Statistical Measures and Distribution Analysis Notes for XAT: Concepts, Formulas, Worked Examples & Practice

Statistical Measures and Distribution Analysis notes for XAT: 36 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Box Plots and Hinge Analysis

Visualize the spread, center, and outliers of any dataset in a single glance.

What you will learn here

Construct and interpret box plots for quick data comparison.
Master the XAT-specific definition of hinges to avoid calculation traps.
Calculate the five-number summary flawlessly for any dataset.
Identify skewness and spread directly from the shape of the box.

Chapter: Statistical Measures and Distribution Analysis

The Anatomy of a Box Plot

A box plot is a standardized way of displaying a dataset based on a five-number summary. It provides a visual snapshot of the data's center, spread, and range.

The Box: Represents the middle 50 percent of the data, stretching from the lower hinge to the upper hinge.

The Median Line: A vertical line inside the box marking the exact center of the dataset.

The Whiskers: Lines extending from the box to the minimum and maximum values, showing the full range.

The Five-Number Summary

Every box plot is built on five critical values, arranged in ascending order.

Minimum — The smallest value in the dataset.

Lower Hinge — The median of the lower half of the data.

Median — The exact middle value of the entire dataset.

Upper Hinge — The median of the upper half of the data.

Maximum — The largest value in the dataset.

These five points divide the data into four equal quarters, each containing 25 percent of the observations.

XAT's Specific Definition of Hinges

XAT uses a very specific method for hinges, often called the position-split method. It differs slightly from standard textbook quartiles.

The Rule

Sort the data in ascending order.
Find the median (the middle value).
Lower Hinge: The median of all values strictly to the left of the median's position.
Upper Hinge: The median of all values strictly to the right of the median's position.

Crucial Note: The median itself is excluded from both halves when finding the hinges.

Statistical Measures and Distribution Analysis: Solved Questions with Step-by-Step Explanations (2 Problems)

Question 1 · Quantitative Aptitude and Data Interpretation (QA & DI) NAT

A dataset consists of $14$ distinct positive integers. When arranged in ascending order, the median is $25$ , the lower hinge is $12$ , and the upper hinge is $38$ .

The minimum value in the dataset is $2$ and the maximum value is $60$ .

The sum of all $14$ integers is exactly $337$ .

What is the maximum possible value of the $6$ th integer in the ascending arrangement?

Correct Answer:

Step-by-Step Solution

Key idea: This is a constrained optimization problem involving the XAT-specific definition of hinges for an even-sized dataset. We must maximize a specific element by minimizing all other elements subject to the ordering and sum constraints.

Step 1: Map the indices and hinge definitions.

For $N = 14$ , the sorted data is $x_{1}, x_{2}, \dots, x_{14}$ .

Median is the average of $x_{7}$ and $x_{8}$ . Given Median $= 25 ⟹ x_{7} + x_{8} = 50$ .
Lower Hinge is the median of the lower half ${x_{1}, \dots, x_{7}}$ . Since there are 7 elements, the median is the 4th element. Thus, $x_{4} = 12$ .
Upper Hinge is the median of the upper half ${x_{8}, \dots, x_{14}}$ . The median of these 7 elements is the 4th element of this subset, which corresponds to $x_{11}$ . Thus, $x_{11} = 38$ .
We are given $x_{1} = 2$ and $x_{14} = 60$ .

Step 2: Set up the sum equation.

The total sum is $337$ .

$x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} + x_{10} + x_{11} + x_{12} + x_{13} + x_{14} = 337$ .

Substitute the known values:

$2 + x_{2} + x_{3} + 12 + x_{5} + x_{6} + 50 + x_{9} + x_{10} + 38 + x_{12} + x_{13} + 60 = 337$ .

$162 + x_{2} + x_{3} + x_{5} + x_{6} + x_{9} + x_{10} + x_{12} + x_{13} = 337$ .

$x_{2} + x_{3} + x_{5} + x_{6} + x_{9} + x_{10} + x_{12} + x_{13} = 175$ .

Step 3: Maximize $x_{6}$ by minimizing the other variables.

