5, the conditions specified by this syllabus. These inequalities do not make every approximation exact; the binomial count remains discrete.
For X distributed as B(100,0.3), np=30 and n(1-p)=70, so the stated conditions hold. The approximating normal Y has mean 30 and variance 21, hence standard deviation sqrt(21). Use that square root in standardisation.
A binomial count's integer bars have width 1. A continuity correction includes the full bars required by the discrete event before taking a continuous normal area. For example X<=24 becomes Y<24.5, while X>=25 becomes Y>24.5.
Apply a continuity correction before standardising 5.5
Other common translations are X<24 to Y<23.5; X>24 to Y>24.5; and 20<=X<=24 to 19.5<Y<24.5. For a single value X=24, use 23.5<Y<24.5. Draw the required integer bars when uncertain rather than attaching 0.5 mechanically.
For B(100,0.3), approximate P(X<=24) by P(Y<24.5). Then z=(24.5-30)/sqrt(21), approximately -1.2002 to 4 d.p. The approximate probability is about 0.115. The exact binomial probability is a different calculation, so label the normal result as an approximation.
Check whether the original question is already about a continuous normal variable. Such a problem needs no binomial continuity correction. When a normal model is itself an approximation to a discrete count, show the binomial parameters, suitability checks, corrected event, normal parameters and standardisation.
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