CUET UG Applied Mathematics Booster Test 2

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CUET UG Applied Mathematics Booster Test 2 - Population & Sample

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Applied Mathematics Booster Test 2 - Population & Sample

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QUESTION 1 OF 20

Let the characteristics of a mini population dataset be represented by the vector:
P ⃗=(3,4)
The parameter representing the true Euclidean norm (magnitude) of this population vector is:
∥P ⃗∥=√(3^2+4^2 )

Answer Review

1 Let the characteristics of a mini population dataset be represented by the vector:
P ⃗=(3,4)
The parameter representing the true Euclidean norm (magnitude) of this population vector is:
∥P ⃗∥=√(3^2+4^2 )

A 7

B 5

C 12

D 25

💡 Short Explanation

Euclidean norm is calculated using Pythagoras theorem √(3² + 4²) = √25 = 5 It represents vector magnitude

✅ Correct Answer Explanation

The Euclidean norm (magnitude) of a vector is defined as √(x² + y²). Substituting values gives √(9 + 16) = √25 = 5. This represents the true geometric length of the population vector.

❓ Why Other Options Are Incorrect

Option A → 7 is not obtained from correct Pythagorean computation
Option C → 12 is arithmetic miscalculation, not geometric norm
Option D → 25 is sum of squares, not square root

🎯 Strategy

Used: Elimination

Application: Incorrect numerical outputs are eliminated using correct formula substitution

Final Logic: Vector magnitude requires square root of sum of squares

🧠 Memory Trick

"Square → add → root = distance"

📚 Class XI – Statistics – Chapter: Measures of Central Tendency – Topic: Vector Magnitude Concept

2 In continuous probability, the total probability over the entire population space Ω is given by:
∫_Ω f(x) dx
This value must always equal:

A The standard deviation σ

B The mean μ

C 1

D 0

💡 Short Explanation

Total probability over full space is always 1 PDF integrates to unity Represents certainty of occurrence

✅ Correct Answer Explanation

For any probability density function, the total area under the curve over the entire sample space is 1, meaning total probability is certain.

❓ Why Other Options Are Incorrect

Option A → σ is spread, not total probability
Option B → μ is central value, not probability total
Option D → 0 would imply impossible event space

🎯 Strategy

Used: Conceptual matching

Application: Used definition of probability axioms

Final Logic: Total probability of full space equals 1

🧠 Memory Trick

"Whole space = Whole chance = 1"

📚 Class XI – Statistics – Chapter: Probability – Topic: Probability Density Function

3 As sample size n increases, the Standard Error decreases. What happens to the sampling distribution curve?

A Area becomes infinite

B Area decreases to 0.5

C Curve becomes flat

D Curve becomes narrower and taller, but total area remains 1

💡 Short Explanation

Larger n reduces variability Distribution becomes tighter around mean Total probability remains constant

✅ Correct Answer Explanation

As sample size increases, standard error decreases, causing sampling distribution to concentrate more around the mean. The curve becomes narrower and taller, but total area remains 1 due to probability conservation.

❓ Why Other Options Are Incorrect

Option A → Area cannot exceed or become infinite
Option B → Total probability never reduces
Option C → Curve does not flatten; it concentrates

🎯 Strategy

Used: Trend analysis

Application: Relationship between sample size and variability used

Final Logic: Increasing sample size tightens distribution

🧠 Memory Trick

"More n → more sharp curve"

📚 Class XI – Statistics – Chapter: Sampling Distribution – Topic: Standard Error

4 Population data: 2,4,6,8,10
Moving average (period 2): 3,5,7,9
Compared to the original population, the standard deviation of the moving average sample is:

A Smaller

B Larger

C Equal

D Zero

💡 Short Explanation

Moving averages smooth fluctuations Reduces variability Lowers standard deviation

✅ Correct Answer Explanation

Moving averages reduce random variation by smoothing adjacent values, thereby decreasing spread. Hence standard deviation becomes smaller than original dataset.

❓ Why Other Options Are Incorrect

Option B → variability does not increase
Option C → smoothing changes dispersion
Option D → variance is not eliminated

🎯 Strategy

Used: Contextual comparison

Application: Compare variability before and after transformation

Final Logic: Smoothing reduces dispersion

🧠 Memory Trick

"Averages smooth = SD shrink"

📚 Class XI – Statistics – Chapter: Measures of Dispersion – Topic: Moving Average

5 To obtain an unbiased estimator of population variance σ², the sample variance formula divides by:

A n²

B n−1

C 0

D n+1

💡 Short Explanation

Degrees of freedom correction required Prevents bias in variance estimate Ensures unbiased estimator

✅ Correct Answer Explanation

Sample variance divides by (n−1) to correct bias caused by using sample mean instead of population mean. This is called Bessel's correction.

