IB Mathematics Analysis and Approaches Topic SL…

Standardizing With z-Scores: Putting Every Normal On One Ruler

One student scores 80 on a test where the class averaged 70; another scores 65 where the class averaged 60.
Raw marks cannot say who did better, because the two tests have different centres and different spreads.
The common unit that fixes this is the z-score, the number of standard deviations a value sits from its mean.
Standardizing collapses any normal distribution onto one reference curve, the standard normal, so different distributions can be compared and a hidden mean or standard deviation can be found.

The z-Score: Distance From The Mean, Measured In Standard Deviations

To standardize a value $x$ from $X \sim N(\mu, \sigma^{2})$, subtract the mean and divide by the standard deviation. $$z = \frac{x - \mu}{\sigma}$$
Subtracting $\mu$ measures how far $x$ is from the centre.
Dividing by $\sigma$ rescales that gap into standard-deviation units, which mean the same on every normal curve.
1. $z = 0$ means $x$ is exactly at the mean.
2. $z = 1$ means $x$ is one standard deviation above the mean, and $z = -2$ means two below.

Note

The formula booklet gives $z = \dfrac{x - \mu}{\sigma}$ under Topic 4 as the standardized normal variable.
A positive $z$ sits above the mean and a negative $z$ below, so the sign already tells you which side you are on.

Example question

Anna scores 80 on a test with mean 70 and standard deviation 5. Ben scores 65 on a different test with mean 60 and standard deviation 4. Use z-scores to decide who performed better relative to their group.

Solution

Standardize each score with $z = \dfrac{x - \mu}{\sigma}$.
Anna: $z = \dfrac{80 - 70}{5} = 2$, so 2 standard deviations above her mean.
Ben: $z = \dfrac{65 - 60}{4} = 1.25$, so 1.25 standard deviations above his mean.
Anna has the higher z-score, so she performed better relative to her group even though the raw-mark gap looks similar.

The Standard Normal, And When You Actually Need z

Standardizing turns $X$ into $Z \sim N(0, 1)$, and the areas match exactly, so $P(X < x) = P(Z < z)$.
Every normal probability is really a standard-normal probability in disguise.

Note

For ordinary probability questions you do not need z at all, because the GDC works straight from $\mu$ and $\sigma$, as in 4.9.
The z-score earns its keep when the mean or standard deviation is unknown, because it bridges a known probability and the missing parameter.

Finding An Unknown Mean Or Standard Deviation

Run the idea in reverse: if you know a probability and the value it applies to, you can recover the z-score.
The equation $z = \dfrac{x - \mu}{\sigma}$ then links the value, the mean and the standard deviation, so you solve it for whichever one is missing.
Turn the given probability into a left-hand area, then find its z-score with $\texttt{invNorm}(\text{area}, 0, 1)$ on the standard normal.
Substitute that $z$, with the known value $x$, into $z = \dfrac{x - \mu}{\sigma}$.
Solve the resulting equation for the unknown mean or standard deviation.

Exam technique

Use $\texttt{invNorm}(\text{area}, 0, 1)$ to get the z-score, because here you work on the standard normal, not the original distribution.
If both the mean and the standard deviation are unknown, you need two probability statements, giving two equations to solve simultaneously.

Example question

The time $X$ taken to complete a task is normally distributed with standard deviation 4 minutes. It is known that 10% of times exceed 30 minutes. Find the mean time $\mu$.

Solution

"10% exceed 30" means $P(X > 30) = 0.10$, so the area to the left of 30 is 0.90.
Standardize: $$z = \texttt{invNorm}(0.90, 0, 1) = 1.28$$
Substitute into $z = \dfrac{x - \mu}{\sigma}$: $$\;1.28 = \dfrac{30 - \mu}{4}$$
Solve: $\mu = 30 - 1.28 \times 4 = 24.9$ minutes (3 s.f.).

Example question

A normal variable $X$ satisfies $P(X < 50) = 0.20$ and $P(X > 70) = 0.10$. Find the mean $\mu$ and standard deviation $\sigma$.

