CUET UG Applied Mathematics Booster Test 2 - Variance and Binomial Distribution -

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CUET UG Applied Mathematics Booster Test 2 - Variance and Binomial Distribution

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Applied Mathematics Booster Test 2 - Variance and Binomial Distribution

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QUESTION 1 OF 20

Two different random variables, X and Y, can share the exact same mean expectation (e.g., E(X) = 2.4 and E(Y) = 2.4) but look completely different on a graph. What statistical measure must be calculated to evaluate the extent to which their values are spread out?

Answer Review

1 Two different random variables, X and Y, can share the exact same mean expectation (e.g., E(X) = 2.4 and E(Y) = 2.4) but look completely different on a graph. What statistical measure must be calculated to evaluate the extent to which their values are spread out?

A Median

B Z-Score

C Variance

D Total Probability

💡 Short Explanation

Mean measures central value Spread is not captured by mean Variance measures dispersion around mean

✅ Correct Answer Explanation

Variance is the correct measure because it quantifies how far data values deviate from the mean on average. Even if two distributions have the same mean (like 2.4), their spread can differ significantly. Variance captures this spread using squared deviations, making it the fundamental tool for comparing variability.

❓ Why Other Options Are Incorrect

Option A → Median measures central position, not spread
Option B → Z-score standardizes values, not overall spread
Option D → Total probability ensures normalization, not dispersion

🎯 Strategy

Used: Elimination

Application: Remove all options not related to variability

Final Logic: Only variance directly measures spread

🧠 Memory Trick

"Mean = center, Variance = scatter"

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Variance and dispersion –

2 Match the specific variance parameter symbols to their definitions.

List I	List II
(A) μ	(I) Variance of X
(B) σ²	(II) Standard Deviation
(C) σ	(III) Theoretical Mean E(X)
(D) E(X²)	(IV) Sum of product of squared values and probabilities

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

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

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

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

💡 Short Explanation

μ is mean σ² is variance σ is standard deviation E(X²) is weighted square expectation

✅ Correct Answer Explanation

Each symbol corresponds to standard statistical notation: μ represents mean expectation E(X), σ² represents variance, σ is its square root (standard deviation), and E(X²) is computed as Σ(x²p). These are foundational NCERT definitions in probability theory.

❓ Why Other Options Are Incorrect

Option B → Mislabels all symbols incorrectly
Option C → Incorrect mapping of mean and variance
Option D → Swaps variance and mean incorrectly

🎯 Strategy

Used: Option Grouping

Application: Match standard formula definitions directly

Final Logic: Only Option A preserves all standard identities

🧠 Memory Trick

"μ mean, σ² spread, σ root spread"

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Mean, variance and standard deviation –

3 In a probability distribution of a discrete random variable X, the variance Var(X) relies on the squared deviations from the mean. Which of the following components are mathematically required to calculate this?

A The corresponding probabilities p_i of each outcome.

B The sum of the product of squared deviations [x_i - E(X)]² and p_i.

C The theoretical mean expectation E(X).

D An infinite number of uncountable outcomes.

💡 Short Explanation

Variance depends on probabilities Uses squared deviations Requires mean value

✅ Correct Answer Explanation

Variance in a discrete distribution is defined as Σ p_i (x_i − μ)². Thus it requires probabilities, squared deviations, and the mean E(X). Option D is irrelevant since distributions are discrete and countable.

❓ Why Other Options Are Incorrect

Option A → Required but not sufficient alone
Option C → Required but incomplete alone
Option D → Incorrect; discrete variables are countable

🎯 Strategy

Used: Option Grouping

Application: Identify components of variance formula

Final Logic: Only A, B, and C together define variance fully

🧠 Memory Trick

"Variance = Probabilities × Squared distance from mean"

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Variance formula –

4 Identify the INCORRECT mathematical expression regarding the variance Var(X).

A Var(X) is denoted by σ².

B Var(X) = E(X²) - [E(X)]².

C Var(X) = [E(X)]² - E(X²).

D E(X²) is calculated as Σ (x_i² * p_i).

💡 Short Explanation

Variance formula is fixed Order matters in subtraction Negative form is invalid

✅ Correct Answer Explanation

Variance is correctly given by E(X²) − [E(X)]². Option C reverses the order, which would produce negative variance, which is impossible since variance is always ≥ 0.

❓ Why Other Options Are Incorrect

Option A → Correct notation
Option B → Correct identity
Option D → Correct definition of E(X²)

🎯 Strategy

Used: Extreme Word Filter

Application: Check sign correctness in formula

Final Logic: Only reversed formula violates definition

🧠 Memory Trick

"Square mean subtracts from square expectation"

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Variance formula –

5 A mixture of deviations creates the total variance of 1.22 for a dataset. To find the standard deviation (σ), what mathematical operation must be applied to the variance?

