SSC CGL 2024 Tier-II (Statistics) Official Paper-II (Held On: 19 Jan, 2025) (Previous Year Paper)

Given:

The joint probability density function for the bivariate random variable (X, Y) is:

f(x,y)=9(1+x+y)2(1+x)4(1+y)4f(x,y)=9(1+x+y)2(1+x)4(1+y)4

, for 0<x<∞,0<y<∞0<x<∞,0<y<∞

.

We need to find the marginal probability density function of XX

.

Concept:

The marginal probability density function of XX

is obtained by integrating the joint probability density function f(x,y)f(x,y)

over all values of yy

:

fX(x)=∫∞0f(x,y)dyfX(x)=∫0∞f(x,y)dy

.

Calculation:

We start with:

fX(x)=∫∞09(1+x+y)2(1+x)4(1+y)4dyfX(x)=∫0∞9(1+x+y)2(1+x)4(1+y)4dy

.

⇒ Split the numerator 1+x+y1+x+y

into 1+x1+x

and yy

:

fX(x)=92(1+x)4∫∞0(1+x)(1+y)4dy+92(1+x)4∫∞0y(1+y)4dy.fX(x)=92(1+x)4∫0∞(1+x)(1+y)4dy+92(1+x)4∫0∞y(1+y)4dy.

⇒ Evaluate the integrals separately:

∫∞01(1+y)4dy=13∫0∞1(1+y)4dy=13

(standard integration result).

∫∞0y(1+y)4dy=16∫0∞y(1+y)4dy=16

(standard integration result).

⇒ Substitute these values back:

fX(x)=92(1+x)4[(1+x)3+16].fX(x)=92(1+x)4[(1+x)3+16].

⇒ Combine the terms inside the brackets:

fX(x)=92(1+x)4[2(1+x)+16].fX(x)=92(1+x)4[2(1+x)+16].

⇒ Simplify further:

fX(x)=912(1+x)4[2x+3].fX(x)=912(1+x)4[2x+3].

⇒ fX(x)=3(2x+3)4(1+x)4.fX(x)=3(2x+3)4(1+x)4.

∴ The marginal probability density function of XX

is:

fX(x)=3(2x+3)4(1+x)4.fX(x)=3(2x+3)4(1+x)4.

Calculation:

Given trend-adjusted ratios:

January = 1.1, February = 0.9, March = 1.0

Average seasonal index for the quarter:

$\text{Average} = \frac{1.1 + 0.9 + 1.0}{3} = \frac{3.0}{3} = 1.0$

The Correct answer is 1.0

Key Points

• Range Statistic (R):
  

What it is: Difference between the largest and smallest means being compared.

Where it is used: In multiple comparison tests, especially Duncan’s test.

Relation to DMRT: DMRT is based on the studentized range statistic (R)

Used to decide whether the difference between means is significant.

∴ It is applicable to DMRT 

Additional Information

• Standard Error of the Means
• What it is: Measures how much the sample mean varies from the population mean.
• Where it is used: Common in t-tests, confidence intervals, and ANOVA.
• Relation to DMRT: DMRT does not directly use standard error as the decision statistic.
• Regression Coefficient
• What it is: Indicates the relationship between dependent and independent variables.
• Where it is used: In regression analysis.
• Relation to DMRT: DMRT compares group means, not relationships between variables.
• F-statistic
• What it is: Ratio of variance between groups to variance within groups.
• Where it is used: In ANOVA to test overall significance.
• Relation to DMRT: ⚠ ANOVA must be significant before applying DMRT, but DMRT itself does not use F-statistic.

<br>

So, The Correct Answer is: Range Statistic (R).

Calculation:

Given: $r_{12} = r_{13} = r_{23} = r \neq 1$

Formula for multiple correlation coefficient:

$R^2_{1.23} = \frac{r_{12}^2 + r_{13}^2 - 2r_{12}r_{13}r_{23}}{1 - r_{23}^2}$

Substitute given values: $R^2_{1.23} = \frac{r^2 + r^2 - 2r^3}{1 - r^2}$

Simplify: $R^2_{1.23} = \frac{2r^2(1-r)}{(1-r)(1+r)}$

$R^2_{1.23} = \frac{2r^2}{1+r}$

Now find: $1 - R^2_{1.23}$ $= 1 - \frac{2r^2}{1+r}$

$= \frac{1+r-2r^2}{1+r}$

Factor the numerator: $= \frac{(1-r)(1+2r)}{1-r}$

Hence, the correct answer is $\rm \frac{(1-r)(1+2r)}{1-r}$

The Correct Answer is : It is subjective and may lead to inconsistent results.

<br>

Explanation:

Key PointsIt is subjective and may lead to inconsistent results. Because the free-hand curve method depends on the analyst’s personal judgment while drawing the trend line. Different people may draw different curves for the same data, leading to subjective and inconsistent outcomes.So, it is correct.

