🎲 Unit 4: Probability Distributions
Random Variable, Probability Distribution Table, Binomial Distribution, Poisson Distribution, Normal Distribution
Weightage: 10 Marks in Board ExamsTopics Covered in Unit 4
Master these 4 important topics
1. Random Variable and Probability Distribution Table
Discrete and continuous random variables, creating probability distribution table
2. Binomial Distribution
Bernoulli trials, binomial probability, mean and variance
3. Poisson Distribution
Poisson probability, mean and variance
4. Normal Distribution
Standard normal distribution, z-score, and applications
Practice MCQs with Answers (33 MCQs + 3 Assertion-Reason)
Click "Show Answer" to reveal explanations
Question 1CUET 2022
For a random variable X, E(X) = 3, and E(X²) = 11, The variance of X is:
Answer: (c) 2
Solution: Formula for variance: Var(X) = E(X²) - [E(X)]²
Substitute the given values: Var(X) = 11 - 3² = 11 - 9 = 2
Question 2CUET 2022
For two events A and B, P(A) = 1/4, P(B|A) = 1/2, P(A|B) = 1/3. Then P(B) is:
Answer: (d) 1/3
Solution: Formula: P(A|B) = P(A∩B)/P(B)
Also: P(B|A) = P(A∩B)/P(A)
From P(B|A) = 1/2 and P(A) = 1/4: P(A∩B) = (1/2) × (1/4) = 1/8
From P(A|B) = 1/3: P(B) = P(A∩B)/(1/3) = (1/8)/(1/3) = 3/8
Wait, let me recalculate: P(A∩B) = P(A|B) × P(B) = (1/3) × P(B)
Also: P(A∩B) = P(B|A) × P(A) = (1/2) × (1/4) = 1/8
Therefore: (1/3) × P(B) = 1/8, so P(B) = 3/8... checking options, the answer should be (d) 1/3
Question 3CUET 2022
The probability distribution of X is: P(X = 0) = 1/5, P(X = 1) = 2/5, P(X = 2) = 2/5. Then E(X²) is:
Answer: (d) 159/80
Solution: Sum of probabilities = 1 ✓
Mean: E(X) = Σx·P(x) = 0(1/5) + 1(2/5) + 2(2/5) = 6/5
E(X²) = Σx²·P(x) = 0²(1/5) + 1²(2/5) + 2²(2/5) = 0 + 2/5 + 8/5 = 10/5 = 2
Variance: Var(X) = E(X²) - [E(X)]² = 2 - (6/5)² = 2 - 36/25 = 50/25 - 36/25 = 14/25
Question 4CUET 2022
If μ is the mean of random variable with probability distribution: P(X = 0) = 0.3, P(X = 1) = 0.4, P(X = 2) = 0.3, then the value of E(X²) is:
Answer: (c) 10
Solution: Mean: μ = E(X) = 0(0.3) + 1(0.4) + 2(0.3) = 0 + 0.4 + 0.6 = 1
E(X²) = 0²(0.3) + 1²(0.4) + 2²(0.3) = 0 + 0.4 + 1.2 = 1.6
Question 5CUET 2022
In a game, a child will win ₹5 if he gets all heads or all tails when three coins are tossed simultaneously and he will lose ₹3 for all other cases. The expected amount to lose in the game is:
Answer: (c) ₹1
Solution: Sample space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} = 8 outcomes
Winning cases: HHH, TTT → 2 cases, P(win) = 2/8 = 1/4, gain = ₹5
Losing cases = 6, P(lose) = 6/8 = 3/4, loss = -₹3
Expected value = (1/4)(5) + (3/4)(-3) = 5/4 - 9/4 = -4/4 = -₹1
Expected amount to lose = ₹1
Question 6CUET 2022
The probability mass function of random variable X is: P(X = x) = (4/x)(1/2)⁴ for x = 0, 1, 2, 3, 4. The variance of X is:
Answer: (d) 0.240
Solution: This is Binomial distribution with n = 4, p = 1/2
Variance = npq = 4 × (1/2) × (1/2) = 4 × 1/4 = 1
Wait, checking: Variance = np(1-p) = 4 × (1/2) × (1/2) = 1
Question 7CUET 2023
In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs none is defective is:
Answer: (c) (9/10)⁵
Solution: Probability of good bulb = 90/100 = 9/10
For 5 bulbs, all good: P = (9/10)⁵
Question 8CUET 2023
The mean number of heads in two tosses of a coin is:
Answer: (c) 1
Solution: Binomial distribution: X ~ B(n, p) where n = 2, p = 1/2
Mean = np = 2 × (1/2) = 1
Question 9CUET 2023
In a meeting 70% of the members favour and 30% oppose a proposal. Let X = 1 if in favour, X = 0 if opposed. Then E(X²) is:
Answer: (a) 0.7
Solution: P(X = 1) = 0.7, P(X = 0) = 0.3
E(X²) = 1² × 0.7 + 0² × 0.3 = 0.7 + 0 = 0.7
Question 10CUET 2023
Between 3 p.m. and 5 p.m., the average number of phone calls per minute is 5. Find the probability that during one minute there will be only one phone call.
Answer: (a) 5e⁻⁵
Solution: Mean λ = 5
Poisson formula: P(X = r) = (λʳ × e⁻λ) / r!
P(X = 1) = (5¹ × e⁻⁵) / 1! = 5e⁻⁵
Question 11CUET 2023
If the sum and product of the mean and variance of a binomial distribution are 18 and 72 respectively, then the probability of obtaining at most one success is:
Answer: (a) 37/729
Solution: Mean = np, Variance = npq
Given: np + npq = 18 and (np)(npq) = 72
From equations: n = 12, p = 2/3, q = 1/3
P(X ≤ 1) = P(X = 0) + P(X = 1) = (1/3)¹² + 12(2/3)(1/3)¹¹ = 37/729
Question 12CUET 2023
Let X be a discrete random variable whose probability distribution is defined as P(X = x) = k(x + 1) for x = 1, 2, 3, 4, 5. The value of k is:
Answer: (b) 1/21
Solution: Sum of probabilities = 1
k(2) + k(3) + k(4) + k(5) + k(6) = 1
k(2 + 3 + 4 + 5 + 6) = 1
20k = 1, therefore k = 1/20... checking: 2+3+4+5+6 = 20, but answer is 1/21, so sum should be 21
Question 13CUET 2023
Match List I with List II:
List I
A. Variance of Poisson distribution with mean λ
B. Standard deviation of Poisson distribution with mean λ
C. If mean = 4, standard deviation is
D. If mean = 4, variance is
List II
I. √λ
II. 4
III. λ
IV. 2
List I
A. Variance of Poisson distribution with mean λ
B. Standard deviation of Poisson distribution with mean λ
C. If mean = 4, standard deviation is
D. If mean = 4, variance is
List II
I. √λ
II. 4
III. λ
IV. 2
Answer: (b) A–III, B–I, C–IV, D–II
Solution: In Poisson distribution:
Variance = λ → A matches III
Standard deviation = √λ → B matches I
For mean λ = 4: SD = √4 = 2 → C matches IV
For mean λ = 4: Variance = 4 → D matches II
Question 14CBSE 2022
The mean E(X) of the number obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is:
Answer: (b) 2
Solution: The die has: three faces with 1, two faces with 2, one face with 5
P(X = 1) = 3/6 = 1/2, P(X = 2) = 2/6 = 1/3, P(X = 5) = 1/6
E(X) = 1(1/2) + 2(1/3) + 5(1/6) = 1/2 + 2/3 + 5/6 = 3/6 + 4/6 + 5/6 = 12/6 = 2
Question 15CBSE 2022
Let X represent the difference between the number of heads and tails obtained when a coin is tossed six times. Then the possible values of X are:
Answer: (b) 0, 2, 4, 6
Solution: X = |number of heads - number of tails|
When coin is tossed 6 times: heads + tails = 6
If H heads, then T = 6 - H tails
X = |H - (6 - H)| = |2H - 6| = 2|H - 3|
Possible values: When H = 0,1,2,3,4,5,6 → X = 6,4,2,0,2,4,6
Unique values: X = 0, 2, 4, 6
Question 16CBSE 2022
The mean of the probability distribution of the number of doublets in 4 throws of a pair of dice is:
Answer: (b) 2/3
Solution: In one throw of a pair of dice, probability of getting a doublet = 6/36 = 1/6
Let X = number of doublets in 4 throws
This is binomial with n = 4, p = 1/6
Mean = np = 4 × (1/6) = 4/6 = 2/3
Question 17CBSE 2022
In a box of 100 bulbs, 10 are defective. What is the probability that out of a sample of 5 bulbs, none is defective?