A Squaring the variance

B Taking the positive square root of the variance

C Dividing the variance by 2

D Multiplying the variance by the mean

💡 Short Explanation

σ = √Var(X) Standard deviation is root of variance Must be non-negative

✅ Correct Answer Explanation

Standard deviation is defined as σ = √Var(X). Since variance measures squared dispersion, taking the square root returns the original scale of data spread.

❓ Why Other Options Are Incorrect

Option A → Wrong transformation
Option C → No relation to definition
Option D → Incorrect formula

🎯 Strategy

Used: Substitution

Application: Apply definition σ = √Var(X)

Final Logic: Only square root converts variance to SD

🧠 Memory Trick

"Variance squared → SD rooted"

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Standard deviation –

6 By definition, the standard deviation σ is a dispersion measure representing the positive square root of the variance. Which of the following constraint regions must always hold true for σ?

A σ < 0

B σ ≥ 0

C σ = -1

D σ ≤ -1

💡 Short Explanation

Square root is non-negative Dispersion cannot be negative σ is always ≥ 0

✅ Correct Answer Explanation

Since σ = √Var(X), and variance is always non-negative, standard deviation must also be non-negative. Therefore, σ ≥ 0 is always true.

❓ Why Other Options Are Incorrect

Option A → Impossible (SD cannot be negative)
Option C → Incorrect fixed value
Option D → Invalid constraint

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Check sign properties of square root

Final Logic: Square root ensures non-negativity

🧠 Memory Trick

"Spread cannot go below zero"

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Standard deviation properties –

7 In a marks distribution application, if E(X²) evaluates to 1323/20 and [E(X)]² evaluates to 400/20, the variance computation Var(X) relies on subtracting these two values. What is the foundational formula applied here?

A Σ(p_i)

B E(X) - E(X²)

C E(X²) - [E(X)]²

D Σ(x_i)

💡 Short Explanation

Variance = E(X²) − mean² Order is fixed Ensures non-negative result

✅ Correct Answer Explanation

The standard variance formula is Var(X) = E(X²) − [E(X)]². This subtracts the square of the mean from the expected square, giving correct dispersion.

❓ Why Other Options Are Incorrect

Option A → Only total probability
Option B → Wrong order
Option D → Not a probability measure

🎯 Strategy

Used: Elimination

Application: Remove non-variance expressions

Final Logic: Only option matching formula is correct

🧠 Memory Trick

"Square first, then subtract mean square"

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Variance computation –

8 When tracking sample variability across different probability distributions, if Var(X) = 0.89 and Var(Y) = 3.5, which distribution has data values clustered more tightly around its moving average (mean)?

A Distribution X

B Distribution Y

C They are clustered equally.

D Cannot be determined without standard deviation.

💡 Short Explanation

Lower variance = less spread Higher variance = more spread Compare directly

✅ Correct Answer Explanation

Variance measures dispersion. Since 0.89 < 3.5, X has less spread, meaning values are closer to the mean compared to Y.

❓ Why Other Options Are Incorrect

Option B → Incorrect; higher variance means more spread
Option C → Values differ
Option D → Variance already sufficient

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Compare magnitudes directly

Final Logic: Lower variance implies tighter clustering

🧠 Memory Trick

"Smaller variance = tighter data"

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Variance comparison –

9 In a Bernoulli trial where drawing a defective basket is a "success", and 6 out of 20 baskets are defective, what are the exact probability values for success (p) and failure (q)?

A p = 1/20, q = 19/20

B p = 3/10, q = 7/10

C p = 6/10, q = 4/10

D p = 1/2, q = 1/2

💡 Short Explanation

Success = defective p = 6/20 q = 1 − p

✅ Correct Answer Explanation

Probability of success is p = 6/20 = 3/10. Hence failure probability q = 1 − 3/10 = 7/10.

❓ Why Other Options Are Incorrect

Option A → Incorrect fraction
Option C → Wrong scaling
Option D → Assumes fairness

🎯 Strategy

Used: Substitution

Application: Direct probability calculation

Final Logic: Convert 6/20 correctly

🧠 Memory Trick

"Defect fraction = success probability"

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

10 A manufacturing unit inspects 2 baskets without replacement from a lot of 20 baskets containing 6 defectives. Why does this vector of trials FAIL to qualify as Bernoulli trials?

A The number of trials is infinite.

B The draws have more than two outcomes.

C The probability of success (drawing a defective basket) changes in the second draw, so trials are not independent.

D The mean npq evaluates to zero.

💡 Short Explanation

No replacement changes probabilities Trials become dependent Bernoulli requires independence

✅ Correct Answer Explanation

Bernoulli trials require constant probability and independence. Without replacement, probabilities change after first draw, violating independence.

