CUET UG Applied Mathematics Booster Test 2

🚀

CUET UG Applied Mathematics Booster Test 2 - Poisson Distribution

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Applied Mathematics Booster Test 2 - Poisson Distribution

1 / 20

QUESTION 1 OF 20

Which of the following real-world scenarios is NOT an appropriate model for a Poisson distribution?

Answer Review

1 Which of the following real-world scenarios is NOT an appropriate model for a Poisson distribution?

A Number of printing errors per in a large book.

B Number of defective baskets drawn without replacement from a finite sample.

C Number of traffic accidents at a specific intersection per month.

D Number of customers arriving at a restaurant per hour.

💡 Short Explanation

�� Poisson applies to rare, independent events in continuous time/space �� Without replacement violates independence �� Finite sample contradicts Poisson assumptions

✅ Correct Answer Explanation

Poisson distribution models random, independent events occurring at a constant rate in time/space. → Option B involves sampling without replacement, making probabilities dependent on previous draws. → Other options describe independent event counts over time/space, suitable for Poisson modeling.

❓ Why Other Options Are Incorrect

�� Option A → Printing errors occur independently per
�� Option C → Accidents over time are classic Poisson events
�� Option D → Customer arrivals are standard Poisson processes

🎯 Strategy

Used: Elimination

Application: Identify violation of independence/finite sampling

Final Logic: Without replacement breaks Poisson conditions

🧠 Memory Trick

"Poisson = PER (/Event/Rate), not sampling without replacement"

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Poisson Distribution –

2 Match the existence conditions and formulas of the Poisson distribution.

List I	List II
(A) Σ P(X = k) from k = 0 to ∞	(I) (λ^k × e^-λ) / k!
(B) Variance	(II) λ
(C) P(X = k)	(III) √λ
(D) Standard Deviation	(IV) 1

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)-(I), (D)-(III)

💡 Short Explanation

�� Total probability equals 1 �� Variance equals λ �� PMF defines probability

✅ Correct Answer Explanation

(A) sum of probabilities = 1 → (IV) → (B) variance of Poisson = λ → (II) → (C) P(X=k) formula → (I) → (D) standard deviation = √λ → (III)

❓ Why Other Options Are Incorrect

�� Option B → mismatches variance and SD
�� Option C → incorrect mapping of mean/variance
�� Option D → incorrect assignment of fundamental definitions

🎯 Strategy

Used: Option Grouping

Application: Match known Poisson identities directly

Final Logic: Standard Poisson properties mapping

🧠 Memory Trick

"1–λ–formula–√λ order"

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Poisson Properties –

3 To accurately apply a Poisson distribution, the "independent events" condition dictates that:
(A) The probability of an arrival is independent of previous arrivals.
(B) One car sale gives no information about when the next sale will happen.
(C) Events must occur systematically at exact 2-minute intervals.
(D) The occurrence of a horse-kick death does not influence the probability of another.

A A,B,D

B A,B

C A,C,D

D A,B,C,D

💡 Short Explanation

�� Events must be independent �� Constant rate assumption applies �� Regular intervals are NOT required

✅ Correct Answer Explanation

A: independence holds → B: no influence of past events → C: probability proportional to interval length → D is false because Poisson is not fixed-interval deterministic

❓ Why Other Options Are Incorrect

�� Option D → includes incorrect condition (fixed intervals)

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Identify true Poisson assumptions

Final Logic: Only independence + constant rate conditions apply

🧠 Memory Trick

"Poisson = Independent + Constant rate, not fixed timing"

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Poisson Conditions –

4 Identify the INCORRECT condition regarding the mean rate in a Poisson distribution framework.

A The rate of occurrence must not change through time.

B The expected mean E(X) must be equal to the variance Var(X).

C The mean rate is derived from an infinite geometric series ratio.

D The mean rate is constant and not based on time.

💡 Short Explanation

�� Poisson rate is constant λ �� Mean equals variance �� No geometric series involved

✅ Correct Answer Explanation

Poisson uses constant rate λ, independent of time. → Mean = variance = λ is valid. → C is incorrect as Poisson has no relation to geometric series.

