CUET UG Applied Mathematics Booster Test 3 - Random Variables and Foundations -

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CUET UG Applied Mathematics Booster Test 3 - Random Variables and Foundations

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Applied Mathematics Booster Test 3 - Random Variables and Foundations

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QUESTION 1 OF 20

A particular river near a small-town floods and overflows twice in every 50-years on an average. If the sample space of flood arrivals is modeled via a Poisson distribution, what is the mean expectation λ for this interval?

Answer Review

1 A particular river near a small-town floods and overflows twice in every 50-years on an average. If the sample space of flood arrivals is modeled via a Poisson distribution, what is the mean expectation λ for this interval?

A 0

B 1

C 2

D 50

💡 Short Explanation

In a Poisson distribution, λ represents the expected number of events in the given interval. The interval considered is 50 years. Since the river floods twice in every 50 years, λ = 2.

✅ Correct Answer Explanation

A Poisson distribution models the number of times an event occurs in a fixed interval of time or space. The parameter λ (lambda) denotes the average or expected number of occurrences during that interval. Here, the given interval is 50 years, and the river floods twice on average during this period. Therefore, λ=2 Hence, the correct answer is Option C. Option A (0): Incorrect because floods do occur. Option B (1): Incorrect since the average is two floods, not one. Option C (2): Correct because λ equals the average number of floods in the specified interval. Option D (50): Incorrect because 50 represents the length of the interval, not the expected number of events.

❓ Why Other Options Are Incorrect

Option A) 0 → Implies no floods occur, contradicting the given information.
Option B) 1 → Underestimates the average number of floods.
Option D) 50 → Represents the duration of the interval, not the Poisson mean.

🎯 Strategy

Used

Substitution

Application: Directly identify the average number of events occurring in the stated interval, since this equals the Poisson parameter λ.

Final Logic: The river floods twice in 50 years, so the Poisson mean is λ = 2, making Option C correct.

🧠 Memory Trick

"Poisson λ = Average events in the given interval."

📚 Class XI – Mathematics – Chapter: Probability – Topic: Poisson Distribution (Mean Parameter λ)

2 Match the Z-score boundaries with their corresponding probabilities of randomly selecting a score in a standard normal distribution.

List I	List II
1. Between -1 and +1	a. 95.45%
2. Between -2 and +2	b. 68.27%
3. Between -3 and +3	c. 34.135%
4. Mean to +1	d. 99.73%

A 1-b, 2-a, 3-d, 4-c

B 1-a, 2-b, 3-c, 4-d

C 1-d, 2-a, 3-b, 4-c

D 1-b, 2-d, 3-a, 4-c

💡 Short Explanation

�� The standard normal distribution follows the empirical (68–95–99.7) rule. �� Larger Z-score intervals cover greater probabilities. �� The area from the mean to +1σ is half of the area between −1σ and +1σ.

✅ Correct Answer Explanation

In a standard normal distribution, the empirical rule states: • Between -1σ and +1σ → 68.27% • Between -2σ and +2σ → 95.45% • Between -3σ and +3σ → 99.73% → Since the normal distribution is symmetric, Area from Mean to +1σ = 68.27% ÷ 2 = 34.135% Therefore, the correct matching is: List I — List II 1. Between -1 and +1 — b. 68.27% 2. Between -2 and +2 — a. 95.45% 3. Between -3 and +3 — d. 99.73% 4. Mean to +1 — c. 34.135% Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Incorrect because it interchanges the probabilities corresponding to ±1σ, ±2σ, and ±3σ.
�� Option C → Incorrect because it wrongly assigns 99.73% to the interval between −1σ and +1σ.
�� Option D → Incorrect because it swaps the probabilities for ±2σ and ±3σ.

🎯 Strategy

Used

Option Grouping

Application:

�� Recall the empirical rule (68–95–99.7) and match each Z-score interval with its corresponding probability.

Final Logic:

�� Only Option A correctly matches all four Z-score intervals with their probabilities.