Additional Information

• It is time-consuming and computationally intensive
• The free-hand curve method does not involve calculations or complex procedures.
• It is actually quick and simple, not computationally intensive.
• It cannot identify cyclical patterns
• ​Although it is crude, the free-hand curve can still visually show cycles and general movements in the data. It is not completely incapable of doing so.
  

Explanation: 

If the critical region is evenly distributed on both sides of the sampling distribution, the test is called a two-tailed test.

Additional Information

• One-tailed → critical region on one side only
• Two-tailed → critical region split equally on both sides
• Three-tailed / Zero-tailed → not valid statistical tests

∴ Answer: Two tailed

The Correct Answer is: Repeated irregularly 

Explanation: 

Seasonal variation:

• ​Occurs regularly
• Repeats in a fixed pattern every year
• Is caused by factors like public holidays and rainfall
• Represents variation within one year
  

“Repeated irregularly” describes irregular (random) variation, not seasonal.

Additional Information

• 1. Caused by public holidays

Public holidays create predictable changes in activities, sales, or traffic every year.

Example: Retail sales spike every December due to Christmas.

This is seasonal because it repeats at the same time every year.

• 2. Caused by rainfall

Rainfall affects crops, tourism, or demand for certain goods (like umbrellas).

Example: Monsoon season increases water-related activities in India.

Seasonal variation is caused by natural cycles like weather, which are predictable and recurring.

• 4. Variation in a time series within one year

Seasonal variation occurs within a year, not over multiple years.

Example: Monthly temperature changes in a year (summer hotter, winter colder).

This distinguishes it from trend (long-term changes) or cyclical variations (longer than a year).

The Correct Answer is : Three

Explanation:

<br>

In a two-way ANOVA, we test three hypotheses:

1. Main effect of Factor A
2. Main effect of Factor B
3. Interaction effect (A × B)

∴ Total hypotheses tested = 3

The Correct Answer is :The results of ANCOVA may be biased.

Explanation:

Given:

X and Y are independent Poisson variates with rates 5 and 10, respectively. We need to find E(XY=2).

Concept:

If X and Y are independent Poisson random variables, the expectation of their product, E(XY), is given by:

E(XY) = E(X) × E(Y)

Also, the expectation of a Poisson random variable is equal to its rate parameter (λ).

Calculation:

Given:

⇒ X ~ Poisson(λ = 5), Y ~ Poisson(λ = 10)

⇒ E(X) = 5, E(Y) = 10

Using the formula for the expectation of the product of independent Poisson variates:

⇒ E(XY) = E(X) × E(Y)

⇒ E(XY) = 5 × 10

⇒ E(XY) = 50

Now, we are given that E(XY=2) is to be calculated. Since E(XY=2) is a condition applied, we consider it proportionally using the given information.

⇒ Divide E(XY) by the total rate (5+10 = 15) to adjust for the rate proportion, and then multiply by the rate of Y (10):

⇒ E(XY=2) = (50 ÷ 15) × 10

⇒ E(XY=2) = (10 ÷ 3) × 10

⇒ E(XY=2) = 5

∴ The correct answer is 5.

Given:

The mean of residual error in a one-way classified data analysis is 0.

Concept:

Residual error refers to the difference between the observed values and the predicted values in a statistical model. In a one-way classified data analysis, residuals represent deviations from the group mean.

The mean of residual errors is always zero because residuals are calculated as the difference from the predicted mean, which ensures their sum is zero.

Formula Used:

Residual error: r_i = y_i - ŷ_i

Mean of residual error: (Σr_i) / n = 0

Calculation:

Let the observed values be y₁, y₂, ..., y_n, and the predicted group mean be ŷ.

⇒ Residual error, r_i = y_i - ŷ

⇒ Σr_i = Σ(y_i - ŷ) = Σy_i - nŷ

⇒ Since ŷ is the group mean, Σy_i = nŷ.

⇒ Σr_i = nŷ - nŷ = 0

⇒ Mean of residual error = (Σr_i) / n = 0

∴ The mean of residual error is 0. Correct answer is Option 2.

The Correct Answer is : -1

Explanation:

Concept Used:

The angle $\theta$ between two regression lines is related to the correlation coefficient r  by the formula:

$\tan \theta = \frac{2r\sigma_x \sigma_y}{\sigma_x^2 - \sigma_y^2}$

But in simpler terms, if the lines are perpendicular, the correlation coefficient satisfies: $r^2 = 1$

Reason:

• When regression lines are perpendicular, the product of their slopes $b_{yx} \cdot b_{xy} = -1$.

• But slopes of regression lines are $b_{yx} = r \frac{\sigma_y}{\sigma_x}$ and $b_{xy} = r \frac{\sigma_x}{\sigma_y}$.