❓ Why Other Options Are Incorrect

�� Option A → correct property
�� Option B → correct identity
�� Option D → correct constancy condition

🎯 Strategy

Used: Extreme Word Filter

Application: Eliminate mathematically unrelated statement

Final Logic: Geometric series is irrelevant to Poisson

🧠 Memory Trick

"Poisson = λ, not series"

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Poisson Mean Rate –

5 If λ = 3, what is P(X=0)?

A 0.15

B 0.05

C 0.00

D 1.00

💡 Short Explanation

�� P(0) = e^-λ �� e^-3 ≈ 0.05 �� Direct substitution

✅ Correct Answer Explanation

P(X=0) = (3^0 e^-3)/0! = e^-3 ≈ 0.05 → Matches option B exactly

❓ Why Other Options Are Incorrect

�� A → incorrect approximation
�� C → impossible value
�� D → only when λ=0

🎯 Strategy

Used: Substitution

Application: Direct formula substitution

Final Logic: e^-3 evaluation

🧠 Memory Trick

"Zero events = e^-λ"

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Poisson Formula –

6 What is the probability of exactly one truck when λ = 1.6?

A 1.6 * e^-1.6

B e^-1.6

C (1.6)^2 * e^-1.6

D 0

💡 Short Explanation

Use P(X=1) formula
λ^1 e^-λ / 1!
Simplifies to λe^-λ

✅ Correct Answer Explanation

P(1) = (1.6^1 e^-1.6)/1 = 1.6 e^-1.6
Matches option A

❓ Why Other Options Are Incorrect

B → missing λ factor
C → corresponds to k=2
D → invalid

🎯 Strategy

Formula application

Application: Apply P(X=k) structure

Final Logic: k=1 substitution

🧠 Memory Trick

“1 event = λe^-λ”

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Poisson Probability –

7 If λ = 0.61, what is the variance?

A 0.00

B 0.37

C 0.61

D 0.78

💡 Short Explanation

�� Poisson variance = λ �� Direct equality �� No computation needed

✅ Correct Answer Explanation

Var(X) = λ = 0.61 → Hence option C is correct

❓ Why Other Options Are Incorrect

�� A → incorrect zero assumption
�� B → incorrect scaling
�� D → incorrect value

🎯 Strategy

Used: Direct property recall

Application: Use Poisson identity

Final Logic: variance equals mean

🧠 Memory Trick

"Poisson twins: mean = variance"

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Poisson Parameters –

8 If λ = 3.2, what is the standard deviation?

A 3.2

B √3.2

C (3.2)^2

D 1.6

💡 Short Explanation

�� SD = √variance �� variance = λ �� So SD = √λ

✅ Correct Answer Explanation

σ = √λ = √3.2 → Option B matches

❓ Why Other Options Are Incorrect

�� A → variance, not SD
�� C → incorrect squaring
�� D → unrelated

🎯 Strategy

Used: Formula mapping

Application: SD identity

Final Logic: √λ rule

🧠 Memory Trick

"Poisson spread = root λ"

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Standard Deviation –

9 When transitioning Binomial to Poisson, what happens to p?

A p → 1

B p = 0.5

C p becomes smaller

D p equals number of trials

💡 Short Explanation

�� Rare events assumption �� n → ∞, p → 0 �� λ fixed

✅ Correct Answer Explanation

Poisson limit requires p → 0 while n → ∞ → Ensures rare event modeling

❓ Why Other Options Are Incorrect

�� A → opposite condition
�� B → irrelevant
�� D → meaningless

🎯 Strategy

Used: Extreme condition filter

Application: Identify limit behavior

Final Logic: rare event condition

🧠 Memory Trick

"Poisson = many trials, tiny probability"

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Binomial Approximation –

10 What must np approach in Poisson approximation?

A Constant

B Zero

C Infinity

D One

💡 Short Explanation

�� λ = np fixed �� Ensures stability �� Poisson limit condition

✅ Correct Answer Explanation

In Poisson approximation, n→∞, p→0 but np = λ remains constant → Hence A is correct

❓ Why Other Options Are Incorrect

�� B → destroys distribution
�� C → divergence
�� D → unnecessary constraint

🎯 Strategy

Used: Dimensional/Limit analysis

Application: Apply limiting condition

Final Logic: λ stability condition

🧠 Memory Trick

"np = λ (always steady)"

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Poisson Limit Theorem –

11 An ice-cream parlour receives customers at an average rate of 4 per minute. What is the correct expression to find the probability of receiving exactly 16 customers in a 4-minute area block?

A (4^16 * e^-4) / 16!

B (16^16 * e^-16) / 16!

C (16^4 * e^-16) / 4!

