CUET PG 2026 Statistics Question Paper with Solutions

Concept:

For any two events \(A\) and \(B\), the probability of their union is given by: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

If \(A\) and \(B\) are independent events, then: \[ P(A \cap B) = P(A)P(B) \]

Thus, the formula becomes: \[ P(A \cup B) = P(A) + P(B) - P(A)P(B) \]

Step 1: {\color{redFind \(P(A \cap B)\) using independence.
\[ P(A \cap B) = P(A)P(B) \]

Substitute the given values:
\[ P(A \cap B) = 0.4 \times 0.5 = 0.20 \]

Step 2: {\color{redApply the union formula.
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

Substitute the values:
\[ P(A \cup B) = 0.4 + 0.5 - 0.20 \]
\[ P(A \cup B) = 0.70 \]
\[ \therefore P(A \cup B) = 0.70 \] Quick Tip: For independent events, remember that the probability of both events occurring together is the product of their probabilities: \[ P(A \cap B) = P(A)P(B) \] This simplifies many probability calculations.

Concept:

For a Poisson distribution with parameter \( \lambda \):


Mean \(= \lambda\)
Variance \(= \lambda\)
The third central moment \( \mu_3 \) is also equal to \( \lambda \)


Thus, \[ \mu_3 = \lambda \]

Step 1: {\color{redIdentify the parameter of the distribution.

The mean of the Poisson distribution is given as: \[ \lambda = 4 \]

Step 2: {\color{redUse the formula for the third central moment.
\[ \mu_3 = \lambda \]

Substituting the value:
\[ \mu_3 = 4 \]
\[ \therefore The third central moment is 4 \] Quick Tip: For a Poisson distribution, several moments have simple forms: \[ Mean = Variance = Third central moment = \lambda \] This property makes calculations involving Poisson moments very straightforward.

Concept:

In statistical estimation, an estimator \( \hat{\theta} \) is said to be unbiased if its expected value equals the true population parameter \( \theta \). Mathematically, this is expressed as:
\[ E(\hat{\theta}) = \theta \]

This means that on average, the estimator correctly estimates the true parameter value.

Step 1: {\color{redUnderstand the definition of an unbiased estimator.

If an estimator \( \hat{\theta} \) satisfies
\[ E(\hat{\theta}) = \theta \]

then the estimator does not systematically overestimate or underestimate the parameter.

Step 2: {\color{redIdentify the property.

The property that ensures the expected value of an estimator equals the population parameter is called Unbiasedness.
\[ \therefore The correct answer is Unbiasedness. \] Quick Tip: An estimator is \textbf{unbiased} if: \[ E(\hat{\theta}) = \theta \] This means the estimator gives the correct value of the parameter on average over many samples.

Concept:

If \( \lambda \) is an eigenvalue of a matrix \(A\), then the eigenvalue of \(A^k\) is \( \lambda^k \).

This follows from the eigenvalue definition: \[ A\mathbf{x} = \lambda \mathbf{x} \]

Multiplying both sides by \(A\):
\[ A^2 \mathbf{x} = A(\lambda \mathbf{x}) = \lambda (A\mathbf{x}) = \lambda(\lambda \mathbf{x}) = \lambda^2 \mathbf{x} \]

Thus, the eigenvalues of \(A^2\) are the squares of the eigenvalues of \(A\).

Step 1: {\color{redIdentify the eigenvalues of \(A\).

Given: \[ \lambda_1 = 2, \qquad \lambda_2 = 3 \]

Step 2: {\color{redFind eigenvalues of \(A^2\).

Square each eigenvalue:
\[ \lambda_1^2 = 2^2 = 4 \]
\[ \lambda_2^2 = 3^2 = 9 \]

Step 3: {\color{redWrite the eigenvalues of \(A^2\).
\[ Eigenvalues of A^2 = 4, 9 \]
\[ \therefore The correct answer is 4, 9. \] Quick Tip: If \( \lambda \) is an eigenvalue of \(A\), then for any positive integer \(k\): \[ Eigenvalues of A^k = \lambda^k \] Thus, simply raise each eigenvalue of \(A\) to the power \(k\).