🧠 Memory Trick

"1σ → 68%, 2σ → 95%, 3σ → 99.7%"

📚 Class XII – NCERT Mathematics – Chapter: Probability – Topic: Standard Normal Distribution (Empirical Rule)

3 Which of the following features are mathematically correct for a Normal Distribution curve?
(A) The total area below the curve is always equal to 1 unit.
(B) The curve has two tails that cross the horizontal axis.
(C) The curve is symmetrical about x = μ.
(D) The mean, median, and mode are exactly the same.
Options format:

A A,B,D

B A,B

C A,C,D

D A,B,C,D

💡 Short Explanation

�� Normal curve is symmetric about mean �� Total probability area equals 1 �� Mean, median, and mode coincide

✅ Correct Answer Explanation

For a normal distribution: Statement A is correct because total probability area under the curve equals 1. Statement C is correct because the curve is symmetric about x=μ. Statement D is correct because mean = median = mode in a normal distribution. Statement B is incorrect because the tails approach the horizontal axis asymptotically but never cross it. Hence, the correct combination is A, C, and D.

❓ Why Other Options Are Incorrect

�� Option A → Includes Statement B, which is false.
�� Option B → Omits correct properties C and D.
�� Option D → Incorrect because tails never intersect the x-axis.

🎯 Strategy

Used

�� Elimination

Application:

�� Remove options containing Statement B since asymptotic tails never touch the axis.

Final Logic:

�� Only Option C contains all valid properties.

🧠 Memory Trick

"Normal Curve: Symmetric and Smooth."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Properties of Normal Distribution –

4 Identify the INCORRECT statement regarding the limiting conversion of a Binomial distribution to a Poisson distribution model.

A The number of trials n approaches infinity.

B The probability of success p becomes very large.

C The mean E(X) = np is kept constant.

D The probability of more than one arrival during an infinitesimally small time interval is 0.

💡 Short Explanation

�� Poisson distribution is a limit of binomial distribution �� n→∞and p→0 �� Mean np remains finite and constant

✅ Correct Answer Explanation

For the Poisson approximation to binomial distribution: n→∞,p→0,np=λ constant Thus: A is correct. C is correct. D is a standard Poisson process property. Statement B is incorrect because p must become very small, not very large. Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → This is a valid Poisson approximation condition.
�� Option C → Constant mean np is essential in Poisson modeling.
�� Option D → Multiple arrivals in a tiny interval are negligible in Poisson processes.

🎯 Strategy

Used

�� Extreme Word Filter

Application:

�� "Very large" contradicts the Poisson condition where probability must be very small.

Final Logic:

�� Poisson limit requires rare events, so p→0.

🧠 Memory Trick

"Poisson = Rare Events."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Poisson Distribution –

5 If the mean and variance of a mixed binomial distribution are 4/3 and 8/9 respectively, what is the probability of success p in each trial?

A 1/3

B 2/3

C 1/2

D 1/4

💡 Short Explanation

�� Mean of binomial distribution = np �� Variance = npq �� Divide variance by mean to obtain q

✅ Correct Answer Explanation

For binomial distribution: μ=np,σ^2=npq Given: np=4/3,npq=8/9 Divide variance by mean: q=8/9/4/3=8/9×3/4=2/3 Since: p+q=1p=1-2/3=1/3 Hence, Option A is correct.

❓ Why Other Options Are Incorrect

�� Option B → Represents q, not p.
�� Option C → Does not satisfy the variance relation.
�� Option D → Gives incorrect mean and variance values.

🎯 Strategy

Used

�� Substitution

Application:

�� Use standard binomial formulas and simplify systematically.

Final Logic:

�� Calculations give p=1/3.

🧠 Memory Trick

"Variance ÷ Mean = q."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Binomial Distribution –

6 For a binomial distribution with n Bernoulli trials, probability of success p, and probability of failure q, the constraint region for Variance is defined strictly as:

A Var = np

B Var = √(npq)

C Var = npq

D Var = p+q

💡 Short Explanation

�� Binomial variance depends on success and failure probabilities �� Both p and q influence spread �� Standard formula is npq

✅ Correct Answer Explanation

For a binomial distribution: Variance=npq where: q=1-p Option A gives the mean. Option B gives standard deviation, not variance. Option D equals 1 because p+q=1. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

�� Option A → Represents mean, not variance.
�� Option B → Represents standard deviation.
�� Option D → Since p+q=1, it cannot represent variance.