D (4^4 * e^-16) / 4!

💡 Short Explanation

�� Poisson uses P(X=k) = (λ^k e^-λ)/k! �� λ scales with time interval �� 4/min × 4 min = 16

✅ Correct Answer Explanation

Rate = 4 customers/minute → Over 4 minutes: λ = 4 × 4 = 16 → For k = 16: P(X=16) = (16^16 e^-16)/16! → Matches option B

❓ Why Other Options Are Incorrect

�� Option A → uses wrong λ (4 instead of 16)
�� Option C → swaps λ and k incorrectly
�� Option D → wrong factorial and exponent structure

🎯 Strategy

Used: Substitution

Application: Compute λ first, then apply formula

Final Logic: Correct scaling gives λ = 16

🧠 Memory Trick

"Rate × time = λ first, then Poisson formula"

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Poisson Application –

12 Integrating traffic flow where λ = 3.2 bicycles/hour, what is the probability that exactly 2 bicycle riders use the track in an hour?

A (10.24 * 0.041) / 2

B (3.2 * 0.041) / 2

C (10.24 * 0.041)

D 0.041 / 2

💡 Short Explanation

�� Use Poisson formula for k=2 �� λ^2 = 3.2^2 = 10.24 �� Divide by 2! = 2

✅ Correct Answer Explanation

P(X=2) = (λ^2 e^-λ)/2! → = (3.2^2 × e^-3.2)/2 → Given e^-3.2 ≈ 0.041 → = (10.24 × 0.041)/2

❓ Why Other Options Are Incorrect

�� Option B → uses λ instead of λ²
�� Option C → misses factorial division
�� Option D → incorrect structure

🎯 Strategy

Used: Formula substitution

Application: Apply k=2 Poisson formula

Final Logic: λ² and divide by 2!

🧠 Memory Trick

"Square for 2 events, divide by 2"

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Poisson Probability –

📄 Passage

Let us revisit the Prussian cavalry historical data to test the adequacy of the Poisson distribution model. The data provided 200 total observations over 20 years. The number of deaths 'k' (0, 1, 2, 3, 4) had corresponding frequencies (109, 65, 22, 3, 1). First, the mean expectation λ was calculated as the weighted average: (0×109 + 1×65 + 2×22 + 3×3 + 4×1) / 200 = 122 / 200 = 0.61. Once λ is established, the Poisson formula P(X=k) = (0.61^k * e^-0.61) / k! is applied. Multiplying these probabilities by the 200 observations yields the Poisson predicted deaths. For instance, P(X=0) was calculated as 0.54, leading to 200 × 0.54 = 108 predicted deaths, which closely matches the 109 actual observed deaths.

Based on the passage, what is the calculated value of the parameter λ?

A 0.54

B 0.61

C 1.22

D 2.00

💡 Short Explanation

�� λ = mean of distribution �� Given calculation provided �� Weighted average result

✅ Correct Answer Explanation

λ = (0×109 + 1×65 + 2×22 + 3×3 + 4×1)/200 → = 122/200 = 0.61 → Hence option B

❓ Why Other Options Are Incorrect

�� Option A → misinterpreted P(0)
�� Option C → doubled value
�� Option D → incorrect scaling

🎯 Strategy

Used: Substitution

Application: Direct passage computation

Final Logic: Weighted mean gives λ

🧠 Memory Trick

"λ = total events ÷ total observations"

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Poisson Parameter Estimation –

📄 Passage

How is the Poisson predicted death count for k=0 calculated?

A By squaring the mean lambda.

B By dividing total observations by frequency.

C By multiplying total observations (200) by P(X=0).

D By integrating standard deviation.

💡 Short Explanation

�� Convert probability to frequency �� Multiply by total observations �� Compare with actual data

✅ Correct Answer Explanation

Expected frequency = N × P(X=k) → Here N = 200 → So predicted deaths = 200 × P(X=0) → Matches option C

❓ Why Other Options Are Incorrect

�� Option A → irrelevant
�� Option B → incorrect method
�� Option D → unrelated concept

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Identify frequency conversion step

Final Logic: Probability → expected count

🧠 Memory Trick

"Probability × total = expected frequency"

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Poisson Application –

15 If λ = 3.2, probability of 2 or less riders requires:

A P(X=0) + P(X=1)

B P(X=0) + P(X=1) + P(X=2)

C 1 - P(X=2)

D P(X=1) + P(X=2) + P(X=3)

