The Normal distribution — standardisation, table precision, and conditional probability

$X \sim N(40, 6^2)$. Find $P(X < 46)$.

• $0.8413$

  Correct. $z = \dfrac{46-40}{6} = 1$, and $\Phi(1) = 0.8413$ is exactly the shape of probability asked for — 'less than' is what the table gives directly. No further arithmetic is needed or correct here.

• B$0.1587$

  This subtracts from 1 when nothing required it. This is the exact over-subtraction pattern a real examiner report records: 'went on to subtract the correct answer from 1, which of course is P(X > 18) and not P(X < 18) which is what was required' — the same error, different numbers.

• C$0.5$

  This treats every probability bound above or below the mean as automatically 50/50. Symmetry is a property of the whole curve about its mean — it says nothing about the area up to a specific point one standard deviation away, which is exactly what this question asks for.

• D$1.1587$

  A probability greater than 1 is impossible on inspection, before any method is even checked — the single cheapest error-catch available on this whole topic. This value comes from adding 1 and 0.1587 instead of subtracting, compounding one error with another.

Traps tested: Subtracted from 1 when not required · Symmetry misapplied as automatic fifty fifty · Arithmetic produces impossible probability

A question needs the z-value satisfying $P(Z > z) = 0.05$. Which value should be written down?

• $1.6449$

  Correct — the exact 4-d.p. value the facts bank verifies from the real formula booklet's Percentage Points table.

• B$1.64$

  The precise, real wrong value an examiner report names directly: 'not using an inaccurate value such as 1.64 instead of 1.6449.' Two decimal places looks reasonable; the mark scheme treats it as a lost accuracy mark.

• C$1.9600$

  This is the correct value for a DIFFERENT row of the same table — the percentage point at $p = 0.025$, not $p = 0.05$. Reading the wrong row of a table you're looking at correctly is its own, separate error from misremembering a value.

• D$0.6449$

  The leading '1' has been dropped. A value under 1 standard deviation from the mean corresponds to a tail probability well above 0.05 — checking the answer against the size of probability asked for would have caught this immediately.

Traps tested: Rounded z value instead of 4dp table value · Wrong row of percentage points table read · Leading digit dropped from table value

$Y \sim N(70, 10^2)$. Given that $P(Y < y) = 0.05$, find the value of $y$.

• $y = 53.6$

  Correct. $P(Y < y) = 0.05$ is a LEFT tail — y lies below the mean — so z must be negative: $z = -1.6449$. $y = 70 + (-1.6449)(10) = 70 - 16.449 = 53.551 \approx 53.6$ (3 s.f.).

• B$y = 86.4$

  This uses the table's value as printed, $z = +1.6449$, without flipping the sign for a left-tail probability. This is the documented sign error, reversed: an October 2021 examiner report records the same family of mistake giving 'a value higher than the upper limit' — an answer checkable as wrong from the question's own shape.

• C$y = 68.4$

  This computes $70 - 1.6449 = 68.3551$ — using the z-value directly as a shift in the original units, without multiplying by $\sigma = 10$ first. Standardisation is $Z = (X-\mu)/\sigma$; reversing it requires multiplying z by $\sigma$ before it can be added to or subtracted from $\mu$.

• D$y = 50.4$

  This uses $z = 1.9600$, the Percentage Points table's entry for $p = 0.025$, in a question that needs $p = 0.05$ — the correct technique, applied to the wrong row.

Traps tested: Sign error reversing standardisation · Z value not scaled by sigma · Wrong row of percentage points table read

$M \sim N(25, 4^2)$. Given that $M > 20$, find $P(M > 28 \mid M > 20)$.

• $0.2534$

  Correct. $P(M > 28)$: $z = 0.75$, $\Phi(0.75) = 0.7734$, so $P(M > 28) = 0.2266$. $P(M > 20)$: $z = -1.25$, so $P(M > 20) = \Phi(1.25) = 0.8944$ by symmetry. Since $\{M>28\} \subset \{M>20\}$, $P(M > 28 \mid M > 20) = 0.2266 / 0.8944 = 0.2534$.

• B$0.2266$

  This is $P(M > 28)$ alone, with no division at all — the given condition $M > 20$ treated as background information rather than the denominator it actually is. This is the single most consistently documented error on this exact skill across every examiner report reviewed for this topic.

• C$2.146$

  This divides by $P(M < 20) = 0.1056$ instead of $P(M > 20) = 0.8944$ — the complement of the given condition, not the condition itself. A probability greater than 1 is impossible on inspection, which is exactly why checking a conditional-probability answer against $[0, 1]$ before moving on is worth the five seconds it costs.

• D$0.4532$

  This divides by $P(M > 25) = 0.5$ — the probability of exceeding the MEAN — rather than $P(M > 20)$, the condition the question actually gives. The mean is a natural but wrong default to reach for; the denominator has to be read from the question, not assumed.

Traps tested: Conditional probability not recognised with normal · Wrong complement used as denominator · Denominator defaulted to the mean instead of the given condition

A 'show that' question asks a student to show that $P(X > 24) = 0.15$ for some $X \sim N(\mu, \sigma^2)$.

Which response is certain to earn full marks?

• Write the standardisation, $z = (24 - \mu)/\sigma$, evaluate it, look up the table value, subtract from 1 if required, and show the resulting figure rounds to 0.15

  Correct. Every intermediate step is visible on the page — the standardisation, the table value, and the final rounding — which is exactly what a 'show that' mark scheme needs something to attach a mark to.

• BWrite '$P(X > 24) = 0.15$' directly, since that's the answer given in the question

  This is the precise, real failure an examiner report records on a genuine 'show that' Normal-distribution question: some students 'simply wrote down 0.1500… with no working at all.' A given answer earns nothing on its own — the working IS what's being marked.

• CWrite the standardisation and the final rounded value, but skip showing the intermediate table value

  Also a documented real failure, from the same examiner report: candidates who 'wrote down the standardisation followed by 0.15 missing out the intermediate step and the accurate answer.' The mark scheme for a show-that part is built around the intermediate step being visible, not just the bookends.

• DUse a calculator to confirm $P(X > 24) = 0.15$ is plausible, and state that this must be the correct value

  Plausibility is not proof, and a 'show that' question specifically asks for the working that gets there, not a check that the given answer seems reasonable. This earns nothing under a mark scheme built around the method, not the destination.

Traps tested: Show that standardisation not shown