❓ Why Other Options Are Incorrect

Option A → Finite condition satisfied
Option B → Only two outcomes exist
Option D → Irrelevant condition

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Check Bernoulli conditions

Final Logic: Dependence breaks Bernoulli assumption

🧠 Memory Trick

"No replacement → no independence"

📚 Class XI – NCERT Mathematics – Chapter: Probability – Topic: Bernoulli trials conditions –

11 Which of the following experimental scenarios strictly satisfies the condition of having a finite number of trials for a Bernoulli area mapping?

A Tossing a coin until a head appears.

B Rolling a die repeatedly until three 5s are obtained.

C Tossing a fair coin exactly 10 times.

D Drawing cards without replacement until an Ace is drawn.

💡 Short Explanation

�� Bernoulli trials require fixed number of trials �� Must be finite and predetermined �� Stopping rules violate condition

✅ Correct Answer Explanation

Bernoulli experiments require a fixed, finite number of trials → Option C fixes n = 10 in advance → Other options involve stopping conditions, making n random → Hence C satisfies the condition

❓ Why Other Options Are Incorrect

�� Option A → Trials continue until success (not fixed n)
�� Option B → Stopping after condition (random n)
�� Option D → Depends on outcome (non-fixed trials)

🎯 Strategy

Used

Condition Filtering

Application: Check whether number of trials is fixed

Final Logic: Only fixed n ensures Bernoulli setup

🧠 Memory Trick

"Bernoulli = fixed trials only"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Bernoulli Trials Conditions –

12 If a bag contains 2 black and 1 red pens, and a pen is drawn, noted, and put back (with replacement), repeating this 3 times makes it a Bernoulli experiment. What is the constant probability of success 'p' if success is drawing a red pen?

A 1/2

B 1/3

C 2/3

D 3/3

💡 Short Explanation

�� Success = red pen �� Total pens = 3 �� Probability stays constant due to replacement

✅ Correct Answer Explanation

Total pens = 3, red pens = 1 → p = 1/3 for each independent trial → Replacement ensures constant probability → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Incorrect ratio
�� Option C → Confuses black vs red
�� Option D → Probability cannot be 1

🎯 Strategy

Used

Direct Calculation

Application: Compute probability from given ratio

Final Logic: p = favourable/total = 1/3

🧠 Memory Trick

"Red = one out of three"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Bernoulli Trials –

📄 Passage

PASSAGE MCQs (COMPULSORY) Let us consider the Binomial distribution denoted by B(n, p). The binomial distribution with n Bernoulli trials and success p defines its key parameters analytically. Specifically, the mean of the distribution (the expected value) is calculated simply as the product of the number of trials and the probability of success, meaning Mean = np. Furthermore, the variance, which measures the spread of the distribution around this mean, is calculated as Variance = npq, where q is the probability of failure (q = 1 - p). From this, the Standard Deviation is immediately derived as √(npq).

Based on the passage parameters, if a distribution B(4, 1/3) is given, what is the theoretical Mean Expectation?

A 1/3

B 4/3

C 8/9

D 4

💡 Short Explanation

�� Mean of binomial = np �� n = 4, p = 1/3 �� Multiply directly

✅ Correct Answer Explanation

E(X) = np → E(X) = 4 × 1/3 = 4/3 → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Only probability value
�� Option C → Variance-related value
�� Option D → Ignores probability factor

🎯 Strategy

Used

Formula Substitution

Application: Apply binomial mean formula

Final Logic: np gives correct expectation

🧠 Memory Trick

"Multiply n and p"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Binomial Mean –

📄 Passage

Based on the passage formulas, what is the calculated Variance for the same distribution B(4, 1/3)?

A 4/9

B 8/9

C 16/9

D 4/3

💡 Short Explanation

�� Variance = npq �� q = 2/3 �� Substitute values

✅ Correct Answer Explanation

q = 1 − p = 2/3 → Var(X) = 4 × (1/3) × (2/3) → = 8/9 → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Missing factor of n
�� Option C → Incorrect multiplication
�� Option D → Confuses mean with variance

🎯 Strategy

Used

Formula Application

Application: Use npq directly

Final Logic: Substitute values correctly

🧠 Memory Trick

"npq = variance"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Binomial Variance –

15 When constructing the binomial distribution table for B(4, 1/3), what is the formula to calculate the exact probability of getting r = 2 successes?