To maximize $x_{6}$ , we must minimize $x_{2}, x_{3}, x_{5}$ and $x_{9}, x_{10}, x_{12}, x_{13}$ .

Minimize $x_{2}, x_{3}$ : Since $x_{1} = 2$ and $x_{4} = 12$ , the smallest distinct integers for $x_{2}, x_{3}$ are $3$ and $4$ .
Minimize $x_{5}$ : Since $x_{4} = 12$ , the smallest integer for $x_{5}$ is $13$ .
Minimize $x_{9}, x_{10}$ : These depend on $x_{8}$ . To minimize them, we must minimize $x_{8}$ . Since $x_{7} + x_{8} = 50$ and $x_{7} < x_{8}$ , the minimum possible value for $x_{8}$ occurs when $x_{7}$ is maximized. However, $x_{7}$ must be strictly greater than $x_{6}$ . Let's assume $x_{6}$ is large, which forces $x_{7}$ to be large. To find the absolute minimum for $x_{9}, x_{10}$ , we look at the constraint $x_{7} < x_{8} ⟹ x_{7} < 25$ . The maximum possible $x_{7}$ is $24$ , which gives $x_{8} = 26$ .

If $x_{8} = 26$ , then $x_{9} \geq 27$ and $x_{10} \geq 28$ .

Minimize $x_{12}, x_{13}$ : Since $x_{11} = 38$ and $x_{14} = 60$ , the smallest distinct integers are $x_{12} = 39$ and $x_{13} = 40$ .

Step 4: Calculate the maximum $x_{6}$ .

Sum of minimized variables (excluding $x_{6}$ ):

$x_{2} + x_{3} + x_{5} + x_{9} + x_{10} + x_{12} + x_{13} = 3 + 4 + 13 + 27 + 28 + 39 + 40 = 154$ .

Substitute back into the sum equation:

$154 + x_{6} = 175 ⟹ x_{6} = 21$ .

Step 5: Verify validity.

If $x_{6} = 21$ , then $x_{7}$ must be at least $22$ .

If $x_{7} = 22$ , then $x_{8} = 28$ .

Then $x_{9} \geq 29, x_{10} \geq 30$ .

The sum of these would be $29 + 30 = 59$ , which is greater than $27 + 28 = 55$ . This would force $x_{6}$ to be smaller than 21.

Therefore, to keep $x_{6}$ at its maximum of 21, we must use the configuration that minimizes the upper variables: $x_{7} = 24, x_{8} = 26$ , which allows $x_{9} = 27, x_{10} = 28$ .

The sequence $2, 3, 4, 12, 13, 21, 24, 26, 27, 28, 38, 39, 40, 60$ is strictly increasing, satisfies all hinge/median conditions, and sums to 337.

Answer: 21

Question 2 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

If the run rate at the end of the 4th over is 8, what is the minimum possible run rate at the end of the 5th over, given that runs scored in any over must be non-negative?

A.
0
B.
6.4
C.
8
D.
10

Correct Answer:

Step-by-Step Solution

Key idea: This is a boundary case question, recognizable because it applies a physical constraint (non-negative runs) to find the minimum possible future run rate.

Step 1: Calculate the total runs after 4 overs. $T_{4} = 4 \times 8 = 32$ .

Step 2: Apply the non-negative constraint. The minimum runs that can be scored in the 5th over is 0.

Step 3: Calculate the minimum total runs after 5 overs. $T_{5} = T_{4} + 0 = 32 + 0 = 32$ .

Step 4: Calculate the minimum run rate after 5 overs. $R R_{5} = \frac{T _{5}}{5} = \frac{32}{5} = 6.4$ .

Answer: 6.4

Statistical Measures and Distribution Analysis Notes for XAT: Concepts, Formulas, Worked Examples & Practice

Box Plots and Hinge Analysis

Box Plots and Hinge Analysis

The Anatomy of a Box Plot

The Anatomy of a Box Plot

The Five-Number Summary

The Five-Number Summary

XAT's Specific Definition of Hinges

XAT's Specific Definition of Hinges

Statistical Measures and Distribution Analysis: Solved Questions with Step-by-Step Explanations (2 Problems)

Step-by-Step Solution

Step-by-Step Solution

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