Statistics

1. a) Dot plot (each dot represents one student):

| Books | Dots | |-------|------| | 2 | $\bullet$ | | 3 | $\bullet\bullet$ | | 4 | $\bullet\bullet\bullet$ | | 5 | $\bullet\bullet\bullet\bullet$ | | 6 | $\bullet\bullet$ | | 7 | $\bullet$ | | 8 | $\bullet$ | | 12 | $\bullet$ |

b) Mean:

$\bar{x} = \frac{2+3+3+4+4+4+5+5+5+5+6+6+7+8+12}{15} = \frac{79}{15} \approx 5.27$

Median: The data is already sorted. With 15 values, the median is the 8th value: $\text{median} = 5$.

Mode: The value that appears most often is $5$ (appears 4 times).

c) The median (or mode, both equal to 5) best represents this data. The value $12$ is much higher than the rest, pulling the mean up. Since the distribution is right-skewed, the median is more resistant to the outlier's influence.

2. a) The data is already sorted. With $n = 20$ values:

• Minimum: $18$
• $Q_1$: median of the lower 10 values ($18, 19, 21, 22, 23, 24, 25, 25, 26, 27$). The median of these is $\frac{23 + 24}{2} = 23.5$.
• Median: $\frac{27 + 28}{2} = 27.5$ (average of 10th and 11th values)
• $Q_3$: median of the upper 10 values ($28, 29, 30, 31, 32, 34, 35, 38, 42, 61$). The median of these is $\frac{32 + 34}{2} = 33$.
• Maximum: $61$

Five-number summary: $\{18,\; 23.5,\; 27.5,\; 33,\; 61\}$

b) $\text{IQR} = Q_3 - Q_1 = 33 - 23.5 = 9.5$

c) Outlier fences:

$\text{Lower fence} = Q_1 - 1.5 \times \text{IQR} = 23.5 - 14.25 = 9.25$ $\text{Upper fence} = Q_3 + 1.5 \times \text{IQR} = 33 + 14.25 = 47.25$

Any value below $9.25$ or above $47.25$ is an outlier. The value $61$ exceeds the upper fence.

Yes, $61$ is an outlier.

3. a) Player A:

$\bar{x}_A = \frac{12+14+15+15+16+17+18+19+20+24}{10} = \frac{170}{10} = 17$

Deviations from the mean: $-5, -3, -2, -2, -1, 0, 1, 2, 3, 7$

$s_A = \sqrt{\frac{(-5)^2+(-3)^2+(-2)^2+(-2)^2+(-1)^2+0^2+1^2+2^2+3^2+7^2}{10-1}}$

$= \sqrt{\frac{25+9+4+4+1+0+1+4+9+49}{9}} = \sqrt{\frac{106}{9}} \approx 3.43$

Player B:

$\bar{x}_B = \frac{8+10+14+16+17+17+18+20+22+28}{10} = \frac{170}{10} = 17$

Deviations from the mean: $-9, -7, -3, -1, 0, 0, 1, 3, 5, 11$

$s_B = \sqrt{\frac{81+49+9+1+0+0+1+9+25+121}{9}} = \sqrt{\frac{296}{9}} \approx 5.74$

b) Both players have the same mean (17 points), but Player A has a smaller standard deviation ($s_A \approx 3.43$ vs. $s_B \approx 5.74$). Player A is more consistent because their scores are more tightly clustered around the mean.

4. a) The distribution peaks in the 70–79 range and tapers off more steeply to the left. It is approximately symmetric with a very slight left skew (the tail on the left side is slightly longer since the 50–59 bin has fewer students than the 90–99 bin, but the difference is minor).

b) Using midpoints: $54.5,\; 64.5,\; 74.5,\; 84.5,\; 94.5$

$\text{Estimated mean} = \frac{2(54.5) + 5(64.5) + 10(74.5) + 8(84.5) + 5(94.5)}{30}$

$= \frac{109 + 322.5 + 745 + 676 + 472.5}{30} = \frac{2325}{30} = 77.5$

c) The median is between the 15th and 16th values. The cumulative frequencies are:

• Through 50–59: 2 students
• Through 60–69: $2 + 5 = 7$ students
• Through 70–79: $7 + 10 = 17$ students

The 15th and 16th values both fall in the 70–79 interval (since cumulative frequency reaches 17 by the end of this interval, and we need the 15th and 16th values). The median is in the 70–79 range.

5. a) The original sum is $20 \times 50 = 1000$. Adding the outlier:

$\text{New mean} = \frac{1000 + 200}{21} = \frac{1200}{21} \approx 57.14$

The mean increased from $50$ to approximately $57.14$ — a shift of over 7 points from a single value.

b) The median will not change significantly. The original median is the average of the 10th and 11th values. After adding $200$, it becomes the 11th value (of 21). Since $200$ is placed at the end of the sorted data, the middle of the data barely shifts — it moves from between the 10th–11th to the 11th value, which is near the original center.

c) The mean uses every value in its calculation (summing all values and dividing by $n$), so a single extreme value directly changes the total sum and shifts the result. The median depends only on the position of the middle value(s) in the sorted data; adding an extreme value at one end shifts the middle position by at most half a step, leaving the median nearly unchanged. This is why the median is called a resistant (or robust) measure of center.