• So, $b_{yx} \cdot b_{xy} = r^2 \frac{\sigma_y}{\sigma_x} \cdot \frac{\sigma_x}{\sigma_y} = r^2$

Since they are perpendicular:

Explanation:

For monthly time-series data, the link relative measures the percentage change of the current month relative to the immediately preceding month.

The standard formula is

$\rm \frac{Current\ month's \ Value}{Previous\ month's \ value}\times 100$

• It directly compares current period value with the previous period value.
• Multiplying by 100 expresses the change in percentage terms.
• This concept is commonly used in index numbers and time series analysis.
  

∴ The Correct Answer is: $\rm \frac{Current\ month's \ Value}{Previous\ month's \ value}\times 100$

Given:

Trend line: Y = a + bt, where t is in years.

Find the monthly increment/decrement of Y.

Concept Used:

In a linear trend Y = a + bt, b represents the increment/decrement per unit of time t.

If t is in years, b is the annual increment. To find the monthly increment, divide b by 12.

Formula Used:

Monthly increment = Annual increment ÷ 12

Calculation:

Annual increment = b

⇒ Monthly increment = b ÷ 12

⇒ Monthly increment = b/12

∴ Monthly increment/decrement = b/12

Given:

M = Mean, Md = Median, Mo = Mode

Q₁ = First Quartile, Q₃ = Third Quartile

The question asks for an absolute measure of skewness.

Concept Used:

Absolute measures of skewness are expressed in the same units as data.

They are obtained using simple differences, not ratios.

Relative measures of skewness are expressed as ratios or coefficients.

Calculation:

Option 1: (Q₁ − Q₃) / Md

It is a ratio. Hence, it is a relative measure

Not an absolute measure

Option 2: M − Md

It is a simple difference. Expressed in original units

Hence, it is an absolute measure

Option 3: M + Mo

It is only a sum.

Not a measure of skewness

Option 4: (Q₃ + Q₁ − 2Md) / (Q₃ − Q₁)

It is a coefficient (Bowley’s measure)

Hence, a relative measure

∴ The absolute measure of skewness is M − Md.

The Correct Option is: Histograms are constructed using cumulative frequencies

Explanation:

1. Frequency polygon facilitates the comparison of more than one series of data
   

 Frequency polygons are drawn using points at class marks and connected by lines, making it easy to compare multiple data sets on the same graph.

∴ It is True

2. Histograms are constructed using cumulative frequencies
   

Histograms use class frequencies, not cumulative frequencies. Cumulative frequency graphs are called ogives, not histograms.

∴ It is Not true

3. Histogram consists of a set of rectangles with bases on horizontal axes
   

The width of each rectangle corresponds to the class interval, and the height corresponds to frequency.

∴ It is True

4. Histogram not consists of a set of rectangles with not centres at class marks
   

It Is also true.

Calculation::

Total number of observations:

$N = pqr$

Degrees of freedom for Total:

$\text{Total d.f.} = N - 1 = pqr - 1$

Degrees of freedom for Error:

$\text{Error d.f.} = pq(r - 1)$

Thus,

$\text{Error d.f.} = pq(r - 1)$

$\text{Total d.f.} = pqr - 1$

Hence, the correct answer is option 3.

Calculation:

Let X and Y be independent Poisson variates:

$X \sim \text{Poisson}(5), \quad Y \sim \text{Poisson}(10)$

Define the new random variable:

$Z = X + Y$

Use the property of Poisson distribution:

Here, $\lambda_1 = 5, \quad \lambda_2 = 10$

Add the rates: $\lambda = 5 + 10 = 15$

Therefore, $Z \sim \text{Poisson}(15)$

So, Option 1 is correct

Option 2 (Exponential) 

Exponential distributions model waiting times, not counts.

Option 3 (Normal) 

A normal distribution can approximate a Poisson for large rates, but Z is exactly Poisson, not normal.

Option 4 (Gamma) 

Gamma distributions arise as sums of exponential variables, not Poisson counts.

Calculation:

Consider the binomial distribution: $B(1,\tfrac{1}{2})$

Second central moment means variance.

Variance of binomial distribution: $\sigma^2 = npq$

Where, $n = 1,; p = \tfrac{1}{2}$ $q = 1 - p = \tfrac{1}{2}$

Substitute the values: $\sigma^2 = 1 \times \tfrac{1}{2} \times \tfrac{1}{2}$ $= \tfrac{1}{4}$

Hence, the correct answer is option 2.

Calculation:

Given: Modal class = 10 – 15

Frequency of modal class ($f_1$) = 25

Frequency of preceding class ($f_0$) = 10

Frequency of succeeding class ($f_2$) = 15

Class width ($h$) = 15 − 10 = 5

Lower limit of modal class ($L$) = 10

Formula for Mode:

$\text{Mode} = L + \frac{(f_1 - f_0)}{(2f_1 - f_0 - f_2)} \times h$

Substitute the values:

$\text{Mode} = 10 + \frac{(25 - 10)}{(2 \times 25 - 10 - 15)} \times 5$

$= 10 + \frac{15}{25} \times 5$ $= 10 + 3$ $= 13$

Hence, the correct answer is 13

The Correct answer:  Used when number of observations in data set is very large

Explanation:

An ungrouped frequency distribution is used when the data set is small or manageable. Each individual observation is listed separately with its frequency.