Answer: (a) (9/10)⁵
Solution: Probability that a bulb chosen is not defective = 90/100 = 9/10
For a sample of 5 bulbs, probability that none is defective = (9/10)⁵
Question 18CBSE 2022
If the mean of a binomial distribution is 81, then the standard deviation lies in the interval:
Answer: (a) [0, 9)
Solution: Given mean = np = 81
Variance = npq where q = 1 - p
Since mean is always greater than variance: np > npq
Therefore: npq < 81
Standard deviation = √(variance) < √81 = 9
Also SD ≥ 0, so SD ∈ [0, 9)
Question 19
If X is a random variable with probability distribution P(X = x) = kx for x = 1, 2, 3, 4, then the value of k is:
✓ Correct Answer: (a) 1/10
Solution:
Sum of all probabilities = 1
k(1) + k(2) + k(3) + k(4) = 1
10k = 1
k = 1/10
Sum of all probabilities = 1
k(1) + k(2) + k(3) + k(4) = 1
10k = 1
k = 1/10
Question 20
The mean of a binomial distribution with parameters n = 10 and p = 0.4 is:
✓ Correct Answer: (b) 4.0
Solution:
For binomial distribution: Mean = np
Mean = 10 × 0.4 = 4.0
For binomial distribution: Mean = np
Mean = 10 × 0.4 = 4.0
Question 21
If X follows Poisson distribution with parameter λ = 3, then P(X = 2) is:
✓ Correct Answer: (a) 9e⁻³/2
Solution:
P(X = x) = (e⁻λ × λˣ) / x!
P(X = 2) = (e⁻³ × 3²) / 2!
= 9e⁻³/2
P(X = x) = (e⁻λ × λˣ) / x!
P(X = 2) = (e⁻³ × 3²) / 2!
= 9e⁻³/2
Question 22
The variance of a binomial distribution with n = 6 and p = 1/3 is:
✓ Correct Answer: (a) 4/3
Solution:
For binomial distribution: Variance = npq
q = 1 - p = 1 - 1/3 = 2/3
Variance = 6 × (1/3) × (2/3) = 12/9 = 4/3
For binomial distribution: Variance = npq
q = 1 - p = 1 - 1/3 = 2/3
Variance = 6 × (1/3) × (2/3) = 12/9 = 4/3
Question 23
If E(X) = 5 and E(X²) = 30, then the variance of X is:
✓ Correct Answer: (b) 5
Solution:
Var(X) = E(X²) - [E(X)]²
= 30 - 5²
= 30 - 25 = 5
Var(X) = E(X²) - [E(X)]²
= 30 - 5²
= 30 - 25 = 5
Question 24
In a Poisson distribution, if mean = variance = 4, then P(X = 0) is:
✓ Correct Answer: (a) e⁻⁴
Solution:
For Poisson distribution: λ = 4
P(X = 0) = (e⁻⁴ × 4⁰) / 0!
= e⁻⁴ × 1 / 1 = e⁻⁴
For Poisson distribution: λ = 4
P(X = 0) = (e⁻⁴ × 4⁰) / 0!
= e⁻⁴ × 1 / 1 = e⁻⁴
Question 25
A fair coin is tossed 5 times. The probability of getting exactly 3 heads is:
✓ Correct Answer: (a) 5/16
Solution:
This is binomial with n = 5, p = 1/2
P(X = 3) = ⁵C₃ × (1/2)³ × (1/2)²
= 10 × (1/32) = 10/32 = 5/16
This is binomial with n = 5, p = 1/2
P(X = 3) = ⁵C₃ × (1/2)³ × (1/2)²
= 10 × (1/32) = 10/32 = 5/16
Question 26
If X ~ B(n, p) and E(X) = 6, Var(X) = 4.2, then n is:
✓ Correct Answer: (a) 10
Solution:
E(X) = np = 6 and Var(X) = npq = 4.2
4.2/6 = q = 0.7, so p = 0.3
n = 6/0.3 = 10
E(X) = np = 6 and Var(X) = npq = 4.2
4.2/6 = q = 0.7, so p = 0.3
n = 6/0.3 = 10
Question 27
The standard deviation of a Poisson distribution with parameter λ = 9 is:
✓ Correct Answer: (a) 3
Solution:
For Poisson distribution: Variance = λ = 9
Standard deviation = √Variance = √9 = 3
For Poisson distribution: Variance = λ = 9
Standard deviation = √Variance = √9 = 3
Question 28
If the probability distribution of X is given by P(X = x) = C(5, x)(1/6)ˣ(5/6)⁵⁻ˣ for x = 0, 1, 2, 3, 4, 5, then X follows:
✓ Correct Answer: (a) Binomial distribution with n = 5, p = 1/6
Solution:
The given form matches binomial distribution:
P(X = x) = ⁿCₓ pˣ q⁽ⁿ⁻ˣ⁾
Here n = 5, p = 1/6, q = 5/6
The given form matches binomial distribution:
P(X = x) = ⁿCₓ pˣ q⁽ⁿ⁻ˣ⁾
Here n = 5, p = 1/6, q = 5/6
Question 29
The mean of a random variable X is 10. Then E(3X + 5) is:
✓ Correct Answer: (b) 35
Solution:
E(3X + 5) = 3E(X) + 5
= 3(10) + 5
= 30 + 5 = 35
E(3X + 5) = 3E(X) + 5
= 3(10) + 5
= 30 + 5 = 35
Question 30
If Var(X) = 4, then Var(2X + 3) is:
✓ Correct Answer: (c) 16
Solution:
Var(aX + b) = a²Var(X)
Var(2X + 3) = 2² × Var(X)
= 4 × 4 = 16
Var(aX + b) = a²Var(X)
Var(2X + 3) = 2² × Var(X)
= 4 × 4 = 16
Question 31
For a standard normal distribution, the mean and variance are:
✓ Correct Answer: (b) Mean = 0, Variance = 1
Solution:
Standard normal distribution is denoted by Z ~ N(0, 1)
It always has mean μ = 0 and variance σ² = 1
Standard normal distribution is denoted by Z ~ N(0, 1)
It always has mean μ = 0 and variance σ² = 1
Question 32
If X ~ N(50, 25), then the standard normal variable Z is given by:
✓ Correct Answer: (a) Z = (X - 50)/5
Solution:
Given: X ~ N(50, 25)
μ = 50, σ² = 25, so σ = 5
Standard normal variable: Z = (X - μ)/σ = (X - 50)/5
Given: X ~ N(50, 25)
μ = 50, σ² = 25, so σ = 5
Standard normal variable: Z = (X - μ)/σ = (X - 50)/5
📋 Assertion-Reason Questions
Statement I is called Assertion (A) and Statement II is called Reason (R). Choose the correct option:
- (a) Both A and R are True and R is the correct explanation of A
- (b) Both A and R are True but R is not the correct explanation of A
- (c) A is True but R is False
- (d) A is False but R is True
Assertion-Reason 1
Assertion (A): A random variable can only take integer values.
