Mathematics 9709 · Probability & Statistics 1

Standardising: z = (x − μ)/σ

Your variable has its own scale: cm, seconds, grams, or marks.
The normal table only speaks standard deviations from the mean.
Standardising is the translator: $z=\dfrac{x-\mu}{\sigma}$ .
That asks: how many standard deviations is $x$ above or below the mean?
Translate to $z$ , then read the area from $\Phi$ .

z = (x − μ)/σ is the whole engine

Every “find the probability / how many” question is: standardise → read Φ → (for a count) × N.

$z=\dfrac{x-\mu}{\sigma}\qquad P(X<x)=\Phi(z)\qquad P(a<X<b)=\Phi(z_b)-\Phi(z_a)$

Drag the raw value and watch the scale change: first subtract the mean, then divide by the standard deviation.

Raw scale: X

Table scale: Z

mean μ = 20standard deviation σ = 4raw value x = 26

start
x = 26

subtract μ
x - μ = 6

divide by σ
z = 1.50

read table
Φ(z) = 0.9332

The coloured area does not change: P(X < x) = P(Z < z) = 0.9332.

Write $X\sim N(\mu,\sigma^2)$ and note $\sigma=\sqrt{\text{variance}}$ .
Sketch, mark $\mu$ and $x$ , shade the region.
$z=\dfrac{x-\mu}{\sigma}$ to 4 significant figures (keep the sign).
Turn it into a $\Phi$ expression (use the §8.1 table).
“How many of $N$ ?” → multiply by $N$ and round to a whole number (no $\approx$ ).

Worked example

Simplest rung: one clean standardise

$X\sim N(11,25)$ , so $\sigma=5$ (the second slot is the variance). Find $P(X<18)$ : $z=\tfrac{18-11}{5}=1.4$ , then $P(X<18)=\Phi(1.4)=0.919$ .

Answer

0.919

Worked example

A negative z, then a right tail

$X\sim N(20,7)$ : $z=\tfrac{16.6-20}{\sqrt7}=-1.285$ , so $P(X\le16.6)=1-\Phi(1.285)=0.0994$ . The sign of $z$ and the “1 minus” for the tail are where the marks live.

Answer

0.0994

Worked example

Estimate a count (between two values)

Lengths $X\sim N(20,3^2)$ ; in a sample of 400, how many between 17 and 24? $z_1=\tfrac{17-20}{3}=-1,\ z_2=\tfrac{24-20}{3}=1.333$ , so $P=\Phi(1.333)-\Phi(-1)=0.9088-0.1587=0.7501$ , then $\times400$ . (9709/52/M/J/23 Q5) (9709/52/O/N/24 Q4)

Answer

\approx 300\text{ items}

Modelling: turn a probability into a count, then chain it

Real questions usually name a context: babies, oil, tomatoes, reaction times.
Often they ask how many out of $N$ fall in a region.
Find the normal probability first, then multiply by $N$ .
One step harder: use that normal probability as the $p$ in a small binomial.
A normal probability can become the success chance in $B(n,p)$ .

Worked example

How many of N, in a real context

Newborn masses $X\sim N(3.35,\,0.0858)$ . Of 1356 babies, estimate how many had mass under 3.5 kg.

$z=\dfrac{3.5-3.35}{\sqrt{0.0858}}=0.512$ , so $P(X<3.5)=\Phi(0.512)=0.6957$ .
Probability into a count: $0.6957\times1356=943.4$ , rounded to a whole baby.

Answer

\approx 943\text{ babies}

Worked example

Percentage in context, then chain to a binomial

Half-litre tins hold oil $X\sim N(506.18,\,2.96^2)$ . (a) What percentage hold under 500 ml? (b) Then find the chance that exactly 1 of 3 randomly chosen tins is under 500 ml.

(a) $z=\dfrac{500-506.18}{2.96}=-2.088$ , so $P(X<500)=1-\Phi(2.088)=0.0184$ , i.e. 1.84%.
(b) Use that as $p$ : let $Y$ count under-500 tins, $Y\sim B(3,\,0.0184)$ .
$P(Y=1)=\binom{3}{1}(0.0184)(0.9816)^2=0.0532$ .

Answer

1.84\%\quad\text{and}\quad0.0532

Examiner note

The standardisation line is the method mark — write

z=\dfrac{x-\mu}{\sigma}

with the actual numbers in (substitute

\mu,\ x,\ \sigma

, not

\sigma^2

). For a count, the whole-number answer (no

\approx

) is its own mark; when chaining, do not round the normal probability before feeding it into the binomial.

Common mistake

Dividing by the variance instead of

\sigma

(the single most common

M0

in the chapter), and dropping the sign of

z

. For a right tail, don't stop at

\Phi(z)

— do

1-\Phi(z)

Common mistake

On a “between” count, doing only one side, or writing the count as

\approx 300

/ “about 300” — the mark needs a clean whole number. On a chain, using a rounded

p

(e.g. 0.02 for 0.0184).

Your turn— tap to reveal the worked answer (9709-style)

Reaction times $X\sim N(0.42,0.05^2)$ . Find $P(X>0.5)$ .

$z=\tfrac{0.5-0.42}{0.05}=1.6$ , so $P(X>0.5)=1-\Phi(1.6)=1-0.9452$ .