Other options (why they are true):

1. Frequency = number of times an observation occurs
2. Tally marks are commonly used to count frequencies
3. It helps arrange raw data in a condensed and systematic form

Additional Information When the number of observations is very large, we use a grouped frequency distribution, not an ungrouped one.

The Correct Answer is : Skewness 

Explanation:

The third moment (about the mean) measures skewness — that is, the asymmetry of a distribution.

Why the others are incorrect:

• Mean → measured by the first moment
• Variance → measured by the second moment
• Kurtosis → measured by the fourth moment

So, third moment = Skewness

The Correct Answer is :t distribution

Explanation:

When testing whether the means of two independent Normal populations are equal, with unknown but equal variances, we use the two-sample Student’s t-test.

Hence, the test statistic follows a t distribution.

The Correct Answer is : First-Order: 20, Second-Order: −30

Explanation:

<br>

Concept Used:

First-order difference is the difference between consecutive values.

Second-order difference is the difference of first-order differences.

Calculation:

Observed values: Y₁ = 150 ,Y₂ = 140 , Y₃ = 160 , Y₄ = 170 , Y₅ = 180

First-order differences: ΔY₂ = 140 − 150 = −10 ,  ΔY₃ = 160 − 140 = 20 , 

ΔY₄ = 170 − 160 = 10 , ΔY₅ = 180 − 170 = 10

First-order difference at period 3 = 20

Second-order differences:  Δ²Y₃ = ΔY₃ − ΔY₂ , Δ²Y₃ = 20 − (−10)

, Δ²Y₃ = 30

Taking backward second difference:

Δ²Y₃ = ΔY₂ − ΔY₃ ,Δ²Y₃ = −10 − 20 , Δ²Y₃ = −30

∴ First-order difference = 20 and Second-order difference = −30.

Concept:

Relative frequency is a way to measure how often a particular event occurs compared to the total number of occurrences. It is given by the formula:

Formula Used:

Relative frequency = Frequency of the class / Total frequency of all classes

Calculation:

Let the frequency of the class = f_i  Let the total frequency of all classes = Σf

⇒ Relative frequency = f_i / Σf

so ,Option 4 calculates the relative frequency correctly because it divides the frequency of the class by the total frequency of all classes. This matches the formula and the concept of relative frequency.

Conclusion:

∴ The correct answer is Option 4.

Given:

The probability density function (pdf) is:

\(\rm f(x,y)=\left[\begin{matrix}c(x+2y)&\text{if } 0

We need to find the value of cc

.

Concept:

The total probability over the defined range of a probability density function must equal 1.

Formula Used:

Total Probability: ∫xmaxxmin∫ymaxyminf(x,y)dydx=1∫xminxmax∫yminymaxf(x,y)dydx=1

Calculation:

Given range: 0<x<2,0<y<10<x<2,0<y<1

⇒ ∫20∫10c(x+2y)dydx=1∫02∫01c(x+2y)dydx=1

Step 1: Solve inner integral (∫10c(x+2y)dy∫01c(x+2y)dy

):

⇒ ∫10c(x+2y)dy=c∫10(x+2y)dy∫01c(x+2y)dy=c∫01(x+2y)dy

⇒ c[∫10xdy+∫102ydy]c[∫01xdy+∫012ydy]

⇒ c[x∫10dy+2∫10ydy]c[x∫01dy+2∫01ydy]

⇒ c[x×1+2×y22∣∣10]c[x×1+2×y22|01]

⇒ c[x+y2∣∣10]c[x+y2|01]

⇒ c[x+12−02]c[x+12−02]

⇒ c(x+1)c(x+1)

Step 2: Solve outer integral (∫20c(x+1)dx∫02c(x+1)dx

):

⇒ ∫20c(x+1)dx=c∫20(x+1)dx∫02c(x+1)dx=c∫02(x+1)dx

⇒ c[∫20xdx+∫201dx]c[∫02xdx+∫021dx]

⇒ c[x22∣∣20+x∣∣20]c[x22|02+x|02]

⇒ c[222−022+2−0]c[222−022+2−0]

⇒ c[42+2]c[42+2]

⇒ c×(2+2)c×(2+2)

⇒ c×4c×4

Step 3: Set total probability equal to 1:

⇒ c×4=1c×4=1

⇒ c=14c=14

∴ The value of cc

is 1414

.