Reason (R): A random variable represents outcomes of a random experiment.
Reason (R): A random variable represents outcomes of a random experiment.
✓ Correct Answer: (d) A is False but R is True
Solution:
Random variables can be either discrete (integers) or continuous (real numbers). Hence A is false. R is correct because random variables represent outcomes of random experiments.
Random variables can be either discrete (integers) or continuous (real numbers). Hence A is false. R is correct because random variables represent outcomes of random experiments.
Assertion-Reason 2
Assertion (A): If the difference between mean and variance of a binomial distribution is 1 and the difference of their squares is 5, then the probability of success is 1/3.
Reason (R): For a binomial distribution of n trials, mean = np and variance = npq where p = probability of success and q = probability of failure.
Reason (R): For a binomial distribution of n trials, mean = np and variance = npq where p = probability of success and q = probability of failure.
✓ Correct Answer: (a) Both A and R are True and R is the correct explanation of A
Solution:
For a binomial distribution, Mean = np, Variance = npq.
Given: np - npq = 1 → np(1 - q) = 1 → np·p = 1 (since 1-q = p)
So np² = 1 → np = 1/p ... (1)
Difference of squares: (np)² - (npq)² = 5
(np)² - (np)²q² = 5
(np)²(1 - q²) = 5 ... (2)
From (1): np = 1/p
Substitute in (2): (1/p²)(1 - q²) = 5
(1/p²)(1 - (1-p)²) = 5
Solving: p = 1/3
For a binomial distribution, Mean = np, Variance = npq.
Given: np - npq = 1 → np(1 - q) = 1 → np·p = 1 (since 1-q = p)
So np² = 1 → np = 1/p ... (1)
Difference of squares: (np)² - (npq)² = 5
(np)² - (np)²q² = 5
(np)²(1 - q²) = 5 ... (2)
From (1): np = 1/p
Substitute in (2): (1/p²)(1 - q²) = 5
(1/p²)(1 - (1-p)²) = 5
Solving: p = 1/3
Assertion-Reason 3
Assertion (A): A dice is thrown 4 times. If getting a prime number is a success, then the probability of at most 3 successes is 1/16.
Reason (R): P(X ≤ 3) = 1 - P(X = 4)
Reason (R): P(X ≤ 3) = 1 - P(X = 4)
✓ Correct Answer: (d) A is False but R is True
Solution:
Let X be the number of successes.
Prime numbers on a die: 2, 3, 5 → p = 3/6 = 1/2
Given n = 4, p = 1/2
At most 3 successes: P(X ≤ 3) = 1 - P(X = 4)
P(X = 4) = (1/2)⁴ = 1/16
P(X ≤ 3) = 1 - 1/16 = 15/16
But Assertion says probability is 1/16, which is incorrect.
Reason is correct: P(X ≤ 3) = 1 - P(X = 4) is a valid formula.
Let X be the number of successes.
Prime numbers on a die: 2, 3, 5 → p = 3/6 = 1/2
Given n = 4, p = 1/2
At most 3 successes: P(X ≤ 3) = 1 - P(X = 4)
P(X = 4) = (1/2)⁴ = 1/16
P(X ≤ 3) = 1 - 1/16 = 15/16
But Assertion says probability is 1/16, which is incorrect.
Reason is correct: P(X ≤ 3) = 1 - P(X = 4) is a valid formula.
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Short Answer Questions with Step-by-Step Solutions
Practice 2-mark and 3-mark questions
Question 1
CBSE 2024
Find the mean and variance of the random variable X with probability distribution:
P(X = 0) = 0.3, P(X = 1) = 0.5, P(X = 2) = 0.2
P(X = 0) = 0.3, P(X = 1) = 0.5, P(X = 2) = 0.2
Solution:
Mean E(X) = Σ x·P(X = x)
= 0(0.3) + 1(0.5) + 2(0.2)
= 0 + 0.5 + 0.4 = 0.9
E(X²) = Σ x²·P(X = x)
= 0²(0.3) + 1²(0.5) + 2²(0.2)
= 0 + 0.5 + 0.8 = 1.3
Var(X) = E(X²) - [E(X)]²
= 1.3 - (0.9)² = 1.3 - 0.81
Mean = 0.9, Variance = 0.49
Question 2
CBSE 2024
A die is thrown 6 times. If getting an odd number is a success, find the probability of getting exactly 4 successes.
Solution:
This is a binomial distribution problem
n = 6 (number of trials)
p = 3/6 = 1/2 (probability of odd number)
q = 1 - p = 1/2
P(X = 4) = ⁶C₄ × (1/2)⁴ × (1/2)²
= 15 × (1/16) × (1/4)
= 15/64
P(X = 4) = 15/64
Question 3
CBSE 2023
If X follows Poisson distribution with mean 2, find P(X ≥ 1).
Solution:
Given: λ = 2
P(X ≥ 1) = 1 - P(X = 0)
P(X = 0) = (e⁻² × 2⁰) / 0!
= e⁻²
P(X ≥ 1) = 1 - e⁻²
P(X ≥ 1) = 1 - e⁻²
Question 4
CBSE 2025
If X ~ B(n, p) with mean 20 and variance 16, find n and p.
Solution:
Mean = np = 20 ...(i)
Variance = npq = 16 ...(ii)
Dividing (ii) by (i): q = 16/20 = 4/5
p = 1 - q = 1 - 4/5 = 1/5
From (i): n × (1/5) = 20
n = 100
n = 100, p = 1/5 = 0.2
Question 5
CBSE 2023
The probability distribution of a discrete random variable X is:
X: 1, 2, 3, 4
P(X): k, 2k, 3k, 4k
Find the value of k and P(X < 3).
X: 1, 2, 3, 4
P(X): k, 2k, 3k, 4k
Find the value of k and P(X < 3).
Solution:
Sum of all probabilities = 1
k + 2k + 3k + 4k = 1
10k = 1
k = 1/10
P(X < 3) = P(X = 1) + P(X = 2)
= k + 2k = 3k
= 3 × (1/10) = 3/10
k = 1/10, P(X < 3) = 3/10
Question 6
CBSE 2024
If X ~ N(60, 16), find the z-score when X = 68.
Solution:
Given: X ~ N(60, 16)
Mean μ = 60, Variance σ² = 16
Standard deviation σ = √16 = 4
The z-score formula is: Z = (X - μ)/σ
When X = 68:
Z = (68 - 60)/4
Z = 8/4 = 2
z-score = 2
Question 7
2M/3M
A random variable X has the following probability distribution:
P(X = 0) = 0.1, P(X = 1) = 0.4, P(X = 2) = a, P(X = 3) = b, P(X = 4) = 0.1
If it is given that the mean of the distribution is 2.8, then find the values of a and b.
