Data Distribution and Variability

1. Ordered data: $2, 3, 4, 4, 5, 5, 6, 7$

Mean:

$\bar{x} = \frac{2+3+4+4+5+5+6+7}{8} = \frac{36}{8} = 4.5$

Median (average of 4th and 5th values):

$\text{Median} = \frac{4+5}{2} = 4.5$

Mode: Both $4$ and $5$ appear twice — the data is bimodal: mode $= 4$ and $5$.

2. Range:

$7 - 2 = 5$

Shape: The data is roughly symmetric — values are spread fairly evenly around the center with no strong pull in either direction. The mean and median are both $4.5$, which supports a symmetric distribution.

3. Data: $\{1, 3, 5, 7, 9\}$

Step 1 — Find the mean:

$\bar{x} = \frac{1+3+5+7+9}{5} = \frac{25}{5} = 5$

Step 2 — Find each absolute deviation from the mean:

| Value | $|value - 5|$ | |-------|--------------| | 1 | 4 | | 3 | 2 | | 5 | 0 | | 7 | 2 | | 9 | 4 |

Step 3 — Average the absolute deviations:

$\text{MAD} = \frac{4+2+0+2+4}{5} = \frac{12}{5} = 2.4$

4. Both classes have the same mean of $7$, so their typical score is the same.

Class A has MAD $= 0.5$, meaning scores were on average only $0.5$ points away from the mean — very consistent.

Class B has MAD $= 2.8$, meaning scores varied widely (on average $2.8$ points from the mean) — much less consistent.

Class A had more consistent scores. The MAD measures the average spread of data around the mean; a smaller MAD means less variability.

5. Original set: $\{10, 12, 14, 16, 18\}$

Original mean: $\frac{10+12+14+16+18}{5} = \frac{70}{5} = 14$

Original median (middle value): $14$

New set with outlier: $\{10, 12, 14, 16, 18, 100\}$

New mean: $\frac{70+100}{6} = \frac{170}{6} \approx 28.3$

New median (average of 3rd and 4th values): $\frac{14+16}{2} = 15$

Effect: The mean jumped from $14$ to $\approx 28.3$ — a large change. The median only changed from $14$ to $15$ — a small change.

The median is more resistant to outliers.

6. Ages ordered: $11, 12, 12, 13, 13, 13, 14, 15$

(a) Shape: The distribution is slightly skewed right (the tail extends toward older ages; most participants cluster around 12–13).

(b) Measure of center: The median is a good choice here. Since the distribution is slightly skewed, the median is more representative than the mean.

Median (average of 4th and 5th values): $\frac{13+13}{2} = 13$

(c) Spread:

$\text{Range} = 15 - 11 = 4$

Mean: $\frac{11+12+12+13+13+13+14+15}{8} = \frac{103}{8} = 12.875$

Absolute deviations: $|11-12.875|=1.875$, $|12-12.875|=0.875$ (×2), $|13-12.875|=0.125$ (×3), $|14-12.875|=1.125$, $|15-12.875|=2.125$

$\text{MAD} = \frac{1.875 + 0.875 + 0.875 + 0.125 + 0.125 + 0.125 + 1.125 + 2.125}{8} = \frac{7.25}{8} \approx 0.91$

Ages are typically within about 0.9 years of the mean age.