❓ Why Other Options Are Incorrect

Option A → over-scales variance incorrectly
Option C → division by zero undefined
Option D → inflates variance incorrectly

🎯 Strategy

Used: Theoretical rule recall

Application: Known statistical correction applied

Final Logic: Degrees of freedom = n−1

🧠 Memory Trick

"Variance fix = n minus 1"

📚 Class XI – Statistics – Chapter: Dispersion – Topic: Variance

6 Given: SEM = σ/√n, σ = 20, SEM = 2
Solve for n.

A 10

B 50

C 100

D 400

💡 Short Explanation

2 = 20/√n √n = 10 n = 100

✅ Correct Answer Explanation

Using SEM formula, rearranging gives √n = 10, hence n = 100.

❓ Why Other Options Are Incorrect

Option A → too small for equation balance
Option B → incorrect square root relation
Option D → gives wrong SEM value

🎯 Strategy

Used: Substitution

Application: Direct formula substitution and solving

Final Logic: Solve algebraically for n

🧠 Memory Trick

"Cross multiply, square simplify"

📚 Class XI – Statistics – Chapter: Sampling – Topic: Standard Error

7 If a sample of size n is drawn with replacement from a population of size N, the probability of selecting one specific ordered sequence is:

A n/N

B 1/((N¦n))

C 1/N!

D 1/N^n

💡 Short Explanation

Each selection independent Probability = 1/N per draw Multiply n times

✅ Correct Answer Explanation

With replacement, each draw has probability 1/N. For n independent draws, probability becomes (1/N)^n.

❓ Why Other Options Are Incorrect

Option A → incorrect scaling
Option B → combination formula misuse
Option C → factorial not relevant

🎯 Strategy

Used: Probability multiplication rule

Final Logic: Independent events multiply

🧠 Memory Trick

"With replacement → power n"

📚 Class XI – Statistics – Chapter: Probability – Topic: Independent Events

8 To make decisions, statisticians use hypotheses:
The null hypothesis \(H_{0}\): no difference or relationship
The alternative hypothesis \(H_{1}\): a real difference exists
Sample data is used to compute statistics and decide whether to reject \(H_{0}\).
Which statements are correct?
I. H₀ represents no difference
II. H₁ represents the opposite claim
III. If test statistic exceeds critical value, reject H₀

A I only

B I and II

C II and III

D I, II, and III

💡 Short Explanation

H₀ = null hypothesis H₁ = alternative hypothesis Decision rule uses critical value

✅ Correct Answer Explanation

All statements correctly define hypothesis testing framework and decision rule in inferential statistics.

❓ Why Other Options Are Incorrect

Option A → ignores correct statements II and III
Option B → excludes rejection rule
Option C → ignores null hypothesis definition

🎯 Strategy

Used: Concept verification

Final Logic: All standard hypothesis rules are correct

🧠 Memory Trick

"H0 no diff, H1 difference, reject if extreme"

📚 Class XI – Statistics – Chapter: Hypothesis Testing – Topic: Null & Alternative Hypothesis

9 To make decisions, statisticians use hypotheses:
The null hypothesis \(H_{0}\): no difference or relationship
The alternative hypothesis \(H_{1}\): a real difference exists
Sample data is used to compute statistics and decide whether to reject \(H_{0}\).
Population data: 2,4,6,8,10
Moving average (period 2): 3,5,7,9
Compared to the original population, the standard deviation of the moving average sample is:

A Smaller

B Larger

C Equal

D Zero

💡 Short Explanation

Smoothing reduces variability Moving average reduces spread SD decreases

✅ Correct Answer Explanation

Moving averages reduce fluctuations, leading to reduced dispersion compared to original data.

❓ Why Other Options Are Incorrect

Option B → variability does not increase
Option C → transformation changes spread
Option D → variance not eliminated

🎯 Strategy

Used: Pattern recognition

Final Logic: Smoothing reduces dispersion

🧠 Memory Trick

"Smooth data = less spread"

📚 Class XI – Statistics – Chapter: Dispersion – Topic: Moving Average

10 From N = 1000, what is the probability a specific item is not selected on the first draw?

A 1/1000

B 1/50

C 50/1000

D 999/1000

💡 Short Explanation

One item selected out of 1000 Remaining 999 not selected Probability = 999/1000

✅ Correct Answer Explanation

Only one item is selected, so probability of not being selected is (999/1000).

