Mathematics 9709 · Probability & Statistics 1

Finding μ and σ from probabilities

This is the same standardising machine, but run backwards.
Now the question gives an area and asks for $\mu$ or $\sigma$ .
The area fixes a $z$ -value.
Read that $z$ from the table backwards.
Then $z=\dfrac{x-\mu}{\sigma}$ becomes an algebra equation.

A probability fixes z; then it's algebra

A probability fixes a z; then z = (x − μ)/σ is algebra. Sketch first so the SIGN of z is right.

The most used critical values: $\Phi^{-1}(0.90)=1.282,\ \Phi^{-1}(0.95)=1.645,\ \Phi^{-1}(0.975)=1.960$ .

Sketch and shade the given area; is the boundary above or below $\mu$ ?
Is the area to the left of the boundary more or less than 0.5? → that fixes the sign of $z$ .
$z=\Phi^{-1}(\text{left area})$ (read the table backwards).
Write $z=\dfrac{x-\mu}{\sigma}$ with the signed $z$ , and solve.
Two unknowns → two such equations; subtract to kill $\mu$ .

Given	Boundary vs μ	z sign
$P(X<x)=0.9$	above μ	+
$P(X<x)=0.2$	below μ	−
$P(X>x)=0.1$	above μ	+
$P(X>x)=0.7$	below μ	−

Worked example

Simplest rung: one unknown σ

$T\sim N(10,\sigma^2)$ , $P(T>14.7)=0.04$ . Then $P(T<14.7)=0.96$ , so $z=\Phi^{-1}(0.96)=1.751$ , and $1.751=\tfrac{14.7-10}{\sigma}$ .

Answer

\sigma=\tfrac{4.7}{1.751}\approx2.68

Worked example

One unknown μ, from a percentile in context

Tree heights $X\sim N(\mu,\,3.6^2)$ ; 75% are under 10 m, so $P(X<10)=0.75$ . Then $z=\Phi^{-1}(0.75)=0.674$ and $0.674=\tfrac{10-\mu}{3.6}$ , so $\mu=10-0.674(3.6)$ .

Answer

\mu\approx7.57\text{ m}

Worked example

The sign trap: a sub-mean value with a big tail

$T\sim N(50,\sigma^2)$ , $P(T>44)=0.70$ . Here 44 is below the mean and 0.70 > 0.5, so $z$ is negative: $z=\Phi^{-1}(0.30)=-0.524$ , giving $\tfrac{44-50}{\sigma}=-0.524$ . (9709/52/M/J/23 Q5) (9709/52/O/N/24 Q4)

Answer

\sigma=\tfrac{6}{0.524}\approx11.5

Worked example

Two unknowns: μ and σ

$P(Y<10)=0.75\Rightarrow 10-\mu=0.674\sigma$ ; $P(Y\ge12)=0.1\Rightarrow 12-\mu=1.282\sigma$ . Subtract: $2=0.608\sigma$ , so $\sigma=3.29$ , then $\mu=10-0.674(3.29)$ .

Answer

\sigma\approx3.29,\ \mu\approx7.78

Examiner note

Write the

z

down explicitly before substituting — schemes award a mark just for a correct

z

in range. Then the standardisation equation is the method mark, and the answer needs signs consistent to give a positive $\sigma$ (a negative SD means a dropped sign).

Common mistake

Using the probability itself as $z$ (e.g. plugging 0.70 where the

z

belongs) — you must invert through the table. And for two unknowns, adding the equations instead of subtracting, so

\mu

never cancels.

The slick one: when only the ratio matters

If $X\sim N(\mu,\sigma^2)$ with $\sigma=\tfrac23\mu$ , find $P(X>0)$ : $z=\dfrac{0-\mu}{\tfrac23\mu}=-1.5$ — the $\mu$ cancels, so $P(X>0)=\Phi(1.5)=0.9332$ whatever $\mu$ is. Sometimes you need only the ratio of $\mu$ to $\sigma$ . (9709/52/O/N/23 Q5)

PreviousStandardising: z = (x − μ)/σ NextNormal approximation to the binomial

Finding μ and σ from probabilities

This is the same standardising machine, but run backwards.
Now the question gives an area and asks for $\mu$ or $\sigma$ .
The area fixes a $z$ -value.
Read that $z$ from the table backwards.
Then $z=\dfrac{x-\mu}{\sigma}$ becomes an algebra equation.

A probability fixes z; then it's algebra

A probability fixes a z; then z = (x − μ)/σ is algebra. Sketch first so the SIGN of z is right.

The most used critical values: $\Phi^{-1}(0.90)=1.282,\ \Phi^{-1}(0.95)=1.645,\ \Phi^{-1}(0.975)=1.960$ .

Sketch and shade the given area; is the boundary above or below $\mu$ ?
Is the area to the left of the boundary more or less than 0.5? → that fixes the sign of $z$ .
$z=\Phi^{-1}(\text{left area})$ (read the table backwards).
Write $z=\dfrac{x-\mu}{\sigma}$ with the signed $z$ , and solve.
Two unknowns → two such equations; subtract to kill $\mu$ .

Given	Boundary vs μ	z sign
$P(X<x)=0.9$	above μ	+
$P(X<x)=0.2$	below μ	−
$P(X>x)=0.1$	above μ	+
$P(X>x)=0.7$	below μ	−

Worked example

Simplest rung: one unknown σ

$T\sim N(10,\sigma^2)$ , $P(T>14.7)=0.04$ . Then $P(T<14.7)=0.96$ , so $z=\Phi^{-1}(0.96)=1.751$ , and $1.751=\tfrac{14.7-10}{\sigma}$ .

Answer

\sigma=\tfrac{4.7}{1.751}\approx2.68

Worked example

One unknown μ, from a percentile in context

Tree heights $X\sim N(\mu,\,3.6^2)$ ; 75% are under 10 m, so $P(X<10)=0.75$ . Then $z=\Phi^{-1}(0.75)=0.674$ and $0.674=\tfrac{10-\mu}{3.6}$ , so $\mu=10-0.674(3.6)$ .

Answer

\mu\approx7.57\text{ m}

Worked example

The sign trap: a sub-mean value with a big tail

Answer

\sigma=\tfrac{6}{0.524}\approx11.5

Worked example

Two unknowns: μ and σ

$P(Y<10)=0.75\Rightarrow 10-\mu=0.674\sigma$ ; $P(Y\ge12)=0.1\Rightarrow 12-\mu=1.282\sigma$ . Subtract: $2=0.608\sigma$ , so $\sigma=3.29$ , then $\mu=10-0.674(3.29)$ .

Answer

\sigma\approx3.29,\ \mu\approx7.78

Examiner note

Write the

z

down explicitly before substituting — schemes award a mark just for a correct

z

in range. Then the standardisation equation is the method mark, and the answer needs signs consistent to give a positive $\sigma$ (a negative SD means a dropped sign).

Common mistake

Using the probability itself as $z$ (e.g. plugging 0.70 where the

z

belongs) — you must invert through the table. And for two unknowns, adding the equations instead of subtracting, so

\mu

never cancels.

The slick one: when only the ratio matters