Concept:

A Normal distribution follows the well-known Empirical Rule (also called the 68–95–99.7 rule). This rule describes how data is distributed around the mean.


About \(68%\) of the data lies within \(1\) standard deviation of the mean.
About \(95%\) of the data lies within \(2\) standard deviations of the mean.
About \(99.7%\) of the data lies within \(3\) standard deviations of the mean.


Step 1: {\color{redIdentify the required interval.

The question asks for the percentage of data within two standard deviations from the mean.
\[ \mu - 2\sigma \le X \le \mu + 2\sigma \]

Step 2: {\color{redApply the Empirical Rule.

According to the empirical rule:
\[ P(\mu - 2\sigma \le X \le \mu + 2\sigma) \approx 95% \]
\[ \therefore Approximately 95% of the data lies within two standard deviations of the mean. \] Quick Tip: Remember the \textbf{68–95–99.7 rule} for Normal distribution: \[ 1\sigma \rightarrow 68%, \qquad 2\sigma \rightarrow 95%, \qquad 3\sigma \rightarrow 99.7% \] It is frequently used in statistics and data analysis.

Concept:

The identity matrix \(I_n\) is a square matrix with ones on the main diagonal and zeros elsewhere. The null matrix (or zero matrix) has all its elements equal to zero.

Key properties:

Adding a null matrix to any matrix does not change the matrix.
\[ A + O = A \]
The rank of the identity matrix \(I_n\) is \(n\) because all rows (and columns) are linearly independent.


Step 1: {\color{redAdd the matrices.

Let the identity matrix be \[ I_3 = \begin{bmatrix} 1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix} \]

The null matrix is \[ O = \begin{bmatrix} 0 & 0 & 0
0 & 0 & 0
0 & 0 & 0 \end{bmatrix} \]

Adding them:
\[ I_3 + O = I_3 \]

Step 2: {\color{redFind the rank of the resulting matrix.

Since the resulting matrix is still the identity matrix \(I_3\), all its rows (and columns) are linearly independent.
\[ Rank(I_3) = 3 \]
\[ \therefore The rank of the matrix is 3. \] Quick Tip: The rank of an identity matrix \(I_n\) is always \(n\), because all rows and columns are linearly independent.

Concept:

In Simple Random Sampling Without Replacement (SRSWOR), each unit in the population has an equal probability of being selected at any draw.

Even though units are not replaced, the probability that any particular unit appears in a specific draw remains the same due to symmetry of selection.

If the population size is \(N\), then the probability that a specific unit is selected in any particular position of the sample is:
\[ P = \frac{1}{N} \]

Step 1: {\color{redConsider the population size.

Let the population contain \(N\) units, and suppose we are interested in the probability that a particular unit (say unit \(i\)) is selected at the second draw.

Step 2: {\color{redConsider the two possible ways this can happen.

The specific unit can appear in the second draw if:


It was not selected in the first draw.
It is selected in the second draw.


Thus,
\[ P(unit i at second draw) = P(not selected in first) \times P(selected in second) \]
\[ = \left(\frac{N-1}{N}\right) \left(\frac{1}{N-1}\right) \]
\[ = \frac{1}{N} \]

Step 3: {\color{redInterpret the result.

Hence, the probability that a particular unit is selected at the second draw is:
\[ \frac{1}{N} \]
\[ \therefore The correct answer is \dfrac{1}{N}. \] Quick Tip: In SRSWOR, every unit has an equal chance of appearing in any position of the sample. Thus, the probability that a specific unit appears in the \(k^{th}\) draw is always: \[ \frac{1}{N} \] where \(N\) is the population size.

Concept:

The coefficient of determination is the square of the correlation coefficient. It represents the proportion of the variation in the dependent variable that is explained by the independent variable.