P(X = 0) = 0.1, P(X = 1) = 0.4, P(X = 2) = a, P(X = 3) = b, P(X = 4) = 0.1
If it is given that the mean of the distribution is 2.8, then find the values of a and b.
Solution:
Total probability = 1: 0.1 + 0.4 + a + b + 0.1 = 1
∴ a + b = 0.4 ... (1)
Mean = E(X) = Σx·P(x) = 2.8
0(0.1) + 1(0.4) + 2(a) + 3(b) + 4(0.1) = 2.8
0.4 + 2a + 3b + 0.4 = 2.8
2a + 3b = 2.0 ... (2)
Solving (1) and (2):
From (1): a = 0.4 - b
Substitute in (2): 2(0.4 - b) + 3b = 2.0
0.8 - 2b + 3b = 2.0
b = 1.2 (This exceeds 1, indicating an error in calculation)
Correct approach: From 2a + 3b = 2.0 and a + b = 0.4
Multiply equation (1) by 2: 2a + 2b = 0.8
Subtract from equation (2): b = 1.2 - 0.8 = ... needs recalculation
Total probability = 1: 0.1 + 0.4 + a + b + 0.1 = 1
∴ a + b = 0.4 ... (1)
Mean = E(X) = Σx·P(x) = 2.8
0(0.1) + 1(0.4) + 2(a) + 3(b) + 4(0.1) = 2.8
0.4 + 2a + 3b + 0.4 = 2.8
2a + 3b = 2.0 ... (2)
Solving (1) and (2):
From (1): a = 0.4 - b
Substitute in (2): 2(0.4 - b) + 3b = 2.0
0.8 - 2b + 3b = 2.0
b = 1.2 (This exceeds 1, indicating an error in calculation)
Correct approach: From 2a + 3b = 2.0 and a + b = 0.4
Multiply equation (1) by 2: 2a + 2b = 0.8
Subtract from equation (2): b = 1.2 - 0.8 = ... needs recalculation
Question 8
2M/3M
If the probability of success in a single trial is 0.01, how many minimum number of Bernoulli trials must be performed in order that the probability of at least one success is 0.95 or more?
(Use log 20 = 1.301 and log 2 = 0.301)
(Use log 20 = 1.301 and log 2 = 0.301)
Solution:
Given: p = 0.01, q = 0.99
Probability of at least one success: P(X ≥ 1) = 1 - P(X = 0) ≥ 0.95
P(X = 0) = (0.99)ⁿ
1 - (0.99)ⁿ ≥ 0.95
(0.99)ⁿ ≤ 0.05
(0.99)ⁿ ≤ 1/20
Taking log both sides:
n·log(0.99) ≤ log(1/20)
n·log(0.99) ≤ -log(20)
n[log(1) - log(100/99)] ≤ -1.301
n[-log(100/99)] ≤ -1.301
Since log(0.99) is negative, dividing reverses inequality:
n ≥ 1.301 / 0.00436 ≈ 298.5
Minimum n = 299 trials
Given: p = 0.01, q = 0.99
Probability of at least one success: P(X ≥ 1) = 1 - P(X = 0) ≥ 0.95
P(X = 0) = (0.99)ⁿ
1 - (0.99)ⁿ ≥ 0.95
(0.99)ⁿ ≤ 0.05
(0.99)ⁿ ≤ 1/20
Taking log both sides:
n·log(0.99) ≤ log(1/20)
n·log(0.99) ≤ -log(20)
n[log(1) - log(100/99)] ≤ -1.301
n[-log(100/99)] ≤ -1.301
Since log(0.99) is negative, dividing reverses inequality:
n ≥ 1.301 / 0.00436 ≈ 298.5
Minimum n = 299 trials
Question 9
2M/3M
A car hire firm has two cars, which it hires out day by day. The number of demands for cars on each day is distributed as a Poisson distribution with mean 1.5. Calculate the probability of days on which some demand is refused.
(Use e⁻¹·⁵ = 0.223)
(Use e⁻¹·⁵ = 0.223)
Solution:
Mean λ = 1.5, Number of cars = 2
Demand is refused when X > 2
P(X > 2) = 1 - P(X ≤ 2)
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
Using Poisson formula: P(X = r) = (λʳ · e⁻λ) / r!
P(X = 0) = (1.5⁰ · e⁻¹·⁵) / 0! = 0.223
P(X = 1) = (1.5¹ · e⁻¹·⁵) / 1! = 1.5 × 0.223 = 0.3345
P(X = 2) = (1.5² · e⁻¹·⁵) / 2! = 2.25 × 0.223 / 2 = 0.2509
P(X ≤ 2) = 0.223 + 0.3345 + 0.2509 = 0.8084
P(X > 2) = 1 - 0.8084 = 0.1916 ≈ 0.19 or 19%
Mean λ = 1.5, Number of cars = 2
Demand is refused when X > 2
P(X > 2) = 1 - P(X ≤ 2)
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
Using Poisson formula: P(X = r) = (λʳ · e⁻λ) / r!
P(X = 0) = (1.5⁰ · e⁻¹·⁵) / 0! = 0.223
P(X = 1) = (1.5¹ · e⁻¹·⁵) / 1! = 1.5 × 0.223 = 0.3345
P(X = 2) = (1.5² · e⁻¹·⁵) / 2! = 2.25 × 0.223 / 2 = 0.2509
P(X ≤ 2) = 0.223 + 0.3345 + 0.2509 = 0.8084
P(X > 2) = 1 - 0.8084 = 0.1916 ≈ 0.19 or 19%
Question 10
2M/3M
A company averages 3 defects per batch. Assuming Poisson distribution, find the probability of exactly 2 defects per batch.
(e⁻³ = 0.049787)
(e⁻³ = 0.049787)
Solution:
Mean λ = 3
For exactly 2 defects:
P(X = 2) = (λ² · e⁻λ) / 2!
= (3² × 0.049787) / 2
= (9 × 0.049787) / 2
= 0.448083 / 2
= 0.224 or 22.4%
Mean λ = 3
For exactly 2 defects:
P(X = 2) = (λ² · e⁻λ) / 2!
= (3² × 0.049787) / 2
= (9 × 0.049787) / 2
= 0.448083 / 2
= 0.224 or 22.4%
Question 11
2M/3M
A river near a small town floods and overflows twice in every 10 years on an average. Assuming that the Poisson distribution is appropriate:
(a) What is the mean expectation?
(b) Calculate the probability of 3 or less overflows in a 10-year interval.
(a) What is the mean expectation?
(b) Calculate the probability of 3 or less overflows in a 10-year interval.