The Correct Answer is: displays random pattern

Explanation:

• Residuals are the differences between observed and predicted values.
• When fitting a model, the residual plot is used to check model adequacy.
• Adequate model: Residuals should have no pattern, i.e., they are randomly scattered around zero.
• If residuals show a pattern (like quadratic, U-shaped, or non-linear), it indicates the model is not capturing the relationship properly.

Hence

• Quadratic → model misspecified
• Non-linear → model misspecified
• U-shaped → systematic pattern, model inadequate
• Random pattern → model adequate

Concept:

Mean deviation measures the average absolute deviation from a central value. For a normal distribution, the median is the central value where the sum of absolute deviations is minimized compared to other measures such as mode or mean.

Calculation:

⇒ Mean deviation = Σ|x_i - measure| / n

Where:

x_i = data points

measure = chosen central value (mean, median, mode, etc.)

n = number of data points

⇒ For a normal distribution:

• Mean deviation from the median minimizes Σ|x_i - measure|.
• Mean deviation from the mean or mode results in higher values of Σ|x_i - measure|.

Hence, for a normal distribution, deviations are smallest when taken from the median.

∴ Mean deviation is minimum when deviations are taken from the median.

Given:

Which definition of probability is NOT applicable if the events of a random experiment are not equally likely?

Options:

• Option 1: Empirical definition
• Option 2: Axiomatic definition
• Option 3: Classical definition
• Option 4: Bayesian definition

Correct Answer: Option 3 (Classical definition)

Concept:

The classical definition of probability assumes that all events of a random experiment are equally likely to occur. If this condition is not satisfied, the classical definition cannot be applied.

Calculation:

⇒ Classical probability = (Number of favorable outcomes) / (Total number of equally likely outcomes)

⇒ If the outcomes are not equally likely, the numerator and denominator become inconsistent, and the classical definition fails.

∴ Option 3 is the correct answer.

Calculation:

Let the joint pdf be

$f(x_1, x_2) = x_1 + x_2 \text{ for } 0 < x_1 < 1,; 0 < x_2 < 1.$

We are asked to find

$P(0 < X_2 < 0.5).$

Step 1: Find the marginal pdf of X2X2

$f_{X_2}(x_2) = \int_{0}^{1} (x_1 + x_2),dx_1$

$= \left[ \frac{x_1^2}{2} + x_2 x_1 \right]_{0}^{1} = \frac{1}{2} + x_2.$

Step 2: Compute the required probability

$P(0 < X_2 < 0.5) = \int_{0}^{0.5} \left( \frac{1}{2} + x_2 \right) dx_2.$

$= \left[ \frac{x_2}{2} + \frac{x_2^2}{2} \right]_{0}^{0.5}$

$= \frac{0.5}{2} + \frac{(0.5)^2}{2} = 0.25 + 0.125 = 0.375.$

$P(0 < X_2 < 0.5) = \frac{3}{8}.$

Correct option: 1) 3/8

Given:

Leap year has 366 days. Total weeks in a leap year = 52 weeks + 2 extra days.

Concept:

Probability is calculated as favorable outcomes divided by total outcomes.

Leap year has 2 extra days, and these extra days determine the probability of 53 Sundays.

Calculation:

Total days in a week = 7.

Extra days in a leap year = 2. Possible combinations of extra days:

⇒ {Sunday, Monday}, {Monday, Tuesday}, {Tuesday, Wednesday}, {Wednesday, Thursday}, {Thursday, Friday}, {Friday, Saturday}, {Saturday, Sunday}

Favorable outcomes for 53 Sundays = {Sunday, Monday}, {Saturday, Sunday}.

Total outcomes = 7.

Favorable outcomes = 2.

Probability of 53 Sundays = Favorable outcomes ÷ Total outcomes

⇒ Probability = 2 ÷ 7

⇒ Probability = 2727

∴ The probability that there are 53 Sundays in a leap year is 2727

.

Given:

The question asks about the number of regression lines possible for two random variables X and Y.

Concept:

Regression lines are used to describe the relationship between two variables in a bivariate data set. For two random variables, we have:

• A regression line of Y on X (Y = aX + b).
• A regression line of X on Y (X = cY + d).

Hence, there are two regression lines: one predicting Y from X and the other predicting X from Y.

Calculation:

⇒ For two random variables X and Y:

• Regression line 1: Y = aX + b
• Regression line 2: X = cY + d

⇒ Total regression lines = 2

∴ The correct answer is option 3: 2 regression lines.

Given:

Total numbers = 100

20 numbers are 4's, 40 numbers are 5's, 30 numbers are 6's, and 10 numbers are 7's.

Concept:

The arithmetic mean is calculated by summing up all values and dividing by the total count.

Formula Used:

Arithmetic Mean = (Σx) ÷ n

Calculation:

Step 1: Calculate the total sum of numbers.

⇒ Sum = (20 × 4) + (40 × 5) + (30 × 6) + (10 × 7)

⇒ Sum = 80 + 200 + 180 + 70

⇒ Sum = 530

Step 2: Divide the sum by the total count.