Solution

First value: $P(X < 50) = 0.20$ gives $z_{1} = \texttt{invNorm}(0.20,0,1) = -0.842$, so $50 = \mu -0.842\sigma$.
Second value: $P(X > 70) = 0.10$ means area 0.90 to the left, so $z_{2} = 1.28$ and $70 = \mu + 1.28\sigma$.
Subtract the equations: $70 - 50 = (1.28 - (-0.842))\sigma$, so $20 = 2.12\,\sigma$.
$\sigma = 9.42$ (3 s.f.).
Back-substitute: $\mu = 50 - (-0.842) \times 9.42 = 57.9$ (3 s.f.).

See A Value Become Its z-Score

Standardizing is a rescaling that slides the mean to 0 and stretches or squeezes the spread to 1, while the shaded probability stays the same.
Watching a value and its area move onto the standard curve makes the formula feel inevitable.

Hint

Standardizing never changes a probability; it only changes the ruler, and that shared z-ruler is what lets you compare distributions and solve for a hidden $\mu$ or $\sigma$.

Active recall

Write down the formula for the z-score and state what a value of $z = -1.5$ tells you about $x$.
Two results come from normal distributions with different means and spreads. How does a z-score let you compare them fairly?
For ordinary normal probability questions, why do you usually not need to standardize at all?
Describe the three steps for finding an unknown mean when you are given a probability and the value it applies to.
Why do you need two probability statements to find both an unknown mean and an unknown standard deviation?

Standardizing With z-Scores: Putting Every Normal On One Ruler

One student scores 80 on a test where the class averaged 70; another scores 65 where the class averaged 60.
Raw marks cannot say who did better, because the two tests have different centres and different spreads.
The common unit that fixes this is the z-score, the number of standard deviations a value sits from its mean.
Standardizing collapses any normal distribution onto one reference curve, the standard normal, so different distributions can be compared and a hidden mean or standard deviation can be found.

The z-Score: Distance From The Mean, Measured In Standard Deviations

To standardize a value $x$ from $X \sim N(\mu, \sigma^{2})$, subtract the mean and divide by the standard deviation. $$z = \frac{x - \mu}{\sigma}$$
Subtracting $\mu$ measures how far $x$ is from the centre.
Dividing by $\sigma$ rescales that gap into standard-deviation units, which mean the same on every normal curve.
1. $z = 0$ means $x$ is exactly at the mean.
2. $z = 1$ means $x$ is one standard deviation above the mean, and $z = -2$ means two below.

Note

The formula booklet gives $z = \dfrac{x - \mu}{\sigma}$ under Topic 4 as the standardized normal variable.
A positive $z$ sits above the mean and a negative $z$ below, so the sign already tells you which side you are on.

Example question

Solution

Standardize each score with $z = \dfrac{x - \mu}{\sigma}$.
Anna: $z = \dfrac{80 - 70}{5} = 2$, so 2 standard deviations above her mean.
Ben: $z = \dfrac{65 - 60}{4} = 1.25$, so 1.25 standard deviations above his mean.
Anna has the higher z-score, so she performed better relative to her group even though the raw-mark gap looks similar.

The Standard Normal, And When You Actually Need z

Standardizing turns $X$ into $Z \sim N(0, 1)$, and the areas match exactly, so $P(X < x) = P(Z < z)$.
Every normal probability is really a standard-normal probability in disguise.

Note

For ordinary probability questions you do not need z at all, because the GDC works straight from $\mu$ and $\sigma$, as in 4.9.
The z-score earns its keep when the mean or standard deviation is unknown, because it bridges a known probability and the missing parameter.

Finding An Unknown Mean Or Standard Deviation

Run the idea in reverse: if you know a probability and the value it applies to, you can recover the z-score.
The equation $z = \dfrac{x - \mu}{\sigma}$ then links the value, the mean and the standard deviation, so you solve it for whichever one is missing.
Turn the given probability into a left-hand area, then find its z-score with $\texttt{invNorm}(\text{area}, 0, 1)$ on the standard normal.
Substitute that $z$, with the known value $x$, into $z = \dfrac{x - \mu}{\sigma}$.
Solve the resulting equation for the unknown mean or standard deviation.