🎯 Strategy

Used

�� Option Grouping

Application:

�� Distinguish between formulas for mean, variance, and standard deviation.

Final Logic:

�� Variance formula for binomial distribution is npq.

🧠 Memory Trick

"Mean = np, Variance = npq."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Binomial Distribution –

7 (Expected Money/Event Index) A die is thrown again and again until three 5s are obtained. The expected probability calculation of obtaining the third 5 precisely on the seventh throw requires the initial occurrence of how many 5s in the first six throws?

A Exactly three

B Exactly two

C Zero

D Exactly one

💡 Short Explanation

The third success must occur on the final throw. Therefore, the first six throws must contain exactly two successes. This follows the negative binomial distribution.

✅ Correct Answer Explanation

To obtain the third 5 exactly on the seventh throw: The seventh throw must be a 5. Among the first six throws, there must already be exactly two 5s. Any other number of 5s would prevent the third success from occurring on the seventh throw. Thus, Option B is correct. Option A: Three 5s in the first six throws means the third success occurred earlier. Option B: Correct because the third success occurs on throw seven. Option C: Too few successes. Option D: Only one success is insufficient.

❓ Why Other Options Are Incorrect

Option A) Exactly three → Third 5 occurs before the seventh throw.
Option C) Zero → Impossible to obtain the third 5 on the seventh throw.
Option D) Exactly one → Only the second 5 could occur on the seventh throw.

🎯 Strategy

Used

Contextual/Tonal Matching

Application: Interpret the phrase "third 5 precisely on the seventh throw" carefully to determine the required earlier outcomes.

Final Logic: Exactly two 5s must occur in the first six throws so the third appears on throw seven.

🧠 Memory Trick

"Third success last ⇒ Two before."

📚 Class XI – Mathematics – Chapter: Probability – Topic: Negative Binomial Distribution (Extension Concept)

8 (Moving Average) A traffic engineer records that the moving average (mean expectation) is 3.2 trucks approaching an intersection every minute. Using the Poisson distribution, the value of λ for this interval is:

A 1.6

B 3.2

C 6.4

D 0.041

💡 Short Explanation

In a Poisson distribution, λ equals the expected number of events. The moving average is the expected number of arrivals. Therefore, λ equals 3.2.

✅ Correct Answer Explanation

For a Poisson distribution, λ=Expected number of events in the interval The engineer observes an average of 3.2 trucks per minute. Hence, λ=3.2 Therefore, Option B is correct. Option A: Half the required mean. Option B: Correct value of λ. Option C: Double the given mean. Option D: Unrelated value.

❓ Why Other Options Are Incorrect

Option A) 1.6 → Not the given average.
Option C) 6.4 → Incorrectly doubles the mean.
Option D) 0.041 → Not supported by the data.

🎯 Strategy

Used

Substitution

Application: Directly identify λ from the given average arrival rate.

Final Logic: Since λ equals the expected number of arrivals, λ = 3.2.

🧠 Memory Trick

"Poisson Mean = λ."

📚 Class XI – Mathematics – Chapter: Probability – Topic: Poisson Distribution (Extension Concept)

9 Rohit aims to achieve the probability of hitting a target at least once to be 0.99. If his success rate per shot is 1/6, the inequality to find the minimum number of countable trials n is:

A 1 - (5/6)^n ≥ 0.99

B (5/6)^n ≥ 0.99

C 1 - (1/6)^n ≥ 0.99

D (1/6)^n ≥ 0.99

💡 Short Explanation

"At least one success" equals one minus the probability of no success. Failure probability per shot is 5/6. The required inequality follows directly.

