CUET UG Applied Mathematics Booster Test 2 - Random Variables and Foundations -

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CUET UG Applied Mathematics Booster Test 2 - Random Variables and Foundations

⏱️ 30 Minutes

❓ 20 Questions

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CUET UG Applied Mathematics Booster Test 2 - Random Variables and Foundations

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QUESTION 1 OF 20

In a manufacturing unit inspection, 2 baskets are drawn without replacement from a lot of 20. Let X denote the number of defective baskets. What are the possible values in the sample space of X?

Answer Review

1 In a manufacturing unit inspection, 2 baskets are drawn without replacement from a lot of 20. Let X denote the number of defective baskets. What are the possible values in the sample space of X?

A 0, 1

B 0, 1, 2

C 1, 2, 3

D Infinite

💡 Short Explanation

X counts defective baskets in 2 draws Minimum = 0 defective, maximum = 2 defective Hence possible values are 0, 1, 2

✅ Correct Answer Explanation

Since 2 baskets are drawn, defective count can vary from none to both defective X = 0 (no defective), X = 1 (one defective), X = 2 (both defective) All values in {0,1,2} are possible outcomes of the random variable

❓ Why Other Options Are Incorrect

Option A → excludes possibility of 2 defective baskets
Option C → not valid since maximum is 2, not 3
Option D → X is discrete and finite, not infinite

🎯 Strategy

Used: Elimination

Application: Eliminated options exceeding possible draws

Final Logic: Maximum defect count is 2, so {0,1,2} is correct

🧠 Memory Trick

"Max = number of draws"

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Random Variables –

2 Match the probability distribution variables to their key identifying parameters.

List I	List II
(A) Binomial Distribution	(I) μ and σ
(B) Poisson Distribution	(II) λ
(C) Normal Distribution	(III) μ = 0, σ = 1
(D) Standard Normal	(IV) n and p

A (A)-(IV), (B)-(II), (C)-(I), (D)-(III)

B (A)-(I), (B)-(II), (C)-(III), (D)-(IV)

C (A)-(IV), (B)-(I), (C)-(II), (D)-(III)

D (A)-(II), (B)-(IV), (C)-(III), (D)-(I)

💡 Short Explanation

Binomial → n, p Poisson → λ Normal → μ, σ Standard normal → μ=0, σ=1

✅ Correct Answer Explanation

Binomial distribution depends on number of trials (n) and probability (p) Poisson depends on rate λ Normal distribution uses mean μ and standard deviation σ Standard normal is standardized μ=0, σ=1

❓ Why Other Options Are Incorrect

Option B → swaps parameter assignments incorrectly
Option C → mismatches all distributions
Option D → incorrect mapping of all variables

🎯 Strategy

Used: Option Grouping

Application: Match known distribution parameters

Final Logic: Each distribution has unique defining parameters

🧠 Memory Trick

"B-N-P-S → n,p; λ; μσ; 0-1"

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Probability Distributions –

3 Which of the following conditions must be met for a collection of trials to be called Bernoulli trials?
(A) The number of trials is finite.
(B) Each trial has exactly two outcomes.
(C) The trials are independent.
(D) The probability of success remains the same in each trial.

A A,B,D

B A,B

C A,B,C

D A,B,C,D

💡 Short Explanation

Bernoulli trials require fixed number of trials Each trial has two outcomes Independence and constant probability required

✅ Correct Answer Explanation

All conditions define Bernoulli process Finite trials ensure bounded experiment Two outcomes define success/failure structure Independence ensures no influence between trials Constant probability maintains consistency

❓ Why Other Options Are Incorrect

Option A → missing independence condition
Option B → incomplete definition
Option C → still incomplete (missing probability constancy)

🎯 Strategy

Used: Elimination

Application: Checked full Bernoulli conditions

Final Logic: All four conditions are necessary

🧠 Memory Trick

"F2IC = Finite, Two outcomes, Independent, Constant p"

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Bernoulli Trials

4 Which of the following statements about the variance of a discrete probability distribution is INCORRECT?

A It is denoted by Var(X) or σ².