A C(4,2) * (1/3)² * (2/3)²

B C(4,2) * (1/3)² * (1/3)²

C C(4,4) * (1/3)⁴ * (2/3)⁰

D C(4,0) * (1/3)⁰ * (2/3)⁴

💡 Short Explanation

�� Binomial probability formula �� Uses r successes and n−r failures �� Standard structure

✅ Correct Answer Explanation

P(r) = C(n,r)pʳqⁿ⁻ʳ → Here n=4, r=2, p=1/3, q=2/3 → Correct substitution gives option A

❓ Why Other Options Are Incorrect

�� Option B → Wrong failure probability
�� Option C → r = 4 case
�� Option D → r = 0 case

🎯 Strategy

Used

Substitution

Application: Plug values into binomial formula

Final Logic: C(4,2)p²q²

🧠 Memory Trick

"Choose, power p, power q"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Binomial Distribution –

16 In the probability distribution table for tossing a coin 3 times (n=3, p=1/2), the probability term for r=1 success is calculated as C(3,1) * (1/2)¹ * (1/2)². What does this evaluate to?

A 1/8

B 3/8

C 5/8

D 7/8

💡 Short Explanation

�� C(3,1) = 3 �� (1/2)³ = 1/8 �� Multiply

✅ Correct Answer Explanation

C(3,1) = 3 → (1/2)¹(1/2)² = (1/2)³ = 1/8 → 3 × 1/8 = 3/8 → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Misses combination factor
�� Option C → Incorrect arithmetic
�� Option D → Overestimation

🎯 Strategy

Used

Calculation Substitution

Application: Compute stepwise

Final Logic: 3 × 1/8 = 3/8

🧠 Memory Trick

"3 ways × 1/8 = 3/8"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Binomial Terms –

17 A company manufactures light bulbs, where the probability of a bulb being defective is 0.05. In a sample of 100 randomly selected bulbs, what is the expected mean number of defective bulbs?

A 0.5

B 5.0

C 50.0

D 95.0

💡 Short Explanation

�� Mean = np �� n = 100 �� p = 0.05

✅ Correct Answer Explanation

E(X) = np = 100 × 0.05 → = 5 → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Underestimates mean
�� Option C → Incorrect scaling
�� Option D → Confuses success/failure

🎯 Strategy

Used

Direct Formula

Application: Apply binomial mean

Final Logic: 100 × 0.05 = 5

🧠 Memory Trick

"5 per 100"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Binomial Mean Applications –

18 Using the same light bulb sample (n=100, p=0.05, q=0.95), what is the calculated variance for the number of defective bulbs?

A 4.75

B 5.00

C 9.50

D 47.50

💡 Short Explanation

�� Variance = npq �� Substitute values �� Multiply stepwise

✅ Correct Answer Explanation

Var(X) = 100 × 0.05 × 0.95 → = 4.75 → Hence option A is correct

❓ Why Other Options Are Incorrect

�� Option B → Ignores q factor
�� Option C → Incorrect multiplication
�� Option D → Overestimation

🎯 Strategy

Used

Substitution

Application: Apply npq formula

Final Logic: 100 × 0.05 × 0.95

🧠 Memory Trick

"Multiply all three: n, p, q"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Binomial Variance –

19 A fair coin is tossed 9 times. To calculate the probability of getting exactly 5 tails, which of the following expressions is correct?

A C(9,4) * (1/2)⁴

B C(9,5) * (1/2)⁹

C C(9,5) * (1/2)⁵

D C(5,9) * (1/2)⁹

💡 Short Explanation

�� Binomial formula �� p = 1/2, q = 1/2 �� Total exponent = 9

✅ Correct Answer Explanation

P(r=5) = C(9,5)(1/2)⁵(1/2)⁴ → = C(9,5)(1/2)⁹ → Hence option B is correct

❓ Why Other Options Are Incorrect

�� Option A → Missing probability power
�� Option C → Incomplete exponent structure
�� Option D → Invalid combination

🎯 Strategy

Used

Formula Simplification

Application: Combine powers of 1/2

Final Logic: (1/2)⁹ total probability

🧠 Memory Trick

"Half powers add up"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Binomial Distribution –

20 In the defective basket experiment with replacement (n=2, p=3/10, q=7/10), what is the probability of drawing exactly zero defective baskets?

A 9/100

B 42/100

C 49/100

D 100/100

💡 Short Explanation

�� Zero successes → qⁿ �� q = 7/10 �� n = 2

✅ Correct Answer Explanation

P(0 defects) = (7/10)² → = 49/100 → Hence option C is correct

❓ Why Other Options Are Incorrect

�� Option A → Incorrect probability setup
�� Option B → Mixed terms
�� Option D → Impossible certainty

🎯 Strategy

Used

Extreme Case Rule

Application: Use qⁿ for zero success

Final Logic: (7/10)² = 49/100

🧠 Memory Trick

"No success → only failure power"

📚 Class XI – NCERT Mathematics (Part I) – Chapter: Probability – Topic: Binomial Applications –

CUET UG Applied Mathematics Booster Test 2 - Variance and Binomial Distribution

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