NIMCET 2027 Statistics Measures Of Central Tendency Previous Year Questions (PYQs) – Page 2 of 4

Statistics Puzzle: Variance under Linear Transformation

Given:

• Y = 10.5 + 2X
• Var(Y) = 100

Formula: If Y = a + bX, then Var(Y) = b² × Var(X)

Apply:

• 100 = 2² × Var(X)
• 100 = 4 × Var(X)
• Var(X) = 100 / 4 = 25

✅ Final Answer: The variance of the marks given by the first judge is 25.

Given two sets:

• Set \( A \) has median \( m_1 = 2 \)
• Set \( B \) has median \( m_2 = 4 \)

What can we say about the median of the combined set \( A \cup B \)?

✅ Answer:

The combined median depends on the size and values of both sets.

Without that information, we only know that:

\[ \text{Combined Median} \in [2, 4] \]

So, the exact median cannot be determined with the given data.

Mean, Median, and Mode Relation

Given:

• Mean = 1
• Median = \( 3^x \)
• Mode = \( 9^x \)

Use empirical formula:

\[ \text{Mode} = 3 \cdot \text{Median} - 2 \cdot \text{Mean} \]

\[ 9^x = 3 \cdot 3^x - 2 \Rightarrow (3^x)^2 = 3 \cdot 3^x - 2 \]

Let \( y = 3^x \), then:

\[ y^2 = 3y - 2 \Rightarrow y^2 - 3y + 2 = 0 \Rightarrow (y - 1)(y - 2) = 0 \]

So, \( y = 1 \text{ or } 2 \Rightarrow 9^x = y^2 = 1 \text{ or } 4 \)

✅ Final Answer: \(\boxed{1 \text{ or } 4}\)

Given: Mean of squares of first n natural numbers is 11.

Use the identity for sum of squares: $$\sum_{k=1}^{n} k^2=\frac{n(n+1)(2n+1)}{6}$$

Mean \(=\) (sum) \( \div n \Rightarrow \) $$\frac{1}{n}\sum_{k=1}^{n}k^2=\frac{(n+1)(2n+1)}{6}=11$$

Solve for \(n\): $$(n+1)(2n+1)=66 $$ $$\;\Rightarrow\; 2n^2+3n+1=66 \;\Rightarrow\; $$ $$n=\frac{-3\pm\sqrt{9+520}}{4}=\frac{-3\pm 23}{4}$$

Only positive integer solution: $$n=\frac{20}{4}=5$$

Answer: \(n=5\).