Honors Algebra II Unit 13 Review: Probability and Statistics

Many probability problems are really counting problems. The Fundamental Counting Principle says that if one choice has $m$ options and the next has $n$ options for each first choice, together they have $m \times n$ outcomes. So 4 shirts and 3 pairs of pants give $4 \times 3 = 12$ outfits.

When you pick $r$ distinct items from $n$ distinct items without replacement, ask whether order matters. If it does, use permutations, $P(n,r) = \frac{n!}{(n-r)!}$. If it doesn't, use combinations, $C(n,r) = \frac{n!}{r!(n-r)!}$. The extra $r!$ divides out the different orderings of the same group.

For equally likely outcomes, $P(A)$ equals favorable outcomes over total outcomes. The complement rule $P(A') = 1 - P(A)$ handles "at least one" questions. The general addition rule is $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$. For independent events, $P(A \text{ and } B) = P(A) \cdot P(B)$. Conditional probability, $P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)}$, is the chance of $A$ once you know $B$ happened, provided $P(B) > 0$.

Worked example. Choose 3 people from a club of 10. For president, vice president and treasurer, order matters, so $P(10,3) = 10 \cdot 9 \cdot 8 = 720$. For a 3-person committee, order doesn't matter, so $C(10,3) = \frac{720}{3!} = 120$. Now roll a fair die 4 times. The chance of getting no 6 is $(5/6)^4 = 625/1296$, so the chance of at least one 6 is $1 - 625/1296 = 671/1296 \approx 0.518$.

Descriptive statistics summarize a dataset with a few numbers for its center and spread. For center, the mean is the sum divided by the count, the median is the middle value in sorted order, and the mode is the most frequent value. Outliers pull the mean toward them, while the median is generally less sensitive to them. That makes the median the better center for skewed data like home prices.

For spread, the range is the maximum minus the minimum. The interquartile range, $\text{IQR} = Q_3 - Q_1$, covers the middle 50% of the data. Standard deviation measures the typical distance from the mean. To find it, square each deviation, add the squares, divide, then take the square root. Divide by $n$ for a whole population ($\sigma$) and by $n-1$ for a sample ($s$).

Shape matters too. In a right-skewed distribution, the long tail is on the right, and the mean is often above the median. A common outlier test flags values more than $1.5 \times \text{IQR}$ below $Q_1$ or above $Q_3$.

Worked example. Find the sample standard deviation of 4, 8, 6, 5, 7. The mean is $30/5 = 6$. The deviations are $-2, 2, 0, -1, 1$, and their squares add to $4+4+0+1+1 = 10$. The sample variance is $10/(5-1) = 2.5$, so $s = \sqrt{2.5} \approx 1.58$. If these five values were the whole population, you would get $\sigma = \sqrt{10/5} \approx 1.41$ instead.

A normal distribution is a symmetric, bell-shaped curve set by two numbers. The mean $\mu$ fixes the center, and the standard deviation $\sigma$ fixes the spread. The total area under the curve is 1, and the area over an interval is the probability that a value lands in it. Heights, test scores and measurement errors often come close to normal.

The Empirical Rule gives quick estimates. About 68% of values fall within $1\sigma$ of the mean, about 95% within $2\sigma$, and about 99.7% within $3\sigma$. Because the curve is symmetric, you can split these in half. For example, about 2.5% of values lie above $\mu + 2\sigma$.

A z-score, $z = \frac{x - \mu}{\sigma}$, tells you how many standard deviations a value sits from the mean. A positive z means above the mean, and a negative z means below. Z-scores let you compare values from different distributions. For a normal distribution, a z-table or a calculator command like normalcdf can find probabilities for intervals even when the cutoffs aren't whole standard deviations.

Worked example. Scores on each test are normally distributed. Test A has mean 70 and $\sigma = 8$. Test B has mean 80 and $\sigma = 10$. You score 86 on A and 90 on B. On A, $z = (86-70)/8 = 2$. On B, $z = (90-80)/10 = 1$, so the 86 is the stronger result. By the Empirical Rule, only about 2.5% of Test A takers score above 86, while about 16% of Test B takers score above 90.