Notes
Spread and the normal curve
- Standard deviation measures typical distance from the mean. Adding a constant doesn't change it; multiplying by $c$ multiplies it by $|c|$.
- A normal distribution is symmetric and bell-shaped. Empirical rule: about 68%, 95% and 99.7% of data fall within 1, 2 and 3 standard deviations of the mean.
| Region | Percent |
|---|---|
| $\mu$ to $\mu + \sigma$ | 34% |
| $\mu + \sigma$ to $\mu + 2\sigma$ | 13.5% |
| beyond $\mu + 2\sigma$ | 2.5% |
z-scores
$\displaystyle z = \frac{x - \mu}{\sigma}$ and $x = \mu + z\sigma$. Use z-scores to compare values from different distributions. $|z| > 2$ is often called unusual.
Collecting data
- Good samples are random: simple random, stratified, cluster, systematic. Convenience and voluntary response samples are biased.
- Surveys and observational studies show association only. A randomized experiment with a control group can show cause and effect.
Margin of error
For a sample proportion, the 95% margin of error is about $\displaystyle \frac{1}{\sqrt{n}}$. The interval is the sample result $\pm$ the margin.
Topics in this unit
Each topic has its own short notes and practice.
Practice
85 is $-1\sigma$ and 130 is $+2\sigma$: $34\% + 47.5\% = 81.5\%$.
$\displaystyle \frac{62 - 50}{8} = 1.5$.
Students at a game are more likely to support sports funding.
$\displaystyle \frac{1}{\sqrt{900}} = \frac{1}{30} \approx 0.033$.
Only a randomized experiment can show cause and effect.
Multiplying by 3 multiplies both the mean and the standard deviation by 3.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A company's phone batteries last a normally distributed number of hours, with mean 40 hours and standard deviation 5 hours.
(a) About what percent of batteries last between 35 and 50 hours?
Show the answer and scoring guide
35 is 1 standard deviation below and 50 is 2 above. $34\% + 47.5\% = 81.5\%$.
Check off the points you earned:
(b) Find the z-score of a battery that lasts 32 hours. Is this unusual?
Show the answer and scoring guide
$\displaystyle z = \frac{32 - 40}{5} = -1.6$. Since $|z| < 2$, it is not unusual.
Check off the points you earned:
(c) The company surveys a random sample of 625 buyers, and 72% say they are satisfied. Find the approximate margin of error and an interval for the percent of all buyers who are satisfied.
Show the answer and scoring guide
Margin of error $\displaystyle \approx \frac{1}{\sqrt{625}} = \frac{1}{25} = 0.04$, or 4%. Interval: 68% to 76%.
Check off the points you earned: