Statistical Measures and Distribution Analysis Practice Questions for XAT: 154+ Solved Questions with Step-by-Step Solutions

Solve 154+ Statistical Measures and Distribution Analysis practice questions for XAT with answers and detailed solutions. Free sample questions below.

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.

Statistical Measures and Distribution Analysis: Solved Questions with Step-by-Step Explanations (5 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

Question 3 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

If a calculation shows that the minimum number of defective items in a batch is $2.4$ , what is the actual minimum number of defective items, given the integer constraint?

A.
2
B.
2.4
C.
3
D.
2.5

Correct Answer:

Step-by-Step Solution

Key idea: This is a boundary case question, recognizable because it presents a decimal result for a count of physical items and asks for the valid integer.

Step 1: Recall the Integer Constraint rule (card c003). You cannot have a fraction of a defective item.

Step 2: For a "minimum" bound, the actual count must be greater than or equal to the raw calculated value.

Step 3: The smallest whole number greater than or equal to $2.4$ is $3$ . (We round up for minimums).

Answer: 3

Question 4 · Quantitative Aptitude and Data Interpretation (QA & DI) MCQ

If the run rate at the end of the 5th over is 6, what is the total number of runs scored in the first 5 overs?

A.
30
B.
35
C.
1.2
D.
11

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct formula question, recognizable because it requires translating a run rate into a total cumulative score.

Step 1: Recall the Golden Translation Formula: $Total Runs = Run Rate \times Number of Overs$ .

Step 2: Substitute the given values: Run Rate = 6, Number of Overs = 5.

Step 3: Calculate: $6 \times 5 = 30$ .

Answer: 30

Question 5 · Quantitative Aptitude and Data Interpretation (QA & DI) NAT

In a 20-over match, the run rate at the end of over 14.3 is reported as 7.8 (correct to one decimal place). The runs scored in each completed over are non-negative integers, and the current over (over 15) has had 3 legal deliveries so far, scoring 2 runs. The run rate is calculated as total runs divided by total overs faced, where partial overs are counted fractionally (e.g., 14.3 overs = 14 + 3/6 = 14.5 overs).

What is the minimum possible total runs scored at the end of over 14?

Correct Answer:

111

Step-by-Step Solution

Key idea: Fractional over precision. Rounded run rate defines an interval. Back-calculate completed runs using partial over data.

Step 1: Decode Fractional Over.

14.3 overs = 14 overs + 3 balls = $14 + 3/6 = 14.5$ overs.

Step 2: Establish Run Rate Interval.

Reported 7.8 (1 d.p.) $⟹$ True RR $\in [7.75, 7.85)$ .

Step 3: Set Up Equation.

Let $R$ = runs at end of over 14.

Current over adds 2 runs.

Total runs at 14.5 overs = $R + 2$ .

Inequality: $7.75 \leq \frac{R + 2}{14.5} < 7.85$ .

Step 4: Solve for R.

$7.75 \times 14.5 \leq R + 2 < 7.85 \times 14.5$

$112.375 \leq R + 2 < 113.825$

$110.375 \leq R < 111.825$

Step 5: Apply Integer Constraint.

$R$ must be integer.

Possible integers in $[110.375, 111.825)$ : Only 111.

(Note: 110 is too small, 112 is too big).

Minimum (and only) possible R = 111.

Answer: 111

Statistical Measures and Distribution Analysis Practice Questions for XAT: 154+ Solved Questions with Step-by-Step Solutions

Box Plots and Hinge Analysis

Box Plots and Hinge Analysis

The Anatomy of a Box Plot

The Anatomy of a Box Plot

Statistical Measures and Distribution Analysis: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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