Read cumulative normal tables and reflect negative values
Locate a standardised value in a left-tail table and use symmetry or subtraction for the required region.
Exam scope and assessment notes
Probability & Statistics 1 (Paper 5, AS & A Level) · Mathematics 9709, 2026–2027, section 5.5. N(μ,σ²) uses variance as its second parameter, with σ > 0; probabilities are areas, not density heights. Show the full standardisation Z = (X − μ)/σ. The MF19 table gives Φ(z) = P(Z ≤ z), with negative values found by symmetry. Model sketches and shaded probabilities may be required. Probability answers use quoted four-decimal table values where supplied; calculations with non-tabulated z use unrounded standardisation and numerical normal probabilities, so table interpolation can differ slightly in the last digits. Inverse examples use the quoted three-decimal MF19 critical values and are approximate. A binomial normal approximation requires both np > 5 and n(1 − p) > 5, and a continuity correction. It remains approximate, including for integer probability thresholds. A continuous normal variable has zero probability at a single exact value; a discrete binomial count does not. Sample means, the Central Limit Theorem and hypothesis testing belong to Statistics 2 and are not covered here. Model assumptions need evidence in real applications. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Use decimal place value.
- Identify left tails, right tails and intervals.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which region does Φ(z) describe in MF19?
What you will learn
- Read Φ(z) as P(Z ≤ z) from the MF19 table.
- Use complements, reflection and cumulative differences.
Read what the table actually gives
For Z ~ N(0,1), the MF19 normal table gives Φ(z) = P(Z ≤ z), the area to the left. For z = 1.23, use row 1.2 and the hundredths column 3 to read 0.8907. This is not the area between zero and 1.23 and not the right tail.
The upper probability is P(Z > 1.23) = 1 − 0.8907 = 0.1093. If a different table convention is supplied, read its heading rather than assuming it is cumulative. The optional ADD columns in MF19 refine a third decimal; probabilities below use the stated entries rather than implying unlimited numerical precision.
Reflect and subtract complete areas
Symmetry gives Φ(−z) = 1 − Φ(z). Thus P(Z ≤ −1.23) = 1 − 0.8907 = 0.1093. The negative threshold lies left of the mean, so its left-tail area must be smaller than one half; this provides a useful sign check.
For −0.50 < Z < 1.20, subtract the lower cumulative area from the upper: Φ(1.20) − Φ(−0.50). Using Φ(1.20) = 0.8849 and Φ(0.50) = 0.6915, the lower area is 0.3085 and the interval is 0.5764. Adding the cumulative values would double-count unwanted regions. Strict and inclusive endpoints agree for a continuous variable.
Pause and explain
If Φ(1.23) = 0.8907, what is Φ(−1.23)?
Worked example
Use the cumulative table to find P(Z > 1.23) when Φ(1.23) = 0.8907.
Show the worked solution
- Identify 0.8907 as the left-tail probability.
- The required right tail is the complementary region.
- Subtract from one: 1 − 0.8907 = 0.1093.
Answer 0.1093 using the quoted table entry.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Read Φ(1.23) = 0.8907 and find P(Z > 1.23).
Give me a hint
The supplied value is the left tail; subtract it from one.
Compare my reasoning
- Locate row 1.2 and hundredths column 3.
- The left-tail value is 0.8907.
- The right tail is 1 − 0.8907 = 0.1093.
Φ(1.23) = 0.8907; P(Z > 1.23) = 0.1093.
Look for these in your work
- Use the correct row and column.
- Identify the required shaded region.
Using Φ(1.23) = 0.8907, calculate P(Z ≤ −1.23).
Give me a hint
Reflect the negative value about zero.
Compare my reasoning
- The standard normal curve is symmetric.
- Φ(−1.23) = 1 − Φ(1.23).
- The probability is 0.1093, less than one half as expected.
0.1093.
Look for these in your work
- Use symmetry rather than a negative probability.
- Check the result agrees with the boundary’s side of zero.
Use Φ(1.20) = 0.8849 and Φ(0.50) = 0.6915 to find P(−0.50 < Z < 1.20).
Give me a hint
Reflect the lower boundary and subtract its cumulative area.
Compare my reasoning
- Φ(−0.50) = 1 − 0.6915 = 0.3085.
- The interval area is Φ(1.20) − Φ(−0.50).
- Subtract 0.8849 − 0.3085 = 0.5764.
0.5764 using the quoted table values.
Look for these in your work
- Reflect the negative lower endpoint.
- Subtract the lower cumulative area only once.
Common mistakes
- Check the table’s area convention.
- Reflect negative z using the complement.
- Subtract lower cumulative area from upper cumulative area for an interval.
What can you explain now?
Is Φ(1.23) the area between zero and 1.23?
Compare with the explanation
No; it includes the entire left tail.
The area from zero to 1.23 is Φ(1.23) − 1/2 instead.
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Make it stick
Read the table heading, locate z, and sketch whether you need its left tail, complement or a difference of cumulative areas.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.