Data, Statistics & Probability

Find the mean, median, and mode of $7,\ 3,\ 9,\ 3,\ 8$.

Order first: $3,\ 3,\ 7,\ 8,\ 9$.

Mean: $\dfrac{3+3+7+8+9}{5} = \dfrac{30}{5} = \mathbf{6}$.

Median: the middle of five ordered numbers is the $3$rd one, so median $= \mathbf{7}$.

Mode: $3$ appears twice, more than any other, so mode $= \mathbf{3}$.

Find the median of $10,\ 4,\ 6,\ 2$. Order them: $2,\ 4,\ 6,\ 10$. There are two middle values, $4$ and $6$, so the median is their mean: $\dfrac{4+6}{2} = \mathbf{5}$. Notice the median need not be a value in the list.

Four friends have $$5$, $$6$, $$7$, and $$8$. Mean $= \dfrac{26}{4} = \$6.50$; median $= \dfrac{6+7}{2} = \$6.50$. Now a fifth friend joins with $$100$ (an outlier). New mean $= \dfrac{126}{5} = \$25.20$, but the ordered list $5,6,7,8,100$ has median $= \mathbf{\$7}$. The single big value pulled the mean way up, while the median barely moved. That is why the median better describes a “typical” friend here.

Test scores were $88,\ 95,\ 72,\ 90,\ 81$. The maximum is $95$ and the minimum is $72$, so the range $= 95 - 72 = \mathbf{23}$.

A class recorded how many pets each student has:

The frequency row tells us $4+6+3+2 = 15$ students answered. The mode is $1$ pet (highest frequency, $6$). The dot plot stacks $4$ dots above $0$, $6$ dots above $1$, and so on. To find the total number of pets, multiply each value by its frequency and add: $0(4)+1(6)+2(3)+3(2) = 0+6+6+6 = 18$ pets, so the mean is $\dfrac{18}{15} = 1.2$ pets.

A circle graph of how $60$ students get to school shows Walk $= 25\%$, Bus $= 50\%$, Car $= 25\%$. Because the whole circle is $100\%$ of $60$ students, the bus slice stands for $50\%$ of $60 = 30$ students, and each $25\%$ slice stands for $15$ students. The slices $25\% + 50\% + 25\% = 100\%$, as they must.

A plant's height was measured each week:

The plant grew every week, so the line rises the whole time. The weekly changes are $5 - 2 = 3$, $6 - 5 = 1$ and $10 - 6 = 4$ cm, so the greatest growth was between Weeks $3$ and $4$. Total growth over the whole period was $10 - 2 = 8$ cm.

A number cube has faces $1,2,3,4,5,6$, so there are $6$ equally likely outcomes. What is $P(\text{even})$? The favorable outcomes are $2,4,6$, so $3$ of them.

What is $P(\text{a }5)$? Only one face works, so $P = \dfrac{1}{6}$.

A bag has $20$ marbles, and $5$ are red. Then

Since $25\%$ is less than $50\%$, drawing red is unlikely. Drawing “a marble” at all is certain: $P = \dfrac{20}{20} = 1 = 100\%$. Drawing a green marble is impossible (there are none): $P = 0$.

You have $3$ shirts and $4$ pairs of shorts. How many different outfits? By the Counting Principle, $3 \times 4 = \mathbf{12}$ outfits.

Flip a coin ($2$ outcomes: H, T) and roll a cube ($6$ outcomes). Total outcomes $= 2 \times 6 = 12$. What is $P(\text{heads and a }3)$? Only one pair works out of $12$, so

In five games a player scored $12,\ 8,\ 15,\ 12,\ 3$ points.

Mean: $\dfrac{12+8+15+12+3}{5} = \dfrac{50}{5} = 10$ points.

Median (order $3,8,12,12,15$): the middle value is $12$.

Mode: $12$ (it appears twice).   Range: $15 - 3 = 12$.

If the coach picks one of these five games at random to rewatch, the probability it is a game where the player scored more than $10$ points (the $12$, $15$, and $12$ games $=3$ games) is $\dfrac{3}{5} = 0.6 = 60\%$ — likely.

Homework is worth $20\%$, quizzes $30\%$, and the final exam $50\%$. A student earns $90$, $80$, and $70$ on these. Multiply each score by its weight (as a decimal) and add:

The plain average would be $\dfrac{90+80+70}{3} = 80$, but because the heavily weighted final was lowest, the weighted grade drops to $77$. The weights $0.20+0.30+0.50 = 1$ must add to $1$ (that is $100\%$).

Find the MAD and standard deviation of $2,\ 4,\ 6,\ 8,\ 10$. First the mean: $\bar{x} = \dfrac{2+4+6+8+10}{5} = \dfrac{30}{5} = 6$. Now each distance from $6$:

MAD: $\dfrac{4+2+0+2+4}{5} = \dfrac{12}{5} = \mathbf{2.4}$.

Standard deviation: the squared distances sum to $16+4+0+4+16 = 40$, so the variance is $\dfrac{40}{5} = 8$ and $\sigma = \sqrt{8} \approx \mathbf{2.83}$. Both numbers describe the “typical” distance of a value from the mean.

For $4,\ 7,\ 8,\ 9,\ 11,\ 12,\ 20$ (already in order, $n=7$), the median is the $4$th value, $Q_2 = 9$. The lower half is $4,7,8$, so $Q_1 = 7$; the upper half is $11,12,20$, so $Q_3 = 12$. Then

Outlier fences: $Q_1 - 1.5(5) = 7 - 7.5 = -0.5$ and $Q_3 + 1.5(5) = 12 + 7.5 = 19.5$. Since $20 > 19.5$, the value $\mathbf{20}$ is an outlier.

From $5$ runners, how many ways can they take gold, silver, and bronze? Order matters, so

From those same $5$ runners, how many ways can we pick $3$ to form a relay team (no ordering)? Now

Each unordered team of $3$ can be ordered in $3! = 6$ ways, and indeed $10 \times 6 = 60$, matching the permutation count.

A game costs $$2$ to play. You roll a number cube: rolling a $6$ wins you $$10$; anything else wins nothing. The expected winnings are

Since the expected win $$1.67$ is less than the $$2$ cost, the game favors the house: on average you lose about $$0.33$ each play.

A bag has $3$ red and $2$ blue marbles ($5$ total). You draw two marbles without putting the first back. Find $P(\text{both red})$. The first draw is red with probability $\dfrac{3}{5}$. Now only $2$ red remain out of $4$ marbles, so the second is red with probability $\dfrac{2}{4}$. These are dependent, so multiply:

If instead you replaced the first marble, the draws would be independent and $P = \dfrac{3}{5} \cdot \dfrac{3}{5} = \dfrac{9}{25}$.

Flip a fair coin $3$ times. Find $P(\text{at least one head})$. Listing all the winning cases is tedious, so use the complement. The only way to get no heads is all tails: $P(\text{TTT}) = \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{1}{8}$. Therefore