✅ Correct Answer Explanation

Success probability per shot: p=1/6 Failure probability: q=5/6 Probability of no success in n shots: (5/6)^n Probability of at least one success: 1-(5/6)^n The requirement is 1-(5/6)^n≥0.99 Hence, Option A is correct. Option A: Correct complement rule. Option B: Represents failure probability. Option C: Incorrect complement. Option D: Represents all successes.

❓ Why Other Options Are Incorrect

Option B) (5/6)^n ≥ 0.99 → Gives probability of no success.
Option C) 1 - (1/6)^n ≥ 0.99 → Incorrect expression.
Option D) (1/6)^n ≥ 0.99 → Probability of all shots being successful.

🎯 Strategy

Used

Elimination

Application: Use the complement rule to identify the only correct probability expression.

Final Logic: At least one success equals 1 − P(no success), giving Option A.

🧠 Memory Trick

"At least one = 1 − None."

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

10 In a Poisson distribution representing distinct values of arrivals, if λ = 30 per hour, the expected number of arrivals vector mapping for a 10-minute interval equates to:

A 30

B 10

C 5

D 3

💡 Short Explanation

The expected number of arrivals is proportional to time. Ten minutes is one-sixth of an hour. Therefore, λ becomes 5.

✅ Correct Answer Explanation

Given, λ = 30 arrivals per hour. Time interval = 10 minutes = 1/6 hour. Therefore, λ_(10 min)=30×1/6=5 Hence, Option C is correct. Option A: Entire hourly expectation. Option B: Incorrect scaling. Option C: Correct proportional expectation. Option D: Incorrect value.

❓ Why Other Options Are Incorrect

Option A) 30 → Applies to one full hour.
Option B) 10 → Incorrect proportional conversion.
Option D) 3 → Underestimates the expected arrivals.

🎯 Strategy

Used

Substitution

Application: Convert the time interval into hours and multiply by the hourly rate.

Final Logic: 30×1/6=5, so Option C is correct.

🧠 Memory Trick

"New λ = Rate × Time."

📚 Class XI – Mathematics – Chapter: Probability – Topic: Poisson Distribution (Extension Concept)

11 From a normal Z-Table, determining the continuous probability area representing P(Z < 1.20) yields approximately:

A 0.5000

B 0.8849

C 0.1151

D 0.9999

💡 Short Explanation

�� Standard normal table gives cumulative probability values �� P(Z<1.20)means area to the left of 1.20 �� Z-table value for 1.20 is approximately 0.8849

✅ Correct Answer Explanation

For a standard normal distribution: P(Z<1.20) represents the cumulative area under the curve to the left of Z=1.20. From the standard normal table: P(Z<1.20)≈0.8849 Thus, Option B is correct. Option A corresponds roughly to the probability at mean Z=0. Option C represents the complementary probability P(Z>1.20). Option D is too close to 1 and corresponds to very large Z-values.

❓ Why Other Options Are Incorrect

�� Option A → 0.5000 is the probability to the left of Z=0, not Z=1.20.
�� Option C → 0.1151 is the right-tail probability since 1-0.8849=0.1151.
�� Option D → This probability is associated with extremely large positive Z-values, not 1.20.

🎯 Strategy

Used

�� Substitution

Application:

�� Substitute the Z-value directly into the standard normal table.

Final Logic:

�� Table lookup for Z=1.20 gives 0.8849.

🧠 Memory Trick

"1.2 → 88% Left Area."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Standard Normal Distribution –

12 The probability density function integral for a standard normal distribution (where μ = 0 and σ = 1) is expressed analytically as f(x) = ?

A (1 / √(2π)) * e^(-x² / 2)

B (1 / (σ√(2π))) * e^(-(x-μ)² / 2)

C (λ^k * e^(-λ)) / k!