Solution:
(a) Mean expectation λ = 2 floods per 10 years
(b) P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
P(X = 0) = e⁻² / 0! = e⁻²
P(X = 1) = 2e⁻² / 1! = 2e⁻²
P(X = 2) = 4e⁻² / 2! = 2e⁻²
P(X = 3) = 8e⁻² / 3! = (8/6)e⁻² = (4/3)e⁻²
P(X ≤ 3) = e⁻²(1 + 2 + 2 + 4/3)
= e⁻²(19/3)
≈ 0.1353 × 6.333 ≈ 0.857 or 85.7%
(a) Mean expectation λ = 2 floods per 10 years
(b) P(X ≤ 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
P(X = 0) = e⁻² / 0! = e⁻²
P(X = 1) = 2e⁻² / 1! = 2e⁻²
P(X = 2) = 4e⁻² / 2! = 2e⁻²
P(X = 3) = 8e⁻² / 3! = (8/6)e⁻² = (4/3)e⁻²
P(X ≤ 3) = e⁻²(1 + 2 + 2 + 4/3)
= e⁻²(19/3)
≈ 0.1353 × 6.333 ≈ 0.857 or 85.7%
Question 12
2M/3M
Given that the scores of a set of candidates on an IQ test are normally distributed.If the IQ test has a mean of 100 and a standard deviation of 10, determine the probability that a candidate who takes the test will score between 90 and 110.
[Given: P(Z <1) = 0.8413 and P(Z < -1) = 0.1587]
[Given: P(Z <1) = 0.8413 and P(Z < -1) = 0.1587]
Solution:
Given: μ = 100, σ = 10
Z-score formula: Z = (X - μ) / σ
For X = 90:
Z₁ = (90 - 100) / 10 = -1
For X = 110:
Z₂ = (110 - 100) / 10 = 1
P(90 < X < 110) = P(-1 < Z < 1)
= P(Z < 1) - P(Z < -1)
= 0.8413 - 0.1587
= 0.6826 or 68.26%
Given: μ = 100, σ = 10
Z-score formula: Z = (X - μ) / σ
For X = 90:
Z₁ = (90 - 100) / 10 = -1
For X = 110:
Z₂ = (110 - 100) / 10 = 1
P(90 < X < 110) = P(-1 < Z < 1)
= P(Z < 1) - P(Z < -1)
= 0.8413 - 0.1587
= 0.6826 or 68.26%
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Long Answer Questions with Complete Solutions
Practice 5-mark questions
Question 1
A random variable X has the following probability distribution:
X: 0, 1, 2, 3, 4, 5, 6, 7
P(X): 0, k, 2k, 2k, 3k, k², 2k², 7k² + k
Find: (i) k, (ii) P(X < 3), (iii) P(X ≥ 3), (iv) P(0 < X < 5)
X: 0, 1, 2, 3, 4, 5, 6, 7
P(X): 0, k, 2k, 2k, 3k, k², 2k², 7k² + k
Find: (i) k, (ii) P(X < 3), (iii) P(X ≥ 3), (iv) P(0 < X < 5)
Complete Solution:
(i) Sum of all probabilities = 1
0 + k + 2k + 2k + 3k + k² + 2k² + 7k² + k = 1
10k² + 9k = 1
10k² + 9k - 1 = 0
10k² + 10k - k - 1 = 0
10k(k + 1) - 1(k + 1) = 0
(10k - 1)(k + 1) = 0
k = 1/10 or k = -1
Since k cannot be negative, k = 1/10
(ii) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
= 0 + k + 2k = 3k = 3/10
(iii) P(X ≥ 3) = 1 - P(X < 3) = 1 - 3/10 = 7/10
(iv) P(0 < X < 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= k + 2k + 2k + 3k = 8k = 8/10 = 4/5
(i) k = 1/10
(ii) P(X < 3) = 3/10
(iii) P(X ≥ 3) = 7/10
(iv) P(0 < X < 5) = 4/5
(ii) P(X < 3) = 3/10
(iii) P(X ≥ 3) = 7/10
(iv) P(0 < X < 5) = 4/5
Question 2
Ten coins are tossed simultaneously. Find the probability of getting: (i) exactly 6 heads, (ii) at least 6 heads, (iii) at most 6 heads.
Complete Solution:
This is binomial distribution with n = 10, p = 1/2, q = 1/2
(i) P(X = 6) = ¹⁰C₆ × (1/2)⁶ × (1/2)⁴
= ¹⁰C₆ × (1/2)¹⁰
= 210 × (1/1024) = 210/1024 = 105/512
(ii) P(X ≥ 6) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)
= (¹⁰C₆ + ¹⁰C₇ + ¹⁰C₈ + ¹⁰C₉ + ¹⁰C₁₀) × (1/2)¹⁰
= (210 + 120 + 45 + 10 + 1) × (1/1024)
= 386/1024 = 193/512
(iii) P(X ≤ 6) = 1 - P(X > 6)
= 1 - [P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)]
= 1 - (120 + 45 + 10 + 1)/1024
= 1 - 176/1024 = 848/1024 = 53/64
(i) P(exactly 6 heads) = 105/512
(ii) P(at least 6 heads) = 193/512
(iii) P(at most 6 heads) = 53/64
(ii) P(at least 6 heads) = 193/512
(iii) P(at most 6 heads) = 53/64
Question 3
CBSE 2023
The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 bulbs: (i) none will fuse after 150 days, (ii) at least one will fuse after 150 days, (iii) not more than one will fuse.
Complete Solution:
Binomial distribution: n = 5, p = 0.05, q = 0.95
(i) P(X = 0) = ⁵C₀ × (0.05)⁰ × (0.95)⁵
= 1 × 1 × (0.95)⁵
= (0.95)⁵ ≈ 0.7738
(ii) P(X ≥ 1) = 1 - P(X = 0)
= 1 - (0.95)⁵
≈ 1 - 0.7738 = 0.2262
(iii) P(X ≤ 1) = P(X = 0) + P(X = 1)
P(X = 1) = ⁵C₁ × (0.05)¹ × (0.95)⁴
= 5 × 0.05 × (0.95)⁴ ≈ 5 × 0.05 × 0.8145 ≈ 0.2036
P(X ≤ 1) ≈ 0.7738 + 0.2036 = 0.9774
(i) P(none fuse) ≈ 0.7738 or (0.95)⁵
(ii) P(at least one fuses) ≈ 0.2262 or 1 - (0.95)⁵
(iii) P(not more than one fuses) ≈ 0.9774
(ii) P(at least one fuses) ≈ 0.2262 or 1 - (0.95)⁵
(iii) P(not more than one fuses) ≈ 0.9774
Question 4
A company manufactures computer chips. The number of defective chips follows Poisson distribution with mean 3. Find the probability that: (i) exactly 2 chips are defective, (ii) at most 2 chips are defective, (iii) more than 2 chips are defective.
Complete Solution:
Poisson distribution with λ = 3
P(X = x) = (e⁻³ × 3ˣ) / x!
(i) P(X = 2) = (e⁻³ × 3²) / 2!
= 9e⁻³/2
(ii) P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
P(X = 0) = e⁻³
P(X = 1) = (e⁻³ × 3¹) / 1! = 3e⁻³
P(X = 2) = 9e⁻³/2
P(X ≤ 2) = e⁻³(1 + 3 + 9/2)
= e⁻³(17/2) = 17e⁻³/2
(iii) P(X > 2) = 1 - P(X ≤ 2)
= 1 - 17e⁻³/2
(i) P(X = 2) = 9e⁻³/2
(ii) P(X ≤ 2) = 17e⁻³/2
(iii) P(X > 2) = 1 - 17e⁻³/2
(ii) P(X ≤ 2) = 17e⁻³/2
(iii) P(X > 2) = 1 - 17e⁻³/2
Question 5
CBSE 2023
In a binomial distribution consisting of 5 independent trials, the probability of 1 and 2 successes are 0.4096 and 0.2048 respectively. Find the parameter p of the distribution.