❓ Why Other Options Are Incorrect

Option A → probability of selection, not rejection
Option B → irrelevant fraction
Option C → incorrect proportion
Option D → correct complement probability

🎯 Strategy

Used: Complement rule

Final Logic: Not selected = 1 − selected probability

🧠 Memory Trick

"Out of 1000, miss = 999"

📚 Class XI – Statistics – Chapter: Probability – Topic: Complement Rule

11 From N=10, choose n=3:
(10¦3)

A 120

B 1000

C 720

D 30

💡 Short Explanation

Combination formula used: nCr 10C3 = 10×9×8 / 3×2×1 Result = 120

✅ Correct Answer Explanation

The number of combinations of selecting 3 items from 10 is given by: 10C3 = 10! / (3!7!) = (10×9×8)/(3×2×1) = 120.

❓ Why Other Options Are Incorrect

Option B → 1000 is unrelated to combination formula
Option C → 720 is 6! not 10C3
Option D → 30 is incorrect partial computation

🎯 Strategy

Used: Formula substitution

Final Logic: Apply nCr directly

🧠 Memory Trick

"nCr = factorial shrink trick"

📚 Class XI – Statistics – Chapter: Permutations & Combinations – Topic: Combination Formula

12 Which mathematical statement is theoretically incorrect regarding sample size n and degrees of freedom Df?

A A higher degree of freedom generally implies larger sample sizes

B The formula for degrees of freedom is Df = n − 1

C A sample size of n ≥ 30 is considered sufficient to apply the Central Limit Theorem

D Decreasing the sample size n mathematically improves the accuracy of population estimates

💡 Short Explanation

Smaller sample reduces accuracy Larger n improves estimation Df = n − 1 is correct

✅ Correct Answer Explanation

Accuracy improves with larger sample sizes due to reduced sampling error. Hence statement D is incorrect because decreasing sample size worsens estimates.

❓ Why Other Options Are Incorrect

Option A → true relationship between df and n
Option B → correct definition of degrees of freedom
Option C → correct CLT rule of thumb

🎯 Strategy

Used: Extreme word filter

Final Logic: Accuracy never improves by reducing sample size

🧠 Memory Trick

"Small sample = big error"

📚 Class XI – Statistics – Chapter: Sampling – Topic: Degrees of Freedom

13 Arrange the stages of statistical inference:
I. Making inferences
II. Sample subset selection
III. Data analysis
IV. Data collection
V. Population identification

A V, IV, II, III, I

B V, II, IV, III, I

C II, V, IV, I, III

D V, III, IV, II, I

💡 Short Explanation

Identify population first Collect data Select sample Analyze data Draw inference

✅ Correct Answer Explanation

Statistical inference begins with identifying the population, followed by data collection, sampling, analysis, and finally inference-making.

❓ Why Other Options Are Incorrect

Option B → sampling before data collection incorrect
Option C → incorrect ordering of analysis and inference
Option D → analysis placed too early

🎯 Strategy

Used: Logical sequencing

Final Logic: Inference follows population → data → sample → analysis

🧠 Memory Trick

"POP → COLLECT → SAMPLE → ANALYZE → INFER"

📚 Class XI – Statistics – Chapter: Sampling Process – Topic: Statistical Inference Steps

14 Assertion (A): A two-tailed t-test is used only when testing if one population parameter is greater than another.
Reason (R): A one-tailed test checks only for inequality without direction.

A Both A and R are false

B A is true, R is false

C Both A and R are true, and R explains A

D A is false, R is true

💡 Short Explanation

Two-tailed test checks both directions One-tailed test has direction Both statements are incorrect

✅ Correct Answer Explanation

A two-tailed test checks for any difference (not only greater). One-tailed tests are directional (greater or lesser), not just inequality. Hence both A and R are false.

❓ Why Other Options Are Incorrect

Option B → both statements incorrect
Option C → incorrect reasoning link
Option D → R is not correct as stated

🎯 Strategy

Used: Concept verification

Final Logic: Both definitions are wrongly stated

🧠 Memory Trick

"Two-tail = both sides"

📚 Class XI – Statistics – Chapter: Hypothesis Testing – Topic: One-tailed vs Two-tailed tests

15 Given:
t₍obt₎ = -3.10, t₍0.025₎ = -2.776

A Accept H₀

B Increase sample size

C Cannot determine

D Reject H₀ since value is more extreme

💡 Short Explanation

|t| observed > critical value Falls in rejection region H₀ rejected

✅ Correct Answer Explanation

Since -3.10 is more extreme than -2.776 in the rejection region, the null hypothesis is rejected.