B Var(X) = E(X²) - [E(X)]².

C A narrower range of sample elements means the variance is high.

D It measures the average degree to which each sample element differs from the mean.

💡 Short Explanation

Variance measures spread Narrow range → low variance So statement C is false

✅ Correct Answer Explanation

Variance increases with dispersion If values are close (narrow range), variance is small Thus statement claiming high variance is incorrect

❓ Why Other Options Are Incorrect

Option A → correct definition
Option B → correct formula
Option D → correct conceptual meaning

🎯 Strategy

Used: Extreme Word Filter

Application: Checked contradiction in "high variance" claim

Final Logic: Narrow spread implies low variance

🧠 Memory Trick

"Narrow → Low variance"

📚 Class XI – NCERT Mathematics – Chapter: Statistics & Probability – Topic: Variance –

5 A coin is tossed thrice. The probability mixture distribution for X (number of heads) shows that P(X=2) is calculated as:

A 1/8

B 3/8

C 4/8

D 7/8

💡 Short Explanation

Total outcomes = 2³ = 8 Ways to get 2 heads = 3 Probability = 3/8

✅ Correct Answer Explanation

Possible sequences: HHT, HTH, THH Each has equal probability So P(X=2) = 3/8

❓ Why Other Options Are Incorrect

Option A → only 1 favorable case (incorrect)
Option C → incorrect total scaling
Option D → not valid probability for this event

🎯 Strategy

Used: Substitution

Application: Count favorable outcomes

Final Logic: 3 favorable outcomes out of 8

🧠 Memory Trick

"3 heads pairs in 3 toss case"

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Binomial Distribution –

6 Let X denote the number of hours a person watches television. If P(X=0)=0.2, P(X=1)=k, P(X=2)=2k, P(X=3)=2k, what is the value of k?

A 0.16

B 4/25

C 1/5

D 8/25

💡 Short Explanation

Total probability = 1 0.2 + 5k = 1 k = 0.16 = 4/25

✅ Correct Answer Explanation

Sum of probabilities must equal 1 0.2 + k + 2k + 2k = 1 → 0.2 + 5k = 1 5k = 0.8 → k = 0.16 = 4/25

❓ Why Other Options Are Incorrect

Option A → correct decimal but fractional form preferred
Option C → incorrect calculation
Option D → exceeds valid probability

🎯 Strategy

Used: Equation balancing

Application: Sum of probabilities = 1

Final Logic: Solve linear equation in k

🧠 Memory Trick

"Total = 1 always"

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Probability Distribution –

7 What is the formula for the theoretical mean E(X) of a discrete distribution?

A Σ (x_i + p_i)

B Σ (x_i * p_i)

C Σ (x_i / p_i)

D Σ (x_i² * p_i)

💡 Short Explanation

Expectation = weighted average Weights = probabilities Multiply and sum

✅ Correct Answer Explanation

Mean of random variable is sum of value × probability Each outcome contributes proportionally to likelihood Hence E(X)=Σ x_i p_i

❓ Why Other Options Are Incorrect

Option A → incorrect addition form
Option C → invalid division structure
Option D → variance-related, not mean

🎯 Strategy

Used: Concept recall

Application: Definition of expectation

Final Logic: Expectation is weighted sum

🧠 Memory Trick

"Mean = value × probability"

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Mathematical Expectation –

8 For Y taking values -1, 0, 4, 5 with probabilities 1/7, 2/7, 3/7, 1/7, what is E(Y)?

A 17/7

B 16/7

C 15/7

D 1

💡 Short Explanation

Multiply values with probabilities Sum all products Compute final expectation

✅ Correct Answer Explanation

E(Y)=(-1)(1/7)+0+4(3/7)+5(1/7) = (-1 + 12 + 5)/7 = 16/7

❓ Why Other Options Are Incorrect

Option A → arithmetic error
Option C → incorrect summation
Option D → not equal to computed expectation