Answer

0.0548

Why a normal probability can be a binomial p

“Under 500 ml” is a fixed, repeatable event with one probability, $p=0.0184$ . Pick 3 tins independently and you are counting successes in 3 trials — a textbook $B(3,p)$ . The normal just supplies $p$ ; the counting is ordinary binomial.

PreviousThe standard normal & z-table NextFinding μ and σ from probabilities

Standardising: z = (x − μ)/σ

Your variable has its own scale: cm, seconds, grams, or marks.
The normal table only speaks standard deviations from the mean.
Standardising is the translator: $z=\dfrac{x-\mu}{\sigma}$ .
That asks: how many standard deviations is $x$ above or below the mean?
Translate to $z$ , then read the area from $\Phi$ .

z = (x − μ)/σ is the whole engine

Every “find the probability / how many” question is: standardise → read Φ → (for a count) × N.

$z=\dfrac{x-\mu}{\sigma}\qquad P(X<x)=\Phi(z)\qquad P(a<X<b)=\Phi(z_b)-\Phi(z_a)$

Drag the raw value and watch the scale change: first subtract the mean, then divide by the standard deviation.

Raw scale: X

Table scale: Z

mean μ = 20standard deviation σ = 4raw value x = 26

start
x = 26

subtract μ
x - μ = 6

divide by σ
z = 1.50

read table
Φ(z) = 0.9332

The coloured area does not change: P(X < x) = P(Z < z) = 0.9332.

Write $X\sim N(\mu,\sigma^2)$ and note $\sigma=\sqrt{\text{variance}}$ .
Sketch, mark $\mu$ and $x$ , shade the region.
$z=\dfrac{x-\mu}{\sigma}$ to 4 significant figures (keep the sign).
Turn it into a $\Phi$ expression (use the §8.1 table).
“How many of $N$ ?” → multiply by $N$ and round to a whole number (no $\approx$ ).

Worked example

Simplest rung: one clean standardise

$X\sim N(11,25)$ , so $\sigma=5$ (the second slot is the variance). Find $P(X<18)$ : $z=\tfrac{18-11}{5}=1.4$ , then $P(X<18)=\Phi(1.4)=0.919$ .

Answer

0.919

Worked example

A negative z, then a right tail

$X\sim N(20,7)$ : $z=\tfrac{16.6-20}{\sqrt7}=-1.285$ , so $P(X\le16.6)=1-\Phi(1.285)=0.0994$ . The sign of $z$ and the “1 minus” for the tail are where the marks live.

Answer

0.0994

Worked example

Estimate a count (between two values)

Answer

\approx 300\text{ items}

Modelling: turn a probability into a count, then chain it

Real questions usually name a context: babies, oil, tomatoes, reaction times.
Often they ask how many out of $N$ fall in a region.
Find the normal probability first, then multiply by $N$ .
One step harder: use that normal probability as the $p$ in a small binomial.
A normal probability can become the success chance in $B(n,p)$ .

Worked example

How many of N, in a real context

Newborn masses $X\sim N(3.35,\,0.0858)$ . Of 1356 babies, estimate how many had mass under 3.5 kg.

$z=\dfrac{3.5-3.35}{\sqrt{0.0858}}=0.512$ , so $P(X<3.5)=\Phi(0.512)=0.6957$ .
Probability into a count: $0.6957\times1356=943.4$ , rounded to a whole baby.

Answer

\approx 943\text{ babies}

Worked example

Percentage in context, then chain to a binomial

Half-litre tins hold oil $X\sim N(506.18,\,2.96^2)$ . (a) What percentage hold under 500 ml? (b) Then find the chance that exactly 1 of 3 randomly chosen tins is under 500 ml.

(a) $z=\dfrac{500-506.18}{2.96}=-2.088$ , so $P(X<500)=1-\Phi(2.088)=0.0184$ , i.e. 1.84%.
(b) Use that as $p$ : let $Y$ count under-500 tins, $Y\sim B(3,\,0.0184)$ .
$P(Y=1)=\binom{3}{1}(0.0184)(0.9816)^2=0.0532$ .

Answer

1.84\%\quad\text{and}\quad0.0532

Examiner note

The standardisation line is the method mark — write

z=\dfrac{x-\mu}{\sigma}

with the actual numbers in (substitute

\mu,\ x,\ \sigma

, not

\sigma^2

). For a count, the whole-number answer (no

\approx

) is its own mark; when chaining, do not round the normal probability before feeding it into the binomial.

Common mistake

Dividing by the variance instead of

\sigma

(the single most common

M0

in the chapter), and dropping the sign of

z

. For a right tail, don't stop at

\Phi(z)

— do

1-\Phi(z)

Common mistake

On a “between” count, doing only one side, or writing the count as

\approx 300

/ “about 300” — the mark needs a clean whole number. On a chain, using a rounded

p

(e.g. 0.02 for 0.0184).

Your turn— tap to reveal the worked answer (9709-style)

Reaction times $X\sim N(0.42,0.05^2)$ . Find $P(X>0.5)$ .

$z=\tfrac{0.5-0.42}{0.05}=1.6$ , so $P(X>0.5)=1-\Phi(1.6)=1-0.9452$ .

Answer

0.0548

Why a normal probability can be a binomial p