Mean, Median, Mode, and Range: Practice

The four numbers must add to $4 \times 15 = 60$. Subtract the three you know.

$60 - (10 + 16 + 20) = 60 - 46 = 14$

With five sorted values the median is the third value, and here that third value is $a$.

$a = \text{median} = 12$

Before, the mean is $\frac{40}{5} = 8$ and the median is $8$. After replacing $11$ with $100$, the sorted data is $3, 8, 8, 10, 100$, so the median is still $8$ but the mean jumps.

$\text{new mean} = \frac{129}{5} = 25.8$

The outlier pulls the mean far but leaves the median fixed.

The sum is the mean times the number of values.

$\text{sum} = 8 \times 7 = 56$

The nine students add to $9 \times 82 = 738$. Adding the tenth score gives a sum of $810$ over $10$ students.

$\text{new mean} = \frac{738 + 72}{10} = \frac{810}{10} = 81$

The four values must add to $4 \times 10 = 40$. Subtract the three you know.

$40 - (8 + 9 + 13) = 40 - 30 = 10$

With six sorted values the median is the average of the third and fourth values, $m$ and $14$.

$\frac{m + 14}{2} = 11.5 \;\Rightarrow\; m + 14 = 23 \;\Rightarrow\; m = 9$

Adding $5$ to each of the $n$ values adds $5n$ to the sum. Dividing the extra $5n$ by $n$ adds $5$ to the mean.

$\text{new mean} = \frac{\text{old sum} + 5n}{n} = \text{old mean} + 5$

The original range is $50 - 10 = 40$. Doubling makes the values $20, 40, 60, 80, 100$, so both the largest and smallest double and the range doubles too.

$\text{new range} = 100 - 20 = 80$

Every value is $3$, so the mean, the median, and the mode are all $3$, while the range is $3 - 3 = 0$.

$\text{mean} = \text{median} = \text{mode} = 3, \qquad \text{range} = 0$

The five scores must add to $5 \times 8 = 40$. Subtract the four she has.

$40 - (7 + 8 + 6 + 9) = 40 - 30 = 10$

The mean is $\frac{40}{5} = 8$, the median is the middle value $8$, and the mode is the most frequent value $8$.

$\text{mean} = \text{median} = \text{mode} = 8$