🎯 Strategy

Used: Substitution

Application: Direct computation of expectation

Final Logic: Weighted sum gives 16/7

🧠 Memory Trick

"Multiply → Add → Divide"

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Expectation –

9 A fair coin is tossed 9 times. The probability of getting exactly five tails is:
**Correct Answer: A) C(9,5) * (1/2)^9**

A C(9,5) * (1/2)^9

B C(9,4) * (1/2)^5

C C(9,5) * (1/2)^4

D C(9,9) * (1/2)^9

💡 Short Explanation

Binomial distribution applies Choose 5 tails from 9 trials Probability factor is (1/2)^9

✅ Correct Answer Explanation

P(X=r)=C(n,r)p^r q^(n-r) Here p=q=1/2 So P(X=5)=C(9,5)(1/2)^9

❓ Why Other Options Are Incorrect

Option B → incorrect combination and power
Option C → wrong exponent
Option D → incorrect combination selection

🎯 Strategy

Used: Formula substitution

Application: Binomial probability formula

Final Logic: Direct binomial evaluation

🧠 Memory Trick

"nCr × (1/2)^n"

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Binomial Distribution –

10 A frequency vector shows frequencies (3, 5, 2, 3, 3, 2, 2). Probability vector element p₁ for mark 14 is:

A 14/20

B 3/20

C 5/20

D 1/20

💡 Short Explanation

Probability = frequency / total Total = 20 For 14, frequency = 3

✅ Correct Answer Explanation

p₁ = 3/20 Total students = 20 So probability of mark 14 = 3/20

❓ Why Other Options Are Incorrect

Option A → uses value instead of frequency
Option C → wrong frequency mapping
Option D → incorrect normalization

🎯 Strategy

Used: Direct proportion

Application: Probability = f/n

Final Logic: 3 out of 20 gives correct probability

🧠 Memory Trick

"f / total"

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Probability Distribution Table –

11 A normal distribution takes infinite outcomes. The total area under its continuous probability distribution normal curve represents the total probability and is equal to:

A 0

B 0.5

C 1

D Infinity

💡 Short Explanation

Total probability under any PDF = 1 Normal distribution is a continuous probability model Area under curve represents probability

✅ Correct Answer Explanation

A probability density function (PDF) like the normal distribution is defined such that the total area under the curve over جميع real numbers equals 1. This ensures that the probability of all possible outcomes combined is complete and normalized. Hence, for a normal distribution, the total area is 1.

❓ Why Other Options Are Incorrect

Option A → 0 is impossible since probability cannot vanish completely
Option B → 0.5 represents only half-area, not total probability
Option D → Infinity contradicts normalization of probability distributions

🎯 Strategy

Used

Elimination
�� Use the fundamental rule of probability normalization

Final Logic:

Total area under any probability distribution must equal 1

🧠 Memory Trick

"PDF = Perfect Distribution Fraction = 1"

📚 Class XII – NCERT Mathematics – Chapter: Probability – Topic: Continuous Random Variable –

12 The integral defining the normal distribution curve is symmetrical about the line:

A x = 0 only

B x = σ

C x = μ

D x = 1

💡 Short Explanation

Normal distribution is symmetric Symmetry axis is mean (μ) Variance does not affect center

✅ Correct Answer Explanation

A normal distribution is perfectly symmetric about its mean μ. The peak of the curve occurs at μ, and both sides mirror each other. Therefore, the axis of symmetry is x = μ.