If the correlation coefficient is \(r\), then:
\[ R^2 = r^2 \]

where \(R^2\) is called the coefficient of determination.

Step 1: {\color{redIdentify the given correlation coefficient.
\[ r = 0.8 \]

Step 2: {\color{redSquare the correlation coefficient.
\[ R^2 = (0.8)^2 \]
\[ R^2 = 0.64 \]

Step 3: {\color{redInterpret the result.

Thus, the coefficient of determination is:
\[ 0.64 \]
\[ \therefore The correct answer is 0.64. \] Quick Tip: The coefficient of determination is simply the square of the correlation coefficient: \[ R^2 = r^2 \] It indicates the proportion of variation in one variable explained by the other.

Concept:

When objects are drawn without replacement, the probability of successive events changes after each draw.
If the total number of objects is \(n\), and \(r\) objects satisfy a condition, the probability can be computed using either:


Multiplication rule of probability, or
Combination formula

\[ P(both red) = \frac{Number of favorable outcomes}{Total number of outcomes} \]

Step 1: {\color{redDetermine the total number of balls.
\[ Total balls = 5 + 7 = 12 \]

Step 2: {\color{redCompute the probability using the multiplication rule.

Probability that the first ball is red:
\[ P(R_1) = \frac{5}{12} \]

After drawing one red ball, remaining red balls \(=4\) and total balls \(=11\).

Probability that the second ball is red:
\[ P(R_2|R_1) = \frac{4}{11} \]

Step 3: {\color{redFind the joint probability.
\[ P(both red) = \frac{5}{12} \times \frac{4}{11} \]
\[ = \frac{20}{132} \]
\[ = \frac{10}{66} \]
\[ \therefore The probability that both balls are red is \frac{10}{66}. \] Quick Tip: For drawing without replacement: \[ P(A \cap B) = P(A) \times P(B|A) \] Alternatively, use combinations: \[ P(both red) = \frac{\binom{5}{2}}{\binom{12}{2}} \]

Concept:

When comparing the means of two samples, the choice of test depends mainly on the sample size and knowledge of population variance.


If the sample size is large (\(n \ge 30\)), the Z-test is generally used.
If the sample size is small (\(n < 30\)) and the population variance is unknown, the Student's t-test is used.


The t-test is specifically designed for small samples and accounts for additional uncertainty in estimating the population variance.

Step 1: {\color{redIdentify the condition given in the question.

The problem states that the samples are small.

Step 2: {\color{redChoose the appropriate statistical test.

For small sample means, the correct test is the Student's t-test.

Step 3: {\color{redState the conclusion.
\[ \therefore The appropriate test for comparing two small sample means is the t-test. \] Quick Tip: Use the \textbf{t-test} when: Sample size is small (\(n < 30\)) Population variance is unknown It is commonly used to test the significance of the difference between two sample means.

Concept:

In hypothesis testing, two types of errors may occur when making a decision about the null hypothesis.


Type I Error: Rejecting the null hypothesis \(H_0\) when it is actually true.
Type II Error: Failing to reject the null hypothesis \(H_0\) when it is actually false.


The probability of committing a Type I error is denoted by \( \alpha \), which is also called the level of significance.

Step 1: {\color{redUnderstand the situation given in the question.

The question states that the null hypothesis \(H_0\) is rejected even though it is actually true.

Step 2: {\color{redIdentify the type of error.

This situation exactly matches the definition of a Type I Error.

Step 3: {\color{redState the conclusion.
\[ \therefore Rejecting a true null hypothesis results in a Type I Error. \] Quick Tip: Remember the two main hypothesis testing errors: \[ Type I Error: Reject H_0 when it is true \] \[ Type II Error: Fail to reject H_0 when it is false \] Type I error probability is the \textbf{significance level} \( \alpha \).

Concept:

The integral \[ \int_{-\infty}^{\infty} e^{-x^2}\,dx \]
is known as the Gaussian integral. It plays a fundamental role in probability theory, especially in the Normal distribution.