Exam technique

Use $\texttt{invNorm}(\text{area}, 0, 1)$ to get the z-score, because here you work on the standard normal, not the original distribution.
If both the mean and the standard deviation are unknown, you need two probability statements, giving two equations to solve simultaneously.

Example question

The time $X$ taken to complete a task is normally distributed with standard deviation 4 minutes. It is known that 10% of times exceed 30 minutes. Find the mean time $\mu$.

Solution

"10% exceed 30" means $P(X > 30) = 0.10$, so the area to the left of 30 is 0.90.
Standardize: $$z = \texttt{invNorm}(0.90, 0, 1) = 1.28$$
Substitute into $z = \dfrac{x - \mu}{\sigma}$: $$\;1.28 = \dfrac{30 - \mu}{4}$$
Solve: $\mu = 30 - 1.28 \times 4 = 24.9$ minutes (3 s.f.).

Example question

A normal variable $X$ satisfies $P(X < 50) = 0.20$ and $P(X > 70) = 0.10$. Find the mean $\mu$ and standard deviation $\sigma$.

Solution

First value: $P(X < 50) = 0.20$ gives $z_{1} = \texttt{invNorm}(0.20,0,1) = -0.842$, so $50 = \mu -0.842\sigma$.
Second value: $P(X > 70) = 0.10$ means area 0.90 to the left, so $z_{2} = 1.28$ and $70 = \mu + 1.28\sigma$.
Subtract the equations: $70 - 50 = (1.28 - (-0.842))\sigma$, so $20 = 2.12\,\sigma$.
$\sigma = 9.42$ (3 s.f.).
Back-substitute: $\mu = 50 - (-0.842) \times 9.42 = 57.9$ (3 s.f.).

See A Value Become Its z-Score

Standardizing is a rescaling that slides the mean to 0 and stretches or squeezes the spread to 1, while the shaded probability stays the same.
Watching a value and its area move onto the standard curve makes the formula feel inevitable.

Hint

Standardizing never changes a probability; it only changes the ruler, and that shared z-ruler is what lets you compare distributions and solve for a hidden $\mu$ or $\sigma$.

Active recall

Write down the formula for the z-score and state what a value of $z = -1.5$ tells you about $x$.
Two results come from normal distributions with different means and spreads. How does a z-score let you compare them fairly?
For ordinary normal probability questions, why do you usually not need to standardize at all?
Describe the three steps for finding an unknown mean when you are given a probability and the value it applies to.
Why do you need two probability statements to find both an unknown mean and an unknown standard deviation?

IB Mathematics Analysis and Approaches SL 4.12—Z Values, Inverse Normal to Find Mean and Standard Deviation Notes

IB Mathematics Analysis and Approaches SL 4.12—Z Values, Inverse Normal to Find Mean and Standard Deviation Notes

Standardizing With z-Scores: Putting Every Normal On One Ruler

The z-Score: Distance From The Mean, Measured In Standard Deviations

The Standard Normal, And When You Actually Need z

Finding An Unknown Mean Or Standard Deviation

See A Value Become Its z-Score

Standardizing With z-Scores: Putting Every Normal On One Ruler

The z-Score: Distance From The Mean, Measured In Standard Deviations

The Standard Normal, And When You Actually Need z

Finding An Unknown Mean Or Standard Deviation

See A Value Become Its z-Score

IB Mathematics Analysis and Approaches SL 4.12—Z Values, Inverse Normal to Find Mean and Standard Deviation Notes

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

IB Mathematics Analysis and Approaches SL 4.12—Z Values, Inverse Normal to Find Mean and Standard Deviation Notes

Number and Algebra

Functions

Geometry & Trigonometry

Statistics & Probability

Calculus

Standardizing With z-Scores: Putting Every Normal On One Ruler

The z-Score: Distance From The Mean, Measured In Standard Deviations

The Standard Normal, And When You Actually Need z

Finding An Unknown Mean Or Standard Deviation

See A Value Become Its z-Score

Standardizing With z-Scores: Putting Every Normal On One Ruler

The z-Score: Distance From The Mean, Measured In Standard Deviations

The Standard Normal, And When You Actually Need z

Finding An Unknown Mean Or Standard Deviation

See A Value Become Its z-Score