❓ Why Other Options Are Incorrect

Option A → x = 0 is only for standard normal case
Option B → σ affects spread, not symmetry axis
Option D → 1 has no statistical relevance here

🎯 Strategy

Used

Contextual/Tonal Matching
�� Match definition of symmetry in normal curve

Final Logic:

Normal distribution symmetry is always centered at mean μ

🧠 Memory Trick

"Normal = Mean is the Mirror"

📚 Class XII – NCERT Mathematics – Chapter: Probability – Topic: Normal Distribution

📄 Passage

PASSAGE MCQs (COMPULSORY) One of the famous historical studies shows an application of Poisson distribution to estimate the numbers of Prussian cavalry soldiers killed due to horse-kicks in a year! As the story goes, the Prussian soldiers monitored 10 cavalry corps over a period of 20 years. The annual number of recorded deaths due to horse-kick 'k' observations showed a total of 200 deaths. The events defined as horse-kicks in such a case are occurring randomly and independently. A discrete probability distribution that expresses the probability of a given number of events occurring over a fixed period of time or space is called a Poisson distribution. The rate of occurrence of events is constant and not based on time.

According to the passage, the mean expectation λ for the number of horse-kick deaths per observation (total 200 observations and 122 deaths recorded) is calculated as:

A 200 / 122

B 122 / 200 = 0.61

C 0

D 1

💡 Short Explanation

Poisson mean λ = average rate λ = total events / total observations 122 / 200 = 0.61

✅ Correct Answer Explanation

In a Poisson distribution, λ represents the average number of occurrences per unit interval. Here, total deaths = 122 and total observations = 200, so λ = 122/200 = 0.61.

❓ Why Other Options Are Incorrect

Option A → reversed ratio
Option C → ignores data
Option D → arbitrary constant

🎯 Strategy

Used

Substitution
�� Direct application of λ formula

Final Logic:

λ is computed as event frequency per observation

🧠 Memory Trick

"Lambda = Loss over Length"

📚 Class XII – NCERT Mathematics – Chapter: Poisson Distribution – Topic: Mean Rate λ –

📄 Passage

Based on the passage, the probability function Strategy Used to model these random, independent horse-kick events over a fixed period is the:

A Normal Distribution

B Binomial Distribution

C Poisson Distribution

D Standard Distribution

💡 Short Explanation

Rare random events over time Independent occurrences Fixed rate process

✅ Correct Answer Explanation

Poisson distribution models the probability of a number of independent events occurring in a fixed interval of time or space, exactly matching horse-kick deaths.

❓ Why Other Options Are Incorrect

Option A → for continuous symmetric data
Option B → for fixed trials only
Option D → not a valid distribution

🎯 Strategy

Used

Contextual/Tonal Matching
�� Match scenario to distribution definition

Final Logic:

Random rare independent events → Poisson model

🧠 Memory Trick

"Poisson = Periodic Possibility of Occurrence"

📚 Class XII – NCERT Mathematics – Chapter: Poisson Distribution – Topic: Real-life Applications –

15 If a normal variate X has a mean of 12 and a standard deviation of 4, the corresponding Z-score for a data point of 20 is:

A 1.5

B 2.0

C 2.5

D 8.0

💡 Short Explanation

Z = (X − μ) / σ (20 − 12) / 4 = 2 Standardization formula

✅ Correct Answer Explanation

Z-score formula converts raw data into standard normal form: Z = (20 − 12) / 4 = 8/4 = 2. Hence correct answer is 2.0.

❓ Why Other Options Are Incorrect

Option A → incorrect arithmetic
Option B → actually correct but option label mismatch (closest correct is 2.0)
Option C → incorrect calculation
Option D → unrelated scale

🎯 Strategy

Used

Substitution
�� Direct use of Z-score formula

Final Logic:

Standardization converts deviation into units of σ

🧠 Memory Trick

"Z = Distance over Deviation"

📚 Class XII – NCERT Mathematics – Chapter: Normal Distribution – Topic: Z-score –

16 In a Poisson distribution, if the mean expectation λ is 2, what is the Variance associated with this distribution?

A 0

B 1

C 2

D 4

💡 Short Explanation

In Poisson, mean = variance λ = 2 So variance = 2

✅ Correct Answer Explanation

A key property of Poisson distribution is that its mean and variance are equal. Therefore if λ = 2, variance = 2.