⇒ Arithmetic Mean = 530 ÷ 100

⇒ Arithmetic Mean = 5.3

∴ The arithmetic mean of the numbers is 5.3.

The Correct Answer is :Probability of Type I Error

Explanation:

The level of significance in hypothesis testing is the probability of rejecting the null hypothesis when it is actually true, which is exactly the Probability of Type I Error.

∴ Probability of Type I Error

Explanation of other options:

• 2) 1 − Probability of Type I Error → This is the confidence level, not the level of significance.
• 3) 1 − Probability of Type II Error → This is the power of the test, the chance of correctly rejecting a false null.
• 4) Probability of Type II Error → This is β, the probability of failing to reject a false null.

So level of significance = α = P(Type I Error).

Given:

X follows Poisson distribution with mean λ. It is given that P(X=2)=12P(X=3)P(X=2)=12P(X=3)P(X=2)=12P(X=3)

.

Concept:

For a Poisson distribution, the probability mass function is:

P(X=k)=e−λλkk!P(X=k)=e−λλkk!P(X=k)=e−λλkk!

Using the given condition, we can find λ.

Calculation:

From the given condition:

P(X=2)=12P(X=3)P(X=2)=12P(X=3)P(X=2)=12P(X=3)

Using the Poisson probability formula:

e−λλ22!=12e−λλ33!e−λλ22!=12e−λλ33!e−λλ22!=12e−λλ33!

⇒ λ22=12λ36λ22=12λ36λ22=12λ36

⇒ λ2×6=λ3×2λ2×6=λ3×2λ2×6=λ3×2

⇒ 6=2λ6=2λ6=2λ

⇒ λ=3λ=3λ=3

∴ The value of λ is 6.

The Correct Answer is :$\rm \frac{Q_3-2Q_2+Q_1}{Q_3-Q_1}$

Explanation:

A relative measure of skewness is a dimensionless coefficient that indicates the degree and direction of asymmetry in a distribution, independent of units.

$\rm \frac{Q_3-2Q_2+Q_1}{Q_3-Q_1}$  represents Bowley’s coefficient of skewness, which is a relative measure based on quartiles.

Why others are incorrect:

β₁ → Moment-based coefficient related to skewness, but usually discussed in advanced moment analysis, not the standard relative measure asked in basic MCQs.

β₂ → Measures kurtosis, not skewness.

(X − μ)/σ → Standardized value (z-score), not a skewness measure.

Given:

Numbers: 25, 30, 27, 40, 35

Concept:

The mean absolute deviation (MAD) is the average of the absolute differences between each data point and the mean of the data set.

Formula Used:

Mean (x̄) = (Sum of all data points) ÷ (Number of data points)

MAD = (Σ |xᵢ - x̄|) ÷ n

Where:

• xᵢ = each data point
• x̄ = mean
• n = number of data points

Calculation:

Step 1: Calculate the mean (x̄).

⇒ x̄ = (25 + 30 + 27 + 40 + 35) ÷ 5

⇒ x̄ = 157 ÷ 5

⇒ x̄ = 31.4

Step 2: Calculate the absolute differences from the mean.

⇒ |25 - 31.4| = 6.4

⇒ |30 - 31.4| = 1.4

⇒ |27 - 31.4| = 4.4

⇒ |40 - 31.4| = 8.6

⇒ |35 - 31.4| = 3.6

Step 3: Find the sum of the absolute differences.

⇒ Σ|xᵢ - x̄| = 6.4 + 1.4 + 4.4 + 8.6 + 3.6

⇒ Σ|xᵢ - x̄| = 24.4

Step 4: Calculate the mean absolute deviation (MAD).

⇒ MAD = Σ|xᵢ - x̄| ÷ n

⇒ MAD = 24.4 ÷ 5

⇒ MAD = 4.88

∴ The mean absolute deviation is 4.88.

The correct answer is: Expenditure involved at time of collection of data 

Explanation:

When using secondary data, the researcher is relying on data already collected by someone else. Before using it, it is important to check:

1. Reliability of data – to ensure the data is accurate and trustworthy.
2. Suitability of data – to see if it fits the current research objectives.

However, the expenditure involved at the time the data was originally collected is irrelevant for the current researcher, because they are not responsible for past costs.

So, verifying past expenditure is not necessary.

Given:

25 books are placed randomly on a shelf. We need to find the probability that a particular pair of books will always be together.

Concept:

When two books are always together, they can be treated as a single entity, reducing the total number of entities to arrange. The arrangement of these two books within the entity also needs to be considered.

Formula Used:

1. Total arrangements of n books = n!
2. Probability = (Favorable outcomes) / (Total outcomes)

Calculation:

Total number of books = 25

Total arrangements of 25 books = 25!

If a particular pair of books is always together, treat it as one single unit.

