KBAT: Normal Distribution Reasoning

Q: Why is 'within 6 cm of the mean' so quick to evaluate?

Because 666 cm is exactly one standard deviation, so the interval is μ±σ\mu\pm\sigmaμ±σ, which standardises to −1<Z<1-1<Z<1−1<Z<1. By symmetry P(−1<Z<1)=1−2P(Z>1)=1−2(0.1587)=0.6826P(-1<Z<1)=1-2P(Z>1)=1-2(0.1587)=0.6826P(−1<Z<1)=1−2P(Z>1)=1−2(0.1587)=0.6826, the familiar 'about 68% within one standard deviation'.

Q: Can I earn marks if my final proportion is wrong?

Yes. Add Math Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line. Correct z-values, valid standardisation equations and a sound simultaneous solution can score well even if an arithmetic slip changes the final figure.

What makes this a KBAT question

A routine normal-distribution question gives you the mean and standard deviation and asks for a probability: standardise, read the table, done. A KBAT question turns that around, it gives you the probabilities and hides the mean or the standard deviation, and expects you to recover them.

That is Kemahiran Berfikir Aras Tinggi, higher-order thinking: standardising with $z=\dfrac{x-\mu}{\sigma}$ and reading the standard normal table are ordinary Form 5 skills, but running the table in reverse, turning a tail probability back into a z-value, and then writing one equation per condition to solve for two unknowns are steps no one lists for you. In Add Math this rewards students who see that each probability statement is really an equation in $\mu$ and $\sigma$ , and who can recognise when a range is symmetric about the mean so a table value can be reused.

The interpretation and the setup, not the arithmetic, carry the difficulty.

One worked problem, in the style of Paper 2

This is an original question written in the style of SPM Paper 2. Try it yourself before reading the solution.

Standard normal values used: $P(Z>1)=0.1587$ and $P(Z>2)=0.0228$ .

Q1[8 marks]

The lengths of metal rods produced by a machine are normally distributed with mean $\mu$ cm and standard deviation $\sigma$ cm. It is found that $15.87\%$ of the rods are longer than $82$ cm, and $2.28\%$ are longer than $88$ cm.

(a) Find the values of $\mu$ and $\sigma$ . (b) A rod is graded 'A' if its length is within $6$ cm of the mean.

Find the proportion of rods graded 'A'.

Show worked solution

Understand. We do not know $\mu$ or $\sigma$ , but we are told two tail probabilities.

Each one, once turned into a z-value, becomes a standardisation equation. Two equations in two unknowns can be solved together.

Part (b) then uses the recovered $\mu$ and $\sigma$ .

Plan. Read each percentage back into a z-value using the standard normal table in reverse.

Write $z=\dfrac{x-\mu}{\sigma}$ for each. Solve the pair for $\mu$ and $\sigma$ .

For (b), turn 'within 6 cm of the mean' into a symmetric interval and use symmetry.

Execute and check. (a) A tail of $15.87\%=0.1587$ above $82$ means $P(Z>z_{1})=0.1587$ , so $z_{1}=1$ .

A tail of $2.28\%=0.0228$ above $88$ means $P(Z>z_{2})=0.0228$ , so $z_{2}=2$ . Standardising each length:

\frac{82-\mu}{\sigma}=1 \quad\text{and}\quad \frac{88-\mu}{\sigma}=2

Clear the denominators to get two linear equations:

82-\mu=\sigma \quad(1)\qquad 88-\mu=2\sigma \quad(2)

Subtract (1) from (2): $(88-\mu)-(82-\mu)=2\sigma-\sigma$ , giving $6=\sigma$ . Then from (1), $\mu=82-\sigma=82-6=76$ .

\mu=76\ \text{cm},\qquad \sigma=6\ \text{cm}

(b) 'Within 6 cm of the mean' means $76-6<X<76+6$ , i.e. $70<X<82$ . Standardise the endpoints with $\sigma=6$ :

z=\frac{70-76}{6}=-1 \qquad\text{and}\qquad z=\frac{82-76}{6}=1

So we need $P(-1<Z<1)$ . By symmetry this is $1$ minus the two equal tails:

P(-1<Z<1)=1-2P(Z>1)=1-2(0.1587)=0.6826

About $68.26\%$ of the rods are graded 'A'.