❓ Why Other Options Are Incorrect

Option A → variance cannot be zero unless λ = 0
Option B → incorrect equality assumption
Option D → incorrect doubling

🎯 Strategy

Used

Elimination
�� Use known Poisson property

Final Logic:

Variance equals λ in Poisson distribution

🧠 Memory Trick

"Poisson twins: Mean = Variance"

📚 Class XII – NCERT Mathematics – Chapter: Poisson Distribution – Topic: Properties –

17 To apply the Z-test correctly on a normal distribution data set, each randomly selected data point must be independent, and the sample size must be:

A Less than 10

B Exactly 20

C Greater than 30

D Infinite

💡 Short Explanation

Large sample approximation Central Limit Theorem Standard condition n > 30

✅ Correct Answer Explanation

For Z-tests, the sample size should generally be large (n ≥ 30) to approximate normality via Central Limit Theorem, ensuring reliability of inference.

❓ Why Other Options Are Incorrect

Option A → too small for Z-test
Option B → arbitrary value
Option D → impractical

🎯 Strategy

Used

Contextual/Tonal Matching
�� Match statistical condition for Z-test

Final Logic:

Z-test requires sufficiently large sample size

🧠 Memory Trick

"Z-test needs Thirty or More"

📚 Class XII – NCERT Mathematics – Chapter: Inferential Statistics – Topic: Z-test Conditions –

18 If 5 baskets are drawn with replacement where the probability of a defective basket is 3/10, the trials qualify as Bernoulli trials because:

A The probability of success varies.

B The drawing is done without replacement.

C The probability of success remains constant in all trials.

D The number of trials is infinite.

💡 Short Explanation

Bernoulli trials need constant probability With replacement ensures independence Two outcomes only

✅ Correct Answer Explanation

Bernoulli trials require fixed probability of success across trials, independence, and two outcomes. With replacement, probability remains constant.

❓ Why Other Options Are Incorrect

Option A → contradicts definition
Option B → wrong condition (should be with replacement)
Option D → irrelevant

🎯 Strategy

Used

Elimination
�� Check Bernoulli conditions

Final Logic:

Constant probability defines Bernoulli trials

🧠 Memory Trick

"Bernoulli = Balanced Repeat Trials"

📚 Class XII – NCERT Mathematics – Chapter: Probability – Topic: Bernoulli Trials –

19 A random variable with a Gaussian distribution whose mean, median, and mode are exactly the same is said to follow a:

A Binomial distribution

B Poisson distribution

C Normal distribution

D Discrete distribution

💡 Short Explanation

Symmetric distribution Mean = Median = Mode Bell-shaped curve

✅ Correct Answer Explanation

In a normal distribution, symmetry ensures that mean, median, and mode coincide at the center.

❓ Why Other Options Are Incorrect

Option A → discrete distribution
Option B → skewed for small λ
Option D → not specific distribution type

🎯 Strategy

Used

Contextual/Tonal Matching
�� Match properties of Gaussian distribution

Final Logic:

Equal central tendencies define normal distribution

🧠 Memory Trick

"Normal = No Gap in Center Measures"

📚 Class XII – NCERT Mathematics – Chapter: Normal Distribution – Topic: Properties –

20 In binomial distributions, predicting the random behaviour of 'r' successes in 'n' Bernoulli trials utilizes the binomial expansion formula of:

A (p - q)^n

B (q + p)^n

C (p + q)^(-n)

D e^(-λ)

💡 Short Explanation

Binomial theorem basis p + q = 1 Expansion gives probabilities

✅ Correct Answer Explanation

The binomial distribution is derived from expanding (p + q)^n, where p is success probability and q = 1 − p.

❓ Why Other Options Are Incorrect

Option A → incorrect sign usage
Option C → invalid exponent
Option D → Poisson form

🎯 Strategy

Used

Substitution
�� Direct binomial theorem application

Final Logic:

Binomial probabilities come from (p + q)^n expansion

🧠 Memory Trick

"Binomial = p plus q power n"

📚 Class XII – NCERT Mathematics – Chapter: Binomial Distribution – Topic: Probability Model –

CUET UG Applied Mathematics Booster Test 2 - Random Variables and Foundations

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