💡 Short Explanation

�� "At most 2" means ≤2 �� Sum all probabilities up to 2 �� Include 0,1,2

✅ Correct Answer Explanation

P(X ≤ 2) = P(0)+P(1)+P(2) → Direct cumulative probability definition → Hence option B

❓ Why Other Options Are Incorrect

�� Option A → misses P(2)
�� Option C → incorrect complement
�� Option D → includes unnecessary values

🎯 Strategy

Used: Extreme Word Filter

Application: Interpret "at most" correctly

Final Logic: cumulative sum up to 2

🧠 Memory Trick

"≤ means sum from 0 upward"

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Cumulative Poisson Probability –

16 What is probability of exactly 3 floods if λ = 2?

A (2^3 * e^-2) / 3!

B (3^2 * e^-3) / 2!

C (8 * e^-3) / 6

D (2^3 * e^-3) / 3!

💡 Short Explanation

�� Use Poisson formula �� k = 3 �� λ = 2

✅ Correct Answer Explanation

P(X=3) = (2^3 e^-2)/3! → Direct substitution gives option A

❓ Why Other Options Are Incorrect

�� Option B → swaps λ and k
�� Option C → wrong exponent
�� Option D → incorrect exponential term

🎯 Strategy

Used: Substitution

Application: Apply formula directly

Final Logic: k=3, λ=2 substitution

🧠 Memory Trick

"Power = k, base = λ"

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Poisson Formula –

17 If 30 passengers/hour, expected rate for last 50 minutes?

A 5

B 20

C 25

D 30

💡 Short Explanation

�� Scale rate with time �� 50 min = 5/6 hour �� λ = 30 × 5/6

✅ Correct Answer Explanation

λ = 30 × (50/60) → = 30 × 5/6 = 25 → Hence option C

❓ Why Other Options Are Incorrect

�� Option A → underestimation
�� Option B → wrong fraction
�� Option D → full hour value

🎯 Strategy

Used: Dimensional/Unit Analysis

Application: Time scaling

Final Logic: proportional reduction

🧠 Memory Trick

"Multiply rate by time fraction"

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Poisson Rate Scaling –

18 Minimum arrivals in remaining 50 minutes if total ≥10?

A 0

B 2

C 8

D 10

💡 Short Explanation

�� Total requirement ≥10 �� First 10 min already 8 �� Need at least 2 more

✅ Correct Answer Explanation

Required total ≥10 → Already 8 in first interval → Remaining needed = 2 → Hence option B

❓ Why Other Options Are Incorrect

�� Option A → insufficient total
�� Option C → exceeds requirement
�� Option D → unnecessary

🎯 Strategy

Used: Subtraction logic

Application: Remaining requirement calculation

Final Logic: 10 − 8 = 2

🧠 Memory Trick

"Total minus done = remaining"

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Poisson Application –

19 Probability of zero errors if λ = 0.2

A e^-0.2

B 0.2 e^-0.2

C 1 - e^-0.2

D e^0.2

💡 Short Explanation

�� P(0) = e^-λ �� λ = 0.2 �� Direct substitution

✅ Correct Answer Explanation

P(0) = e^-0.2 → Standard Poisson result

❓ Why Other Options Are Incorrect

�� Option B → corresponds to k=1
�� Option C → complement error
�� Option D → wrong sign

🎯 Strategy

Used: Formula recall

Application: k=0 shortcut

Final Logic: P(0)=e^-λ

🧠 Memory Trick

"Zero event = exponential decay"

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Poisson Special Case –

20 Variance when average calls = 4.5 per 5 minutes?

A 2.25

B 4.5

C 9.0

D √4.5

💡 Short Explanation

�� Poisson variance = λ �� λ = mean rate �� Direct equality

✅ Correct Answer Explanation

Variance = λ = 4.5 → Poisson property: mean = variance

❓ Why Other Options Are Incorrect

�� Option A → half value incorrect
�� Option C → squared incorrect
�� Option D → SD confusion

🎯 Strategy

Used: Direct property recall

Application: mean-variance identity

Final Logic: variance equals mean

🧠 Memory Trick

"Poisson twin rule: mean = variance"

📚 Class XI – NCERT Mathematics Part II – Chapter: Probability Distributions – Topic: Poisson Parameters –

CUET UG Applied Mathematics Booster Test 2 - Poisson Distribution

Test Complete!

Answer Review