Check. Since $6$ cm is exactly one standard deviation, the Grade A interval is $\mu\pm\sigma$ , and the well-known 'about $68\%$ within one standard deviation' agrees with $0.6826$ .

Substituting back, $(82-76)/6=1$ and $(88-76)/6=2$ reproduce the two given tails, so $\mu$ and $\sigma$ are consistent with the data.

Finding a sensible first step

When a normal question hides the mean or standard deviation, the reliable first step is to convert every probability into a z-value before you write anything else. Read the standard normal table in reverse: a tail area of $0.1587$ corresponds to $z=1$ , and $0.0228$ to $z=2$ .

Each z-value then gives you one standardisation equation, $z=\dfrac{x-\mu}{\sigma}$ , linking the known length to the unknown parameters. With two conditions you have two equations, and clearing the denominators makes them linear and easy to solve simultaneously.

Getting the z-values right is worth more than any later algebra, because a wrong sign or a misread tail poisons both equations. Only once the parameters are found do you move on to the next part, and there, always check whether a range is symmetric about the mean, because symmetry lets you reuse a single table value.

What markers reward

Marking is analytic, so method marks are awarded line by line. On a normal-distribution reasoning question a marker looks for:

Each percentage converted to a z-value by reverse table reading: $0.1587\to z=1$ , $0.0228\to z=2$ .
A standardisation equation written for each condition: $\frac{82-\mu}{\sigma}=1$ and $\frac{88-\mu}{\sigma}=2$ .
The denominators cleared to give the linear pair $82-\mu=\sigma$ and $88-\mu=2\sigma$ .
The simultaneous solution $\sigma=6$ , $\mu=76$ , with working shown.
Recognition that 'within 6 cm' is the symmetric interval $70<X<82$ , i.e. $\mu\pm\sigma$ .
Use of symmetry, $P(-1<Z<1)=1-2P(Z>1)=0.6826$ , with a clear final proportion.

How a teacher helps

The normal distribution improves fastest when a student stops treating the table as one-directional. In a one-to-one lesson our teachers drill the reverse reading, probability to z-value, until it is automatic, then show how each condition becomes an equation you can solve for a hidden parameter.

We spend time on sketching the curve and shading the region first, because a quick diagram fixes the sign of $z$ and reveals when symmetry lets you reuse a table value. Teachers at spmaddmath.com.my are experienced, and lessons are online and taught in English.

Message us on WhatsApp to arrange a one-hour paid class from RM50/hr at the teacher's rate.

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Frequently asked questions

How do I get a z-value from a percentage?

Read the standard normal table in reverse. If the question says $15.87\%$ lie above a value, then $P(Z>z)=0.1587$ , and the table gives $z=1$ .

Likewise $P(Z>z)=0.0228$ gives $z=2$ . Sketching the curve first tells you whether the z-value is positive or negative.

Why do two conditions let me find both

\mu

and

\sigma

Each probability statement becomes one standardisation equation, $z=\frac{x-\mu}{\sigma}$ , with the same two unknowns $\mu$ and $\sigma$ . Two independent equations in two unknowns can be solved together, here clearing the denominators gives $82-\mu=\sigma$ and $88-\mu=2\sigma$ , so $\sigma=6$ and $\mu=76$ .

Why is 'within 6 cm of the mean' so quick to evaluate?

Because $6$ cm is exactly one standard deviation, so the interval is $\mu\pm\sigma$ , which standardises to $-1<Z<1$ . By symmetry $P(-1<Z<1)=1-2P(Z>1)=1-2(0.1587)=0.6826$ , the familiar 'about 68% within one standard deviation'.

Can I earn marks if my final proportion is wrong?

Yes. Add Math Paper 2 is 2 hours 30 minutes and 100 marks, and marking is analytic, method marks are awarded line by line.

Correct z-values, valid standardisation equations and a sound simultaneous solution can score well even if an arithmetic slip changes the final figure.

Source:SRC-DSKP-EN SRC-FORMAT

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

KBAT: Reasoning with the Normal Distribution