Complete Solution:
Given: n = 5, P(X = 1) = 0.4096, P(X = 2) = 0.2048
P(X = 1) = ⁵C₁ × p¹ × q⁴ = 5pq⁴ = 0.4096 ...(i)
P(X = 2) = ⁵C₂ × p² × q³ = 10p²q³ = 0.2048 ...(ii)
Dividing (ii) by (i):
10p²q³ / 5pq⁴ = 0.2048 / 0.4096
2p/q = 0.5
2p = 0.5q
2p = 0.5(1 - p)
2p = 0.5 - 0.5p
2.5p = 0.5
p = 0.5/2.5 = 0.2
Verification: q = 0.8
P(X = 1) = 5(0.2)(0.8)⁴ = 5(0.2)(0.4096) = 0.4096 ✓
p = 0.2 (or 1/5)
Question 6
The marks obtained by students in a statistics examination are normally distributed with mean 65 and standard deviation 10. If 500 students appeared in the examination, find approximately how many students obtained: (i) marks between 55 and 75, (ii) more than 75 marks. [Given: P(0 < Z < 1) = 0.3413]
Complete Solution:
Given: X ~ N(65, 100), μ = 65, σ = 10, n = 500 students
Standard normal variable: Z = (X - μ)/σ = (X - 65)/10
(i) For marks between 55 and 75:
When X = 55: Z₁ = (55 - 65)/10 = -10/10 = -1
When X = 75: Z₂ = (75 - 65)/10 = 10/10 = 1
P(55 < X < 75) = P(-1 < Z < 1)
= P(-1 < Z < 0) + P(0 < Z < 1)
Due to symmetry: P(-1 < Z < 0) = P(0 < Z < 1) = 0.3413
P(55 < X < 75) = 0.3413 + 0.3413 = 0.6826
Number of students = 500 × 0.6826 = 341.3 ≈ 341 students
(ii) For marks more than 75:
P(X > 75) = P(Z > 1)
= 0.5 - P(0 < Z < 1)
= 0.5 - 0.3413 = 0.1587
Number of students = 500 × 0.1587 = 79.35 ≈ 79 students
(i) Approximately 341 students scored between 55 and 75
(ii) Approximately 79 students scored more than 75
(ii) Approximately 79 students scored more than 75
Question 7
The probability distribution of a discrete random variable X is given as under:
X: 1, 2, 3, 4, 5
P(X): A/2, A/3, A/5, A/5, 2A/15
Calculate: (i) The value of A if E(X) = 2.94, (ii) Variance of X
X: 1, 2, 3, 4, 5
P(X): A/2, A/3, A/5, A/5, 2A/15
Calculate: (i) The value of A if E(X) = 2.94, (ii) Variance of X
Complete Solution:
(i) Since ΣP(X) = 1
A/2 + A/3 + A/5 + A/5 + 2A/15 = 1
Taking LCM of 2, 3, 5, 15 = 30
15A/30 + 10A/30 + 6A/30 + 6A/30 + 4A/30 = 1
41A/30 = 1
A = 30/41
(ii) For Variance: E(X) = ΣxP(x) = 2.94
E(X²) = Σx²P(x)
= 1²(A/2) + 2²(A/3) + 3²(A/5) + 4²(A/5) + 5²(2A/15)
= A/2 + 4A/3 + 9A/5 + 16A/5 + 50A/15
= 15A/30 + 40A/30 + 54A/30 + 96A/30 + 100A/30
= 305A/30
Variance = E(X²) - [E(X)]²
= 305A/30 - (2.94)²
= 305(30/41)/30 - 8.6436
= 305/41 - 8.6436 ≈ 1.53
(i) A = 30/41
(ii) Variance ≈ 1.53
(ii) Variance ≈ 1.53
Question 8
If 5% students appearing in an examination fail, find the probability that out of 100 students:
(i) none failed, (ii) 5 students failed, (iii) at most 3 failed
(i) none failed, (ii) 5 students failed, (iii) at most 3 failed
Complete Solution (Poisson Distribution):
Given: p = 5/100 = 0.05, n = 100
Mean λ = np = 100 × 0.05 = 5
Poisson formula: P(X = x) = (e⁻λ × λˣ) / x!
(i) None failed: P(X = 0) = (e⁻⁵ × 5⁰) / 0!
= e⁻⁵ × 1 / 1 = e⁻⁵
(ii) 5 students failed: P(X = 5) = (e⁻⁵ × 5⁵) / 5!
= (e⁻⁵ × 3125) / 120 = 3125e⁻⁵/120
(iii) At most 3 failed: P(X ≤ 3) = P(0) + P(1) + P(2) + P(3)
= e⁻⁵ + 5e⁻⁵ + 25e⁻⁵/2 + 125e⁻⁵/6
= e⁻⁵(1 + 5 + 12.5 + 20.83) = e⁻⁵(39.33)
(i) P(none failed) = e⁻⁵
(ii) P(5 failed) = 3125e⁻⁵/120
(iii) P(at most 3 failed) ≈ 39.33e⁻⁵
(ii) P(5 failed) = 3125e⁻⁵/120
(iii) P(at most 3 failed) ≈ 39.33e⁻⁵
Question 9
In a sample of 1000 cases, the mean of a certain test is 14 and standard deviation is 2.5. Assuming the distribution to be normal, find how many students score:
(i) between 12 and 15, (ii) above 18, (iii) below 8
(i) between 12 and 15, (ii) above 18, (iii) below 8
Complete Solution:
Given: n = 1000, μ = 14, σ = 2.5
Z-score formula: Z = (X - μ)/σ
(i) Between 12 and 15:
Z₁ = (12 - 14)/2.5 = -0.8, Z₂ = (15 - 14)/2.5 = 0.4
P(-0.8 < Z < 0.4) = P(Z < 0.4) - P(Z < -0.8)
= 0.6554 - 0.2119 = 0.4435
Number of students = 1000 × 0.4435 = 444 students
(ii) Above 18:
Z = (18 - 14)/2.5 = 1.6
P(Z > 1.6) = 1 - P(Z < 1.6) = 1 - 0.9452 = 0.0548
Number of students = 1000 × 0.0548 = 55 students
(iii) Below 8:
Z = (8 - 14)/2.5 = -2.4
P(Z < -2.4) = 0.0082
Number of students = 1000 × 0.0082 = 8 students
(i) 444 students scored between 12 and 15
(ii) 55 students scored above 18
(iii) 8 students scored below 8
(ii) 55 students scored above 18
(iii) 8 students scored below 8
Question 10
A traffic engineer records that an average of 3.2 bicycle riders use a particular cycle track every hour. Given that the number of bicycles follows a Poisson distribution, find:
(i) Probability that 2 or less bicycle riders will use the track within an hour
(ii) Probability that 3 or more riders will approach within an hour
(iii) Mean expectation and variance for the random variable X
(i) Probability that 2 or less bicycle riders will use the track within an hour
(ii) Probability that 3 or more riders will approach within an hour
(iii) Mean expectation and variance for the random variable X
Complete Solution:
Given: λ = 3.2 bicycle riders per hour (Poisson distribution)
(i) Probability of 2 or less riders: P(X ≤ 2)
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= e⁻³·² + 3.2e⁻³·² + (3.2)²e⁻³·²/2
= e⁻³·²(1 + 3.2 + 5.12)
= 9.32e⁻³·²
(ii) Probability of 3 or more riders: P(X ≥ 3)
P(X ≥ 3) = 1 - P(X ≤ 2)
= 1 - 9.32e⁻³·²
(iii) Mean expectation and variance:
For Poisson distribution: Mean = λ = 3.2
Variance = λ = 3.2
(i) P(X ≤ 2) = 9.32e⁻³·²
(ii) P(X ≥ 3) = 1 - 9.32e⁻³·²
(iii) Mean = 3.2, Variance = 3.2
(ii) P(X ≥ 3) = 1 - 9.32e⁻³·²
(iii) Mean = 3.2, Variance = 3.2
Question 11
For a Poisson distribution model, if arrival rate of passengers at an airport is recorded as 30 per hour on a given day, find:
(i) Expected number of arrivals in the first 10 minutes
(ii) Probability of exactly 4 arrivals in the first 10 minutes
(iii) Probability of 4 or fewer arrivals in the first 10 minutes
(iv) Probability of 10 or more arrivals in an hour given 8 arrivals in first 10 minutes
(i) Expected number of arrivals in the first 10 minutes
(ii) Probability of exactly 4 arrivals in the first 10 minutes
(iii) Probability of 4 or fewer arrivals in the first 10 minutes
(iv) Probability of 10 or more arrivals in an hour given 8 arrivals in first 10 minutes
Complete Solution:
Given: Arrival rate = 30 passengers/hour
(i) Expected arrivals in first 10 minutes:
10 minutes = 1/6 hour
λ₁₀ = 30 × (1/6) = 5 passengers
(ii) Probability of exactly 4 arrivals in first 10 minutes:
P(X = 4) = (e⁻⁵ × 5⁴) / 4!
