6.1 Measures of Central Tendency and Data Representation

Data representation and central tendency are crucial topics on the SHSAT Math section. These questions test your ability to extract values from visual displays and solve equations based on statistical parameters. To maximize your score, you must be able to perform standard calculations and manipulate formulas to solve for unknown values.

Mean, Median, Mode, and Range

Central tendency refers to the 'center' of a data set. The SHSAT tests four basic statistical measures:

• Mean (Arithmetic Average): The mean is the sum of all data values divided by the number of values. $\text{Mean} = \frac{\text{Sum of all values}}{\text{Total count of values}}$
• Median: The median is the middle value in a sorted list of numbers. To find the median, you must first arrange the values in ascending order. If there is an odd number of values, the median is the middle number. If there is an even number of values, the median is the average of the two middle numbers.
• Mode: The mode is the value that occurs most frequently. A data set can have one mode, more than one mode, or no mode at all (if all values occur with equal frequency).
• Range: The range measures the spread of the data. It is the difference between the greatest and least values in the set. $\text{Range} = \text{Maximum value} - \text{Minimum value}$

Working Backward from the Mean

A very common SHSAT question style provides the mean of a group of numbers and asks you to find a missing value or compute a new mean after values are added or removed. The key to solving these problems is the Sum Formula: $\text{Sum} = \text{Mean} \times \text{Number of values}$

Whenever you see the word 'mean' or 'average' in a word problem, immediately calculate the total sum of the values.

Case 1: Finding a Missing Score

Suppose a student has scored 82, 86, and 91 on their first three exams. What score must they earn on the fourth exam to have a mean score of 88?

1. Calculate the current sum: $\text{Current Sum} = 82 + 86 + 91 = 259$.
2. Calculate the required sum: To get a mean of 88 on 4 exams, the total sum must be $88 \times 4 = 352$.
3. Find the missing score: Subtract the current sum from the required sum. $\text{Fourth Score} = 352 - 259 = 93$.

Case 2: Altering the Dataset

The mean weight of 6 people is 150 pounds. If a 7th person joins the group, the new mean weight is 152 pounds. What is the weight of the 7th person?

1. Calculate the original sum: $\text{Original Sum} = 150 \times 6 = 900\text{ lbs}$.
2. Calculate the new sum: With 7 people, the sum is $152 \times 7 = 1,064\text{ lbs}$.
3. Find the 7th person's weight: $\text{Weight} = 1,064 - 900 = 164\text{ lbs}$.

Interpreting Tables and Graphs

The SHSAT presents data in several formats. You must be comfortable translating visual elements into numerical data.

Frequency Tables

A frequency table lists data values alongside the number of times each value occurs. To calculate the mean from a frequency table, you must find the sum of all individual items by multiplying each value by its frequency, adding these results, and dividing by the sum of all frequencies.

$\text{Mean Score} = \frac{850}{10} = 85$

To find the median from this table of 10 items, locate the 5th and 6th values. Since the first 3 values are 75 and the next 4 are 85, both the 5th and 6th values are 85. The median is therefore 85.

Bar Graphs and Histograms

Bar graphs display categorical data using rectangular bars, where the height or length of each bar represents the value or frequency of that category. Histograms are similar but show continuous numerical data grouped into range intervals (or 'bins') along the x-axis. When analyzing a bar graph or histogram on the SHSAT, always begin by identifying the units and increments on the horizontal and vertical axes. Make sure to distinguish between totals and percentages, as a common trap involves confusing the frequency of a category with its proportion of the whole. For example, if a histogram shows the number of students who scored in various ranges, you may need to add the heights of multiple bars to find the total sample size before computing percentages.

Circle Graphs (Pie Charts)

Circle graphs show how a whole is divided into sectors. Each sector represents a fraction, percentage, or angle of the total.

• The entire circle equals $100%$, which is also equivalent to $360^\circ$.
• To find the value of a sector when given its central angle, set up a proportion: $\frac{\text{Sector Angle}}{360^\circ} = \frac{\text{Sector Value}}{\text{Total Value}}$ For instance, if a circle graph represents 1,200 total students and the sector for 'Music' has a central angle of $72^\circ$, the number of students in music is $\frac{72}{360} \times 1,200 = \frac{1}{5} \times 1,200 = 240$ students.

Box Plots (Box-and-Whisker Plots)

A box plot displays the distribution of a dataset along a number line. It divides the data into four equal parts (quartiles) using five key boundaries:

1. Minimum: The lowest value in the dataset (endpoint of the left whisker).
2. First Quartile ($Q_1$): The median of the lower half of the data (left edge of the box). It represents the 25th percentile.
3. Second Quartile ($Q_2$ or Median): The middle value of the dataset (line inside the box). It represents the 50th percentile.
4. Third Quartile ($Q_3$): The median of the upper half of the data (right edge of the box). It represents the 75th percentile.
5. Maximum: The highest value in the dataset (endpoint of the right whisker).

The Interquartile Range (IQR) is the difference between $Q_3$ and $Q_1$, representing the spread of the middle 50% of the data. A critical fact for the SHSAT is that each of the four regions (Minimum to $Q_1$, $Q_1$ to Median, Median to $Q_3$, $Q_3$ to Maximum) contains exactly 25% of the data points, regardless of the physical length of the segment on the chart.

SHSAT Strategy and Common Pitfalls

• The Weighted Average Trap: If one class of 10 students averages 80% on a test and another class of 20 students averages 90%, the combined average is not $\frac{80 + 90}{2} = 85%$. Because the classes are different sizes, you must find the total points: $\text{Total Points} = (10 \times 80) + (20 \times 90) = 800 + 1800 = 2600$. Divide by the total number of students (30) to get the correct average: $\frac{2600}{30} \approx 86.7%$.
• Outlier Sensitivity: The mean is highly sensitive to outliers (extreme values), while the median is not. If a dataset has a few extremely high values, the mean will increase significantly, but the median will remain relatively stable.
• Box Plot Spacing: Do not confuse the length of a box plot section with the number of data points. A long whisker indicates that the data points in that quartile are widely spread out, not that there are more of them.