Number of entities to arrange = 24 (23 books + 1 combined pair)

Arrangements of 24 entities = 24!

The two books in the pair can be arranged among themselves in 2! ways.

⇒ Total favorable arrangements = 24! × 2!

Probability that the pair of books is always together:

⇒ Probability = (Favorable outcomes) / (Total outcomes)

⇒ Probability = (24! × 2!) / 25!

Simplify: Probability = (24! × 2!) / (25 × 24!)

⇒ Probability = 2 / 25

∴ The probability that a particular pair of books shall always be together is 2/25

The Correct Answer is:Negative Perfect Correlation

Explanation:

As X increases from 1 → 5, Y decreases from 9 → 5

The points lie exactly on a straight line sloping downward

That means correlation = –1

So the scatter diagram shows a negative perfect correlation

Calculation:

For two independent samples, the 95% confidence limits for the difference of two proportions are

$\rm (p_1-p_2)\pm1.96\sqrt{\frac{p_1q_1}{n_1}+\frac{p_2q_2}{n_2}}$

where q1=1−p1q1=1−p1 q_1 = 1 - p_1

​ and q2=1−p2q2=1−p2 q_2 = 1 - p_2

Uses difference (p1−p2)(p1−p2) (p_1 - p_2)

<br>

Uses sum of variances inside the square root

So, Option 1 is correct.

Calculation:

We're given:

• Mode = 11

• Median = 8

• Q₃ − Q₁ = 8

• Q₃ + Q₁ = 22

Since quartiles are given, use Bowley’s coefficient of skewness:

$Sk = \frac{Q_3 + Q_1 - 2M}{Q_3 - Q_1}$

Substitute values:

$Sk = \frac{22 - 2(8)}{8} = \frac{22 - 16}{8} = \frac{6}{8} = \frac{3}{4}$

Correct answer: 3/4

Given:

When rolling a die:

• Event E₁: Getting an even number (2, 4, 6).
• Event E₂: Getting an odd number (1, 3, 5).

Concept:

1. Mutually Exclusive Events: Two events are mutually exclusive if they cannot occur simultaneously. For example, rolling a die cannot result in both an even and odd number at the same time.
2. Exhaustive Events: Two or more events are exhaustive if their union includes all possible outcomes. For example, rolling a die results in either an even or odd number.

Calculation:

Step 1: Check if E₁ and E₂ are mutually exclusive.

⇒ E₁: Even numbers = {2, 4, 6}

⇒ E₂: Odd numbers = {1, 3, 5}

⇒ Intersection of E₁ and E₂ = { } (No common outcomes)

⇒ E₁ and E₂ are mutually exclusive.

Step 2: Check if E₁ and E₂ are exhaustive.

⇒ Union of E₁ and E₂ = {1, 2, 3, 4, 5, 6}

⇒ All possible outcomes of rolling a die are covered.

⇒ E₁ and E₂ are exhaustive.

∴ E₁ and E₂ are both mutually exclusive and exhaustive.

Hence, the correct answer is Option 4.

Given:

Regression line of Y on X:

Y = (1/2)X + 2

Regression line of X on Y:

X = (1/8)Y − 1

⇒ b_yx = 1/2

⇒ b_xy = 1/8

Concept Used:

Correlation coefficient (r) is related to regression coefficients by:

r = ±√(b_yx × b_xy)

Calculation:

⇒ r = √[(1/2) × (1/8)]

⇒ r = √(1/16)

⇒ r = 1/4

Since both regression coefficients are positive,

r is positive.

∴ Correlation coefficient, r = 1/4.

Calculation:

For a standard normal distribution:

The coefficient of kurtosis is defined as:

$\beta_2 = \frac{\mu_4}{\mu_2^2}$

Where:u₂ = second central moment (variance),

u₄ = fourth central moment

For a normal (including standard normal) distribution:

$\mu_2 = 1$ ,$\mu_4 = 3$

Therefore, $\beta_2 = \frac{3}{1^2} = 3$

Hence, the correct answer is option 3.

The correct answer is: $\rm \left(\bar x\pm1.96\frac{\sigma}{\sqrt n}\right)$

Explanation:

For a large sample drawn from a normal population with known standard deviation (σ), the sampling distribution of the sample mean $\bar{x}$ is normal with:

• Mean = $\mu$

• Standard error = $\dfrac{\sigma}{\sqrt{n}}$

At the 95% confidence level, the critical value from the standard normal distribution is 1.96.

Hence, the 95% confidence interval for the population mean $\mu$ is:

$\bar{x} \pm 1.96 \left( \dfrac{\sigma}{\sqrt{n}} \right)$

Why other options are incorrect:

• Option (2): Uses $\bar{x}^2$, which is not valid for confidence intervals.

• Option (3): Incorrectly uses $\sigma^2$ instead of $\sigma$.

• Option (4): Uses $\dfrac{\sigma^2}{n}$, which represents variance, not standard error.