D C(n,r) * p^r * q^(n-r)

💡 Short Explanation

�� Standard normal distribution has mean 0 and standard deviation 1 �� Its PDF is a special case of the normal distribution formula �� Formula is symmetric about x=0

✅ Correct Answer Explanation

The general normal distribution density function is: f(x)=1/σ√(2π)e^(-(x-μ)^2/2σ^2) For the standard normal distribution: μ=0,σ=1 Substituting these values: f(x)=1/√(2π)e^(-x^2/2) Hence, Option A is correct. Option B is the general normal distribution formula, not simplified for standard normal form. Option C is the Poisson probability formula. Option D is the Binomial probability formula.

❓ Why Other Options Are Incorrect

�� Option B → Correct general formula, but incomplete because σ^2 is missing in the exponent denominator.
�� Option C → Represents Poisson distribution probability mass function.
�� Option D → Represents Binomial distribution probability expression.

🎯 Strategy

Used

�� Option Grouping

Application:

�� Identify which formulas belong to normal, Poisson, and binomial distributions.

Final Logic:

�� Only Option A is the exact standard normal density function.

🧠 Memory Trick

"Standard Normal → μ=0, σ=1."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Standard Normal Distribution Function –

📄 Passage

Sonal and Anannya are playing a game by throwing a die alternatively till one of them gets a '1' and wins the game. Getting a 1 on a single throw of the die is considered a success. Clearly the trials are Bernoulli's with an infinite possible number of n. The probability of success p is 1/6 and the probability of failure q is 5/6. Sonal starts the game by throwing the die first. Her probability to win in the first throw is 1/6. She will get to try winning next in the third throw; when Sonal fails in the first throw and then Anannya fails to win in the second throw. This creates an infinite geometric series for calculating their final winning probabilities.

Based on the discrete sequence of turns in the passage, what is the probability that Sonal wins exactly on her second turn (which is the third overall throw of the die)?

A 1/6

B (5/6)² × (1/6)

C (5/6) × (1/6)

D (1/6)³

💡 Short Explanation

�� Sonal must fail first �� Anannya must also fail next �� Sonal succeeds on the third throw

✅ Correct Answer Explanation

To win on the third overall throw: 1. Sonal fails first throw → probability 5/6 2. Anannya fails second throw → probability 5/6 3. Sonal succeeds on third throw → probability 1/6 Therefore: (5/6)^2×1/6 Hence, Option B is correct. Option A gives only one successful throw probability. Option C ignores one failure event. Option D incorrectly assumes three successes.

❓ Why Other Options Are Incorrect

�� Option A → Represents winning on the first throw only.
�� Option C → Missing the second failure probability.
�� Option D → Assumes all three throws are successes, which is impossible here.

🎯 Strategy

Used

�� Elimination

Application:

�� Count required failures before the successful event.

Final Logic:

�� Two failures followed by one success gives (, ).

🧠 Memory Trick

"Fail, Fail, Win."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Bernoulli Trials & Geometric Probability –

📄 Passage

Based on the passage, the total overall probability of Sonal winning the entire game equates to evaluating an infinite geometric series S = a / (1-r). What is the final probability that Sonal wins?

A 5/11

B 6/11

C 1/6

D 1

💡 Short Explanation

�� Sonal wins on odd-numbered turns �� Probabilities form an infinite geometric series �� Sum evaluates to 6/11

✅ Correct Answer Explanation

Sonal can win on: 1^(st),3^(rd),5^(th),… Probability series: 1/6+(5/6)^21/6+(5/6)^41/6+⋯ This is a geometric series with: a=1/6,r=(5/6)^2=25/36 Using: S=a/1-r S=1/6/1-25/36=1/6/11/36=6/11 Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Incorrect simplification of geometric series.
�� Option C → Only the probability of first-turn success.
�� Option D → Winning probability cannot be exactly 1 because both players have chances.

🎯 Strategy

Used

�� Substitution

Application:

�� Substitute values into the geometric series sum formula.

Final Logic:

�� Infinite series simplifies exactly to 6/11.