❓ Why Other Options Are Incorrect

Option A → incorrect decision
Option B → irrelevant to test decision
Option C → sufficient data provided

🎯 Strategy

Used: Critical value comparison

Final Logic: Observed statistic lies in rejection region

🧠 Memory Trick

"More extreme → reject"

📚 Class XI – Statistics – Chapter: Hypothesis Testing – Topic: t-test decision rule

16 Assertion (A): Non-probability sampling methods are prone to selection bias.
Reason (R): These methods rely on convenience or voluntary participation instead of randomization.

A Both A and R are false

B A is true, R is false

C Both A and R are true, and R is the correct explanation

D A is false, R is true

💡 Short Explanation

No randomization causes bias Convenience/voluntary selection is non-random R explains A

✅ Correct Answer Explanation

Non-probability sampling lacks random selection, leading to selection bias. Hence both statements are correct and causally linked.

❓ Why Other Options Are Incorrect

Option A → both statements are true
Option B → R is correct
Option D → A is correct

🎯 Strategy

Used: Causal reasoning

Final Logic: Lack of randomness causes bias

🧠 Memory Trick

"No random → bias grows"

📚 Class XI – Statistics – Chapter: Sampling Methods – Topic: Bias in Sampling

17 n = 60, x̄ = 49.6, μ = 50, S ≈ 1
t = (x̄ − μ)/(S/√n)

A -3.09

B 0.00

C -3.098 ≈ -3.10

D 3.09

💡 Short Explanation

Difference = -0.4 Standard error small t ≈ -3.10

✅ Correct Answer Explanation

t = (49.6 − 50) / (1/√60) ≈ -0.4 / 0.129 ≈ -3.098.

❓ Why Other Options Are Incorrect

Option A → rounded but less precise
Option B → mean ≠ sample mean
Option D → wrong sign

🎯 Strategy

Used: Substitution

Final Logic: Direct formula evaluation

🧠 Memory Trick

"Negative deviation → negative t"

📚 Class XI – Statistics – Chapter: t-test – Topic: One-sample t-test

18 If bias leads to incorrect rejection of a true null hypothesis, then correct statistical significance implies:

A Sample was biased

B Absolute certainty of population parameter

C Result due to random error

D Strong evidence that the relationship is real

💡 Short Explanation

Statistical significance indicates evidence Not absolute truth Supports alternative hypothesis

✅ Correct Answer Explanation

Significance means evidence against H₀, not certainty. It supports that an observed relationship is likely real under sampling assumptions.

❓ Why Other Options Are Incorrect

Option A → bias not implied
Option B → statistics never give certainty
Option C → not necessarily random error

🎯 Strategy

Used: Conceptual distinction

Final Logic: Significance = evidence, not certainty

🧠 Memory Trick

"Significant ≠ certain"

📚 Class XI – Statistics – Chapter: Hypothesis Testing – Topic: Statistical Significance

19 Which are unbiased sampling methods?
I. Stratified random sampling
II. Snowball sampling
III. Simple random sampling
IV. Convenience sampling

A I and II

B I and III

C III and IV

D II and IV

💡 Short Explanation

Stratified sampling is unbiased Simple random sampling is unbiased Others are biased methods

✅ Correct Answer Explanation

Stratified and simple random sampling ensure random selection, reducing bias. Snowball and convenience sampling introduce selection bias.

❓ Why Other Options Are Incorrect

Option A → snowball is biased
Option C → convenience sampling is biased
Option D → both are biased methods

🎯 Strategy

Used: Classification elimination

Final Logic: Only random methods are unbiased

🧠 Memory Trick

"Random = reliable"

📚 Class XI – Statistics – Chapter: Sampling Methods – Topic: Unbiased Sampling

20 Match List I with List II:

List I	List II
Sampling Error	a. x̄ − μ
Degrees of Freedom	b. n − 1
Standard Error of Mean	c. σ/√n

A 1–a, 2–b, 3–c

B 1–c, 2–b, 3–a

C 1–b, 2–a, 3–c

D 1–a, 2–c, 3–b

💡 Short Explanation

Sampling error = difference of means DF = n − 1 SEM formula standard

✅ Correct Answer Explanation

Sampling error is x̄ − μ, degrees of freedom is n − 1, and standard error is σ/√n.

❓ Why Other Options Are Incorrect

Option B → mismatches all definitions
Option C → incorrect pairing
Option D → incorrect mapping

🎯 Strategy

Used: Direct matching

Final Logic: Standard statistical definitions applied

🧠 Memory Trick

"Error, DF, SEM = x̄, n−1, σ/√n"

📚 Class XI – Statistics – Chapter: Sampling – Topic: Error & Estimation

CUET UG Applied Mathematics Booster Test 2 - Population & Sample

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