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

Statistical Measures and Distribution Analysis short notes for XAT: 4 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

The Hinge Calculation Checklist

Summary

The Hinge Calculation Checklist

Final Checklist for XAT Hinge Problems:

Sort the Data: Always arrange in ascending order first.

Locate the Median Index: Find the exact middle position.

Split Strictly: Divide at the median index. Do not include the median in either half.

Calculate Lower Hinge: Find the median of the left part.

Calculate Upper Hinge: Find the median of the right part.

Verify the Split: Both halves plus the median must equal the original dataset size.

The Defective Items Checklist

Final Checklist for Defective Item Problems:

1Count Inclusively: Use $End - Start + 1$ to find the exact number of boxes.
2Calculate Total Items: Multiply box count by capacity per box type.
3Find Raw Bounds: Apply percentage bounds to find raw defective counts.
4Apply Integer Constraint: Round up the minimum and round down the maximum.
5Aggregate Totals: Use frequency tables to multiply and sum defectives.
6Verify Validity: Double-check that all final counts are non-negative integers.

Quick Formula Recap

Formula Recap

Total Runs

T_{k} = k \times R R_{k}

Runs in Over

k

R_{k} = T_{k} - T_{k - 1}

Required RR

\frac{Target - T _{k}}{Rem. Overs}

Lower Bound

R R_{k} \geq \frac{k - 1}{k} R R_{k - 1}

Final Exam Checklist

Exam Checklist

✓

Converted fractional overs (base-6)?

✓

Checked Cumulative vs. Marginal?

✓

Are all over scores

\geq 0

✓

Are over scores integers?

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 Short Notes for XAT: Concepts, Formulas, Worked Examples & Practice

The Hinge Calculation Checklist

The Hinge Calculation Checklist

The Defective Items Checklist

The Defective Items Checklist

Quick Formula Recap

Formula Recap

Final Exam Checklist

Exam Checklist

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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