Statistics and Probability

1. a) Not statistical. There is only one specific answer — it does not anticipate variability in the data.

b) Statistical. You expect a variety of different ages; the data will vary from student to student.

c) Not statistical. It asks for a single measurement at one moment in time.

d) Statistical. Rainfall varies from month to month; the answer involves data with variability.

2. First, order the data: $6, 7, 8, 8, 8, 9, 10$.

Mean:

$\text{Mean} = \frac{6 + 7 + 8 + 8 + 8 + 9 + 10}{7} = \frac{56}{7} = 8$

Median: The middle value (4th value) in the ordered list is $8$.

Mode: The most frequent value is $8$ (appears 3 times).

3. Tree A:

$\text{Mean} = \frac{5 + 6 + 7 + 6 + 5 + 7 + 6}{7} = \frac{42}{7} = 6 \text{ cm}$

$\text{Range} = 7 - 5 = 2 \text{ cm}$

Tree B:

$\text{Mean} = \frac{3 + 5 + 8 + 10 + 4 + 9 + 3}{7} = \frac{42}{7} = 6 \text{ cm}$

$\text{Range} = 10 - 3 = 7 \text{ cm}$

Both data sets have the same mean ($6$ cm), so their centers are equal. Tree B has a much greater range ($7$ vs $2$), so Tree B has greater spread — its data is more spread out from the center.

4. The data is already in order: $12, 15, 18, 20, 22, 25, 30$.

Median: The middle value (4th of 7 values) is $20$.

Range:

$\text{Range} = 30 - 12 = 18$

The center of the distribution is around $20$ (the median). The spread is $18$ units, meaning the data spans from $12$ to $30$. The distribution is roughly symmetric around $20$.

5. First, order the data: $68, 72, 73, 76, 82, 85, 88, 90, 91, 95$.

Mean:

$\text{Mean} = \frac{68 + 72 + 73 + 76 + 82 + 85 + 88 + 90 + 91 + 95}{10} = \frac{820}{10} = 82$

Median: With 10 values, the median is the average of the 5th and 6th values:

$\text{Median} = \frac{82 + 85}{2} = \frac{167}{2} = 83.5$

Range:

$\text{Range} = 95 - 68 = 27$

The mean ($82$) and median ($83.5$) are close, suggesting the data is approximately symmetric. Most scores cluster in the $72$–$95$ range with a spread of $27$ points. The distribution is roughly symmetric with no extreme outliers.