A well-known result from calculus states:
\[ \int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi} \]

This result is usually derived by squaring the integral and converting it into polar coordinates.

Step 1: {\color{redDefine the integral.

Let
\[ I = \int_{-\infty}^{\infty} e^{-x^2}\,dx \]

Step 2: {\color{redSquare the integral.
\[ I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2}\,dx\right) \left(\int_{-\infty}^{\infty} e^{-y^2}\,dy\right) \]
\[ = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}\,dx\,dy \]

Step 3: {\color{redConvert to polar coordinates.

Using \(x^2 + y^2 = r^2\) and \(dx\,dy = r\,dr\,d\theta\):
\[ I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} r\,dr\,d\theta \]
\[ = 2\pi \int_{0}^{\infty} e^{-r^2} r\,dr \]

Let \(u=r^2\), \(du=2r\,dr\):
\[ \int_{0}^{\infty} e^{-r^2} r\,dr = \frac{1}{2} \]

Thus,
\[ I^2 = 2\pi \times \frac{1}{2} = \pi \]

Step 4: {\color{redTake the square root.
\[ I = \sqrt{\pi} \]
\[ \therefore \int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi} \] Quick Tip: The Gaussian integral is a standard result in calculus and statistics: \[ \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi} \] It is closely related to the Normal distribution.

Concept:

A negatively skewed distribution (also called left-skewed distribution) has a longer tail on the left side. In such distributions, smaller values pull the mean toward the left.

Because of this effect:


The mean is pulled most toward the tail.
The median lies between the mean and the mode.
The mode remains near the peak of the distribution.


Thus, the relationship becomes:
\[ Mean < Median < Mode \]

Step 1: {\color{redUnderstand the effect of skewness.

In a negatively skewed distribution, extreme low values pull the mean toward the left side.

Step 2: {\color{redDetermine the order of central tendencies.

Since the mean is affected the most by extreme values:
\[ Mean < Median < Mode \]

Step 3: {\color{redState the conclusion.
\[ \therefore The correct relationship is Mean < Median < Mode. \] Quick Tip: Remember the order of central tendencies: \[ Negatively skewed: Mean < Median < Mode \] \[ Positively skewed: Mean > Median > Mode \] \[ Symmetrical distribution: Mean = Median = Mode \]

Concept:

The limit
\[ \lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n \]

is a well-known fundamental limit in calculus. It defines the mathematical constant \(e\), which is the base of natural logarithms.
\[ e \approx 2.71828 \]

Thus,
\[ \lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n = e \]

Step 1: {\color{redIdentify the standard limit form.

The expression given in the question exactly matches the standard definition of \(e\):
\[ e = \lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n \]

Step 2: {\color{redApply the known result.

Since the expression is the same as the definition of \(e\),
\[ \lim_{n \to \infty}\left(1+\frac{1}{n}\right)^n = e \]

Step 3: {\color{redState the final result.
\[ \therefore The value of the limit is e. \] Quick Tip: A very important limit in calculus is: \[ \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = e \] This limit is used in compound interest, exponential growth, and many areas of mathematics.

Concept:

In a Chi-square test for independence using a contingency table, the degrees of freedom are calculated using the formula:
\[ df = (r-1)(c-1) \]

where:

\(r\) = number of rows
\(c\) = number of columns


Step 1: {\color{redIdentify the number of rows and columns.

Given contingency table:
\[ 3 \times 4 \]

Thus,
\[ r = 3, \qquad c = 4 \]

Step 2: {\color{redApply the formula for degrees of freedom.
\[ df = (r-1)(c-1) \]
\[ df = (3-1)(4-1) \]
\[ df = (2)(3) \]
\[ df = 6 \]

Step 3: {\color{redState the conclusion.
\[ \therefore The degree of freedom is 6. \] Quick Tip: For a Chi-square test using a contingency table: \[ df = (r-1)(c-1) \] where \(r\) is the number of rows and \(c\) is the number of columns.