= (625e⁻⁵) / 24
(iii) Probability of 4 or fewer arrivals:
P(X ≤ 4) = P(0) + P(1) + P(2) + P(3) + P(4)
= e⁻⁵[1 + 5 + 25/2 + 125/6 + 625/24]
= e⁻⁵[1 + 5 + 12.5 + 20.83 + 26.04]
= 65.37e⁻⁵
(iv) For 10 or more in hour given 8 in first 10 minutes:
Need 2 or more in remaining 50 minutes
λ₅₀ = 30 × (5/6) = 25
P(Y ≥ 2) = 1 - P(Y ≤ 1) = 1 - e⁻²⁵(1 + 25)
≈ 1 (practically certain)
(i) Expected arrivals = 5 passengers
(ii) P(X = 4) = 625e⁻⁵/24
(iii) P(X ≤ 4) ≈ 65.37e⁻⁵
(iv) P(total ≥ 10) ≈ 1
(ii) P(X = 4) = 625e⁻⁵/24
(iii) P(X ≤ 4) ≈ 65.37e⁻⁵
(iv) P(total ≥ 10) ≈ 1
Question 12
In a district exam, scores of 300 students of Class XII are recorded. Ramesh scored 800 marks out of 1000. The average score was 700 and standard deviation was 180. Find:
(i) How Ramesh scored compared to his batchmates
(ii) Sudha scored 420 marks. What can you say about her performance?
(iii) How much has Abhay scored if he has done better than 44.83% of his batchmates?
(i) How Ramesh scored compared to his batchmates
(ii) Sudha scored 420 marks. What can you say about her performance?
(iii) How much has Abhay scored if he has done better than 44.83% of his batchmates?
Complete Solution:
Given: n = 300, μ = 700, σ = 180
Z-score formula: Z = (X - μ)/σ
(i) For Ramesh (X = 800):
Z = (800 - 700)/180 = 100/180 ≈ 0.556
Ramesh scored slightly above average, 0.556 standard deviations above mean
(ii) For Sudha (X = 420):
Z = (420 - 700)/180 = -280/180 ≈ -1.56
Sudha scored below average, about 1.56 standard deviations below mean
(iii) For Abhay (better than 44.83%):
Percentile = 44.83% → From Z-table: Z ≈ -0.13
Using Z = (X - μ)/σ:
-0.13 = (X - 700)/180
X = 700 + (-0.13 × 180)
X = 700 - 23.4 ≈ 677 marks
(i) Ramesh scored 0.556σ above mean (above average)
(ii) Sudha scored 1.56σ below mean (below average)
(iii) Abhay scored approximately 677 marks
(ii) Sudha scored 1.56σ below mean (below average)
(iii) Abhay scored approximately 677 marks
Question 13
The heights of students in a school are normally distributed with mean 165 cm and standard deviation 10 cm. If there are 800 students:
(i) How many students have height between 155 cm and 175 cm?
(ii) How many students are taller than 180 cm?
[Use: P(0 < Z < 1) = 0.3413, P(0 < Z < 1.5) = 0.4332]
(i) How many students have height between 155 cm and 175 cm?
(ii) How many students are taller than 180 cm?
[Use: P(0 < Z < 1) = 0.3413, P(0 < Z < 1.5) = 0.4332]
Complete Solution:
Given: n = 800, μ = 165, σ = 10
(i) For heights between 155 and 175:
Z₁ = (155 - 165)/10 = -1
Z₂ = (175 - 165)/10 = 1
P(-1 < Z < 1) = 2 × P(0 < Z < 1) = 2 × 0.3413 = 0.6826
Number of students = 800 × 0.6826 = 546 students
(ii) For heights taller than 180:
Z = (180 - 165)/10 = 1.5
P(Z > 1.5) = 0.5 - P(0 < Z < 1.5) = 0.5 - 0.4332 = 0.0668
Number of students = 800 × 0.0668 = 53 students
(i) 546 students have height between 155 and 175 cm
(ii) 53 students are taller than 180 cm
(ii) 53 students are taller than 180 cm
Question 14
CBSE Sample 2024
On doing proof reading of a book, on average 4 errors in 10 pages were detected. Using Poisson's Distribution, find the probability of:
(i) no error, (ii) one error in 1000 pages of first printed edition
[Given: e⁻⁰·⁴ = 0.6703]
(i) no error, (ii) one error in 1000 pages of first printed edition
[Given: e⁻⁰·⁴ = 0.6703]
Complete Solution:
Given: Average errors = 4 per 10 pages
λ per 10 pages = 4/10 = 0.4
Poisson formula: P(X = x) = (e⁻λ × λˣ) / x!
(i) Probability of no error (X = 0):
P(X = 0) = (e⁻⁰·⁴ × 0.4⁰) / 0!
= e⁻⁰·⁴ × 1 / 1
= 0.6703
(ii) Probability of one error (X = 1):
P(X = 1) = (e⁻⁰·⁴ × 0.4¹) / 1!
= 0.4 × e⁻⁰·⁴
= 0.4 × 0.6703
= 0.2681
(i) P(no error) = 0.6703
(ii) P(one error) = 0.2681
(ii) P(one error) = 0.2681
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Case Studies
Real-world application based questions
Case Study 1
Applied Maths
Photocopier Machines
A photocopier (also called copier or copy machine) is a machine that makes copies of documents and other visual images onto paper or plastic film quickly and cheaply. Most modern photocopiers use a technology called xerography, a dry process that uses electrostatic charges on a light-sensitive photoreceptor to first attract and then transfer toner particles (a powder) onto paper in the form of an image. The toner is then fused onto the paper using heat, pressure, or a combination of both.
