Notes
Center and spread
- Center: mean and median. Use the median when data are skewed or have outliers.
- Spread: range, IQR ($Q_3 - Q_1$) and standard deviation (typical distance from the mean).
- Outlier rule: below $Q_1 - 1.5 \cdot \text{IQR}$ or above $Q_3 + 1.5 \cdot \text{IQR}$.
Displays and categorical data
- Dot plots, histograms and box plots show shape. A right tail means the mean is above the median.
- Two-way tables: use conditional relative frequencies to check for association.
Scatter plots
- A line of fit $\hat{y} = mx + b$ predicts $y$. Residual $=$ actual $-$ predicted.
- Correlation $r$ measures strength and direction of a linear relationship; it does not prove causation.
Topics in this unit
Each topic has its own short notes and practice.
Practice
The sum is 70 and $70 \div 5 = 14$.
Lower half $1, 3, 4, 6$ has median 3.5. Upper half $8, 9, 12, 15$ has median 10.5. IQR $= 10.5 - 3.5 = 7$.
The mean is pulled above the median by large values, which happens in right-skewed data.
$\displaystyle \frac{20}{50} = 0.40 = 40\%$.
$-1.5(12) + 90 = -18 + 90 = 72$.
An $r$ value near 0 means the points do not follow a line.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A teacher records seven quiz scores: $55, 68, 72, 75, 80, 84, 91$.
(a) Find the mean and median.
Show the answer and scoring guide
Sum $= 525$, so the mean is $\displaystyle \frac{525}{7} = 75$. The median is the middle value, 75.
Check off the points you earned:
(b) Find the range and the IQR.
Show the answer and scoring guide
Range $= 91 - 55 = 36$. $Q_1 = 68$ and $Q_3 = 84$, so IQR $= 84 - 68 = 16$.
Check off the points you earned:
(c) Use the $1.5 \cdot \text{IQR}$ rule to decide whether there are outliers.
Show the answer and scoring guide
$1.5 \cdot 16 = 24$. Lower fence: $68 - 24 = 44$. Upper fence: $84 + 24 = 108$. All scores are between 44 and 108, so there are no outliers.
Check off the points you earned: