Standardising a normal variable | CAIE 9709 Probability & Statistics 1 Notes

Probability & Statistics 1

Mathematics9709

Subtract the mean, then divide by the standard deviation

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

Symbols

$z$ = signed standardised value (none)
$x$ = raw boundary (same as X)
$\mu$ = mean of X (same as X)
$\sigma$ = standard deviation of X (same as X)

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.

$x-\mu$ measures signed distance from the centre.
Dividing by $\sigma$ expresses that distance in standard deviations.
$z<0$ is below the mean and $z>0$ is above it.

Common mistake

X\sim N(\mu,\sigma^2)

, take the positive square root of the second parameter before standardising.

Keep the same tail after changing scale

Worked example

A direct probability

Obi's walking time is normal with mean 14.8 minutes and standard deviation 1.5 minutes. Find $P(X>15.6)$ .

z=\dfrac{15.6-14.8}{1.5}=0.5333

P=P(Z>0.5333)

=1-\Phi(0.5333)

Answer

P(X>15.6)=0.297

(9709/53/O/N/25 Q7(a))

Examiner note

The syllabus requires full standardisation working. Direct Normal CD with the raw mean and standard deviation is a fast calculator check, not a replacement for the written

z

-step.

Standardise both ends of an interval

Standardising changes the horizontal scale, not the shaded region. For $a<X<b$ , standardise $a$ and $b$ , then subtract the two left areas.

Worked example

A region between two values

Let $X\sim N(50,6^2)$ . Find $P(44<X<59)$ .

z_1=\dfrac{44-50}{6}=-1

z_2=\dfrac{59-50}{6}=1.5

P=\Phi(1.5)-\Phi(-1)

Answer

P(44<X<59)=0.775

Key idea

A raw boundary must use the same unit as the mean and standard deviation before you standardise. For example, 165 cm is 1.65 m.

Your turnIndependent check [4 marks]

Adult heights $X$ , measured in metres, satisfy $X\sim N(1.72,0.0064)$ . Find the probability that a height exceeds 165 cm.

Show worked answer

Convert 165 cm to 1.65 m before standardising. The standard deviation is $\sqrt{0.0064}=0.08$ m.

z=\dfrac{1.65-1.72}{0.08}=-0.875

P=P(Z>-0.875)

Answer

P(X>1.65)=0.809

Subtract the mean, then divide by the standard deviation

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

Symbols

$z$ = signed standardised value (none)
$x$ = raw boundary (same as X)
$\mu$ = mean of X (same as X)
$\sigma$ = standard deviation of X (same as X)

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.

$x-\mu$ measures signed distance from the centre.
Dividing by $\sigma$ expresses that distance in standard deviations.
$z<0$ is below the mean and $z>0$ is above it.

Common mistake

X\sim N(\mu,\sigma^2)

, take the positive square root of the second parameter before standardising.

Keep the same tail after changing scale

Worked example

A direct probability

Obi's walking time is normal with mean 14.8 minutes and standard deviation 1.5 minutes. Find $P(X>15.6)$ .

z=\dfrac{15.6-14.8}{1.5}=0.5333

P=P(Z>0.5333)

=1-\Phi(0.5333)

Answer

P(X>15.6)=0.297

(9709/53/O/N/25 Q7(a))

Examiner note

The syllabus requires full standardisation working. Direct Normal CD with the raw mean and standard deviation is a fast calculator check, not a replacement for the written

z

-step.

Standardise both ends of an interval

Standardising changes the horizontal scale, not the shaded region. For $a<X<b$ , standardise $a$ and $b$ , then subtract the two left areas.

Worked example

A region between two values

Let $X\sim N(50,6^2)$ . Find $P(44<X<59)$ .

z_1=\dfrac{44-50}{6}=-1

z_2=\dfrac{59-50}{6}=1.5

P=\Phi(1.5)-\Phi(-1)

Answer

P(44<X<59)=0.775

Key idea

A raw boundary must use the same unit as the mean and standard deviation before you standardise. For example, 165 cm is 1.65 m.

Your turnIndependent check [4 marks]

Adult heights $X$ , measured in metres, satisfy $X\sim N(1.72,0.0064)$ . Find the probability that a height exceeds 165 cm.

Show worked answer

Convert 165 cm to 1.65 m before standardising. The standard deviation is $\sqrt{0.0064}=0.08$ m.

z=\dfrac{1.65-1.72}{0.08}=-0.875

P=P(Z>-0.875)

Answer

P(X>1.65)=0.809