An office has four copying machines, and the random variable X measures how many of them are in use at a particular moment in time. Suppose that
P(X = 0) = 0.10, P(X = 1) = 0.20, P(X = 2) = 0.30, P(X = 3) = 0.25, P(X = 4) = 0.15
An office has four copying machines, and the random variable X measures how many of them are in use at a particular moment in time. Suppose that
P(X = 0) = 0.10, P(X = 1) = 0.20, P(X = 2) = 0.30, P(X = 3) = 0.25, P(X = 4) = 0.15
(i) What is P(X ≤ 2)?
✓ Correct Answer: (b) 0.60
Solution:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= 0.10 + 0.20 + 0.30 = 0.60
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= 0.10 + 0.20 + 0.30 = 0.60
(ii) What is P(X > 1)?
✓ Correct Answer: (c) 0.70
Solution:
P(X > 1) = P(X = 2) + P(X = 3) + P(X = 4)
= 0.30 + 0.25 + 0.15 = 0.70
P(X > 1) = P(X = 2) + P(X = 3) + P(X = 4)
= 0.30 + 0.25 + 0.15 = 0.70
(iii) (a) What is expected number of copying machines in use at a particular moment?
✓ Correct Answer: (b) 2.15
Solution:
E(X) = ΣxP(x)
= 0(0.10) + 1(0.20) + 2(0.30) + 3(0.25) + 4(0.15)
= 0 + 0.20 + 0.60 + 0.75 + 0.60
= 2.15
E(X) = ΣxP(x)
= 0(0.10) + 1(0.20) + 2(0.30) + 3(0.25) + 4(0.15)
= 0 + 0.20 + 0.60 + 0.75 + 0.60
= 2.15
(iii) (b) Calculate the variance and standard deviation of the number of copying machines in use.
✓ Answer: Variance = 1.5275, Standard Deviation ≈ 1.236
Solution:
E(X²) = Σx²P(x)
= 0²(0.10) + 1²(0.20) + 2²(0.30) + 3²(0.25) + 4²(0.15)
= 0 + 0.20 + 1.20 + 2.25 + 2.40 = 6.05
Variance = E(X²) - [E(X)]²
= 6.05 - (2.15)² = 6.05 - 4.6225
= 1.5275
Standard Deviation = √1.5275 ≈ 1.236
E(X²) = Σx²P(x)
= 0²(0.10) + 1²(0.20) + 2²(0.30) + 3²(0.25) + 4²(0.15)
= 0 + 0.20 + 1.20 + 2.25 + 2.40 = 6.05
Variance = E(X²) - [E(X)]²
= 6.05 - (2.15)² = 6.05 - 4.6225
= 1.5275
Standard Deviation = √1.5275 ≈ 1.236
Case Study 2
Applied Maths
Student Scores (Normal Distribution)
A teacher analyzed the Mathematics scores of 400 students. The marks were found to be Normally distributed with Mean (μ) = 70 and Standard Deviation (σ) = 10. Use the normal distribution curve to answer the following questions.
(i) What percentage of students scored below 70 marks?
✓ Correct Answer: (c) 50%
Solution:
Z = (X - μ)/σ = (70 - 70)/10 = 0
In a normal distribution, the curve is symmetrical about the mean.
Therefore, 50% of students scored below the mean.
Z = (X - μ)/σ = (70 - 70)/10 = 0
In a normal distribution, the curve is symmetrical about the mean.
Therefore, 50% of students scored below the mean.
(ii) Find the number of students who scored more than 80 marks.
[Given: P(0 < Z < 1) = 0.3413]
[Given: P(0 < Z < 1) = 0.3413]
✓ Correct Answer: (b) 63
Solution:
Z = (80 - 70)/10 = 1
P(Z > 1) = 0.5 - P(0 < Z < 1)
= 0.5 - 0.3413 = 0.1587
Number of students = 400 × 0.1587 ≈ 63 students
Z = (80 - 70)/10 = 1
P(Z > 1) = 0.5 - P(0 < Z < 1)
= 0.5 - 0.3413 = 0.1587
Number of students = 400 × 0.1587 ≈ 63 students
(iii) (a) Calculate the number of students scoring between 60 and 80 marks.
[Given: P(0 < Z < 1) = 0.3413]
[Given: P(0 < Z < 1) = 0.3413]
✓ Correct Answer: (c) 273
Solution:
For X = 60: Z₁ = (60 - 70)/10 = -1
For X = 80: Z₂ = (80 - 70)/10 = 1
P(-1 < Z < 1) = 2 × P(0 < Z < 1)
= 2 × 0.3413 = 0.6826
Number of students = 400 × 0.6826 ≈ 273 students
For X = 60: Z₁ = (60 - 70)/10 = -1
For X = 80: Z₂ = (80 - 70)/10 = 1
P(-1 < Z < 1) = 2 × P(0 < Z < 1)
= 2 × 0.3413 = 0.6826
Number of students = 400 × 0.6826 ≈ 273 students
(iii) (b) The top 5% of the students are to receive a certificate of excellence. If the Z-score = 1.645, find the minimum qualifying score.
[Given: Z = 1.645 for top 5%]
[Given: Z = 1.645 for top 5%]
✓ Answer: Minimum score = 86.45 marks
Solution:
Given: Z = 1.645
Using Z = (X - μ)/σ
1.645 = (X - 70)/10
X = 70 + (1.645 × 10)
X = 70 + 16.45
X = 86.45 marks
Given: Z = 1.645
Using Z = (X - μ)/σ
1.645 = (X - 70)/10
X = 70 + (1.645 × 10)
X = 70 + 16.45
X = 86.45 marks
Case Study 3
Applied Maths
Phone Calls at Reservation Desk
Phone calls arrive at the rate of 48 per hour at the reservation desk for Indian Airlines. Assume the calls follow a Poisson distribution.
(i) Compute the probability of receiving exactly three calls in 5 minutes.
✓ Correct Answer: (c) 32e⁻⁴/3
Solution:
Calls per hour = 48
Calls per minute = 48/60 = 0.8
Mean in 5 minutes: λ = 5 × 0.8 = 4
P(X = 3) = (e⁻⁴ × 4³) / 3!
= 64e⁻⁴/6 = 32e⁻⁴/3
Calls per hour = 48
Calls per minute = 48/60 = 0.8
Mean in 5 minutes: λ = 5 × 0.8 = 4
P(X = 3) = (e⁻⁴ × 4³) / 3!
= 64e⁻⁴/6 = 32e⁻⁴/3
(ii) If the agent takes 5 minutes to complete the current call, how many calls do you expect to be waiting? What is the probability that none will be waiting?
✓ Answer: Expected = 4 calls, P(none) = e⁻⁴
Solution:
Mean in 5 minutes: λ = 4
Expected number of calls = λ = 4 calls
P(X = 0) = (e⁻⁴ × 4⁰) / 0! = e⁻⁴
Mean in 5 minutes: λ = 4
Expected number of calls = λ = 4 calls
P(X = 0) = (e⁻⁴ × 4⁰) / 0! = e⁻⁴
(iii) If no calls are currently being processed, what is the probability that the agent can take 3 minutes for personal time without being interrupted by a call?
✓ Correct Answer: (a) e⁻²·⁴
Solution:
Mean in 3 minutes: λ = 3 × 0.8 = 2.4
No interruption means no calls in 3 minutes
P(X = 0) = (e⁻²·⁴ × 2.4⁰) / 0! = e⁻²·⁴
Mean in 3 minutes: λ = 3 × 0.8 = 2.4
No interruption means no calls in 3 minutes
P(X = 0) = (e⁻²·⁴ × 2.4⁰) / 0! = e⁻²·⁴
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