🧠 Memory Trick

"Odd Turns → Geometric Sum."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Geometric Series in Probability –

15 If a normal distribution has a mean traveling time of 38.8 minutes and a standard deviation of 11.4 minutes, what is the calculated standard normal deviate (Z-Score) for a student whose travel time is 65 minutes? (Approximate to two decimal places)

A 1.25

B 2.30

C -1.25

D 3.05

💡 Short Explanation

�� Use Z-score formula �� Compare observation with mean �� Positive value indicates above-average travel time

✅ Correct Answer Explanation

Z-score formula: Z=X-μ/σ x μ σ z=x-μ/σ≈1.2 Φ(z)≈88.5% Substitute values: X=65,μ=38.8,σ=11.4Z=65-38.8/11.4=26.2/11.4≈2.30 Hence, Option B is correct.

❓ Why Other Options Are Incorrect

�� Option A → Underestimates the standardized deviation.
�� Option C → Negative Z-score would imply value below mean.
�� Option D → Overestimates the deviation from mean.

🎯 Strategy

Used

�� Substitution

Application:

�� Substitute numerical values into the Z-score formula directly.

Final Logic:

�� Computation gives approximately 2.30.

🧠 Memory Trick

"Z = Difference ÷ Spread."

📚 Class XII – NCERT Mathematics Class 12 – Chapter: Probability – Topic: Standard Normal Deviate –

16 Given a Z-Score of 5 associated with a standard normal probability curve, if the mean μ is 12 and σ is 4, what was the original data point value?

A 20

B 26

C 32

D 40

💡 Short Explanation

The Z-score formula is Z=X-μ/σ. Rearrange to find the original value: X=μ+Zσ. Substitute the given values to obtain the answer.

✅ Correct Answer Explanation

The Z-score formula is: Z=X-μ/σ Rearranging, X=μ+Zσ Substitute the given values: X=12+(5×4)=12+20=32 Hence, the original data point is 32. Option A (20): Incorrect because it does not satisfy the Z-score formula. Option B (26): Incorrect because it results in a Z-score of 3.5. Option C (32): Correct since it gives a Z-score of 5. Option D (40): Incorrect because it corresponds to a Z-score of 7.

❓ Why Other Options Are Incorrect

Option A) 20 → Gives Z=20-12/4=2, not 5.
Option B) 26 → Gives Z=26-12/4=3.5, not 5.
Option D) 40 → Gives Z=40-12/4=7, not 5.

🎯 Strategy

Used

Substitution

Application: Substitute the given values into the Z-score formula and solve directly.

Final Logic: Since X=μ+Zσ=12+(5×4)=32, the correct answer is Option C.

🧠 Memory Trick

"Original = Mean + (Z × SD)"

📚 Class XI – Mathematics – Chapter: Statistics – Topic: Standard Normal Distribution (Z-Score)

17 In the hypothesis testing of normal distribution sets, ensuring the sample sizes are equal if at all possible and data points are independent is a prerequisite for which specific statistical test mentioned in the text?

A Binomial Limit Test

B Bernoulli Independence Test

C Z-test

D Poisson variance test

💡 Short Explanation

The Z-test compares means under normality assumptions. Independent observations are essential. Equal sample sizes are often preferred for balanced comparisons.

✅ Correct Answer Explanation

The Z-test is commonly Strategy Used for hypothesis testing involving normally distributed data, particularly when the population variance is known or the sample size is sufficiently large. Its assumptions include: Independent observations. Normally distributed population (or large samples). Equal sample sizes are often desirable for balanced comparisons, though not always mandatory. Therefore, Option C is correct. Option A: Not a standard statistical test. Option B: Not a recognized hypothesis-testing procedure. Option C: Correct statistical test. Option D: Not Strategy Used for this purpose.

❓ Why Other Options Are Incorrect

Option A) Binomial Limit Test → Not a standard hypothesis test.
Option B) Bernoulli Independence Test → Not a recognized statistical test.
Option D) Poisson variance test → Not Strategy Used for comparing normal distribution means.

🎯 Strategy

Used

Elimination

Application: Eliminate non-standard statistical tests and identify the recognized hypothesis test.

Final Logic: Only the Z-test is a standard hypothesis test fitting the stated assumptions.

🧠 Memory Trick

"Z = Normal + Large Sample."