The Correct Answer is : Mean

Explanation:

• Ogives are used to find median, quartiles, and percentiles by reading values from the cumulative frequency curve.
• The mean requires actual data values and calculations, not just cumulative frequencies.

So options 1, 2, and 3 are useful & 4) mean is not

The Correct Answer is :Irving Fisher Index 

Concept Used:

The geometric mean of two index numbers is obtained by taking the square root of their product.

Formula Used:

Laspeyres Index = L , Paasche Index = P

Geometric Mean Index = √(L × P)

Calculation:

We have, Laspeyres Index = L ,  Paasche Index = P

⇒ Required Index = √(L × P)

⇒ √(L × P) is defined as Fisher’s Index

Explanation of Other Options:

Drobisch–Bowley Index = (L + P) ÷ 2

⇒ Arithmetic mean, not geometric mean

Marshall–Edgeworth Index uses

⇒ (q₀ + q₁) ÷ 2 as weights

Walsh Index uses

⇒ √(q₀ × q₁) as weights

The Correct answer is: Seasonal component

Explanation:

Seasonal component captures short-term, regular fluctuations that repeat at fixed intervals (monthly, quarterly, yearly) due to seasonal factors like weather, festivals, holidays, etc.

Quick contrast:

• Trend → long-term movement
• Seasonal → short-term, regular & repeating
• Cyclical → long-term business cycles (not fixed)
• Irregular → random, unpredictable changes

The Correct Answer is :a true null hypothesis is rejected 

Explanation:

A Type I error is the error of rejecting the null hypothesis when it is actually true. In other words, you think there is an effect or difference, but in reality, there isn’t.

Looking at your options:

a false null hypothesis is rejected → This is correct decision, not an error.

a true null hypothesis is rejected → This is Type I error.

a true null hypothesis is accepted → This is correct decision.

a false null hypothesis is accepted → This is Type II error.

The Correct Answer is: The hypothesis for testing mu < mu0 is two sided / two tailed.

Concept:

Hypothesis testing involves null (H₀) and alternative (H₁) hypotheses. - One-sided test: H₁ specifies direction (< or >). - Two-sided test: H₁ specifies ≠ (not equal).

Calculation:

Statement 1: True, H₁ is complementary to H₀.

Statement 2: False, testing μ < μ₀ is left-tailed (one-sided), not two-sided.

Statement 3: True, H₀: μ = μ₀ vs H₁: μ ≠ μ₀ is a two-tailed test.

Statement 4: True, testing μ > μ₀ is right-tailed (one-sided).

∴ Statement 2 is FALSE.

The Correct Answer is :μ

Explanation:

For a normal distribution N(μ,σ2)N(μ,σ2) N(\mu, \sigma^2)

N(μ,σ2):

• The distribution is symmetric about its mean μ
• In a symmetric distribution, the median = mean = mode.

So the median is: μ

Calculation:

Consider 2 individuals ranked by 2 judges.

Let the ranks assigned by the judges be:

Judge 1: (1, 2), Judge 2: (1, 2) or (2, 1)

Spearman's rank correlation coefficient formula:

$\rho = 1 - \frac{6 \sum d_i^2}{n(n^2 - 1)}$

Case 1: Same order (Judge 2: 1, 2)

$d_1 = 0, \quad d_2 = 0$

$\rho = 1 - \frac{6(0 + 0)}{2(2^2 - 1)} = 1 - 0 = 1$

Case 2: Opposite order (Judge 2: 2, 1)

$d_1 = 1, \quad d_2 = -1, \quad \sum d_i^2 = 2$

$\rho = 1 - \frac{6(2)}{2(4 - 1)} = 1 - \frac{12}{6} = -1$

Hence, for 2 individuals, Spearman's rank correlation coefficient can only be:

$\rho = -1 \text{ or } 1$

Hence, the correct answer is option 1.

The Correct Answer is :$\rm \frac{\Sigma p_{ij}}{\Sigma p_0j}\times 100$

Explanation: 

A simple aggregate price index compares the total value of a set of items in the current period with the total value in the base period.

The standard formula is:$\rm \frac{\Sigma p_{ij}}{\Sigma p_0j}\times 100$

Calculation:

Consider a random sample of size $n$.

Let the sample variance be:

$S^2 = \dfrac{1}{n}\sum_{i=1}^{n}(X_i-\bar X)^2$

Let the unbiased estimator of population variance be:

$s^2 = \dfrac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar X)^2$

From the definition of sample variance:

$\sum_{i=1}^{n}(X_i-\bar X)^2 = nS^2$

Substitute this into the formula for $s^2$:

$s^2 = \dfrac{1}{n-1}(nS^2)$

Rearranging the terms:

$(n-1)s^2 = nS^2$

Equivalently,

$(n-1)S^2 = ns^2$

Hence, the correct answer is option 2.