📚 Class XI – Mathematics – Chapter: Statistics – Topic: Normal Distribution and Z-Test (Extension Concept)

18 For experimental trials forming a Poisson distribution, one of the existence conditions requires that the infinite sum Σ (λ^k** * e^(-λ)) / k! from k=0 to ∞ evaluates exactly to:**

A 0

B 1

C λ

D e

💡 Short Explanation

A probability distribution must have total probability equal to 1. The Poisson probability mass function satisfies this condition. Hence the infinite sum equals 1.

✅ Correct Answer Explanation

The Poisson distribution is defined as P(X=k)=λ^ke^(-λ)/k! For any probability distribution, ∑_(k=0)^∞(P(X=k)=1) Thus, ∑_(k=0)^∞(λ^ke^(-λ)/k!)=1 Therefore, Option B is correct. Option A: Violates the probability axiom. Option C: λ is the mean, not the total probability. Option D: e is a mathematical constant, not the required sum.

❓ Why Other Options Are Incorrect

Option A) 0 → Total probability can never be zero.
Option C) λ → Represents the mean parameter only.
Option D) e → Not the normalization value.

🎯 Strategy

Used

Conceptual Recall

Application: Recall that every valid probability distribution sums to one.

Final Logic: Since total probability equals 1, Option B is correct.

🧠 Memory Trick

"Probability always totals ONE."

📚 Class XI – Mathematics – Chapter: Probability – Topic: Probability Distribution (Extension: Poisson Distribution)

19 If the probability of success in a Bernoulli event trial is denoted by p, and probability of failure by q, then what is the expected variance if n such trials are performed?

A np

B √(npq)

C npq

D n/p

💡 Short Explanation

A binomial distribution models repeated Bernoulli trials. Mean = np. Variance = npq.

✅ Correct Answer Explanation

For a binomial random variable with: Number of trials = n Success probability = p Failure probability = q=1-p The variance is npq Hence Option C is correct. Option A: Represents the mean. Option B: Represents the standard deviation. Option C: Represents the variance. Option D: Is not a variance formula.

❓ Why Other Options Are Incorrect

Option A) np → Mean, not variance.
Option B) √(npq) → Standard deviation.
Option D) n/p → Not a valid statistical formula.

🎯 Strategy

Used

Formula Recall

Application: Recall the standard formulas for the binomial distribution.

Final Logic: Variance of a binomial distribution is npq.

🧠 Memory Trick

"Mean = np, Variance = npq."

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

20 Random behavior modeled by standard normal distribution fundamentally scales any normally distributed variable to a universal baseline where the mean μ and standard deviation σ are defined specifically as:

A μ = 1, σ = 0

B μ = 0, σ = 1

C μ = 0, σ = 0

D μ = 100, σ = 10

💡 Short Explanation

A standard normal distribution is obtained through standardization. Its mean is zero. Its standard deviation is one.

✅ Correct Answer Explanation

A standard normal distribution is defined by converting any normal variable using Z=X-μ/σ The transformed distribution always has Mean =0 Standard deviation =1 Therefore, Option B is correct. Option A: Invalid because standard deviation cannot be zero. Option B: Correct definition. Option C: Standard deviation cannot be zero. Option D: Represents another normal distribution, not the standard normal distribution.

❓ Why Other Options Are Incorrect

Option A) μ = 1, σ = 0 → Zero standard deviation is impossible for a standard normal distribution.
Option C) μ = 0, σ = 0 → Standard deviation cannot be zero.
Option D) μ = 100, σ = 10 → Describes a normal distribution, not a standard normal distribution.

🎯 Strategy

Used

Conceptual Recall

Application: Recall the defining characteristics of the standard normal distribution.

Final Logic: The standard normal distribution always has mean = 0 and standard deviation = 1, making Option B correct.

🧠 Memory Trick

"Standard Normal → 0 Mean, 1 SD (0–1 Rule)."

📚 Class XI – Mathematics – Chapter: Statistics – Topic: Standard Normal Distribution

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