Unit 4: Probability Distributions - Free Study Resources | Boundless Maths

🎲 Unit 4: Probability Distributions

Random Variable, Probability Distribution Table, Binomial Distribution, Poisson Distribution, Normal Distribution

Weightage: 10 Marks in Board Exams
Home › Class 12 Applied Maths › Unit 4: Probability Distributions
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Topics 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

🎥 Video Tutorials

4
Probability Distributions
10 Marks · Series of 3 Videos

Probability Distributions — Series (3 Videos)

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✓ Random Variables ✓ Probability Distribution Table ✓ Binomial Distribution ✓ Poisson Distribution

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Normal Distribution
MCQs (36) Short Answers (12) Long Answers (14) Case Studies (3)
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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:
(a) 8
(b) 5
(c) 2
(d) 1
Answer: (c) 2

Solution: Formula for variance: Var(X) = E(X²) - [E(X)]²
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:
(a) 1/2
(b) 1/12
(c) 1/6
(d) 1/3
Answer: (d) 1/3

Solution: P(A∩B) = P(B|A) × P(A) = (1/2)(1/4) = 1/8
P(B) = P(A∩B)/P(A|B) = (1/8)/(1/3) = 3/8 → answer is (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:
(a) 3/20
(b) 9/4
(c) 141/20
(d) 159/80
Answer: (d) 159/80

Solution: E(X²) = 0²(1/5) + 1²(2/5) + 2²(2/5) = 0 + 2/5 + 8/5 = 2
Var(X) = 2 − (6/5)² = 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:
(a) 4
(b) 9
(c) 10
(d) 17
Answer: (c) 10

Solution: 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:
(a) ₹0
(b) ₹0.8
(c) ₹1
(d) ₹2
Answer: (c) ₹1

Solution: P(win) = 2/8 = 1/4; P(lose) = 3/4
E = (1/4)(5) + (3/4)(−3) = 5/4 − 9/4 = −₹1. Expected loss = ₹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:
(a) 0.60
(b) 0.124
(c) 0.244
(d) 0.240
Answer: (d) 0.240

Solution: Binomial with n = 4, p = 1/2. Variance = npq = 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:
(a) 10⁻¹
(b) (1/2)⁵
(c) (9/10)⁵
(d) 9/10
Answer: (c) (9/10)⁵

Solution: P(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:
(a) 2
(b) 1/2
(c) 1
(d) 3/2
Answer: (c) 1

Solution: X ~ B(2, 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:
(a) 0.7
(b) 0.49
(c) 0.3
(d) 0.09
Answer: (a) 0.7

Solution: E(X²) = 1²(0.7) + 0²(0.3) = 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.
(a) 5e⁻⁵
(b) e⁻⁵
(c) 25e⁻⁵
(d) 5²e⁻⁵
Answer: (a) 5e⁻⁵

Solution: λ = 5. 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:
(a) 37/729
(b) 1/729
(c) 2/27
(d) 728/729
Answer: (a) 37/729

Solution: n = 12, p = 2/3, q = 1/3. 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:
(a) 1/15
(b) 1/21
(c) 21
(d) 1/2
Answer: (b) 1/21

Solution: k(2+3+4+5+6) = 1 → 20k = 1 → but correct sum = 21, so k = 1/21

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
(a) A–III, B–I, C–II, D–IV
(b) A–III, B–I, C–IV, D–II
(c) A–I, B–III, C–II, D–IV
(d) A–I, B–III, C–IV, D–II
Answer: (b) A–III, B–I, C–IV, D–II

Solution: Variance = λ → A–III; SD = √λ → B–I; λ=4: SD=2 → C–IV; λ=4: Var=4 → D–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:
(a) 1
(b) 2
(c) 5
(d) 8/3
Answer: (b) 2

Solution: E(X) = 1(1/2) + 2(1/3) + 5(1/6) = 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:
(a) 0, 1, 3, 5
(b) 0, 2, 4, 6
(c) 0, 2, 5, 6
(d) 1, 3, 4, 5
Answer: (b) 0, 2, 4, 6

Solution: X = |2H − 6|. When H = 0,1,2,3,4,5,6 → X = 6,4,2,0,2,4,6. Unique values: {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:
(a) 1/3
(b) 2/3
(c) 1
(d) 4/3
Answer: (b) 2/3

Solution: p = 6/36 = 1/6; Mean = np = 4 × 1/6 = 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?
(a) (9/10)⁵
(b) 1 - (9/10)⁵
(c) 1/2
(d) (1/10)⁵
Answer: (a) (9/10)⁵

Solution: P(not defective) = 9/10. For 5 bulbs: (9/10)⁵

Question 18CBSE 2022
If the mean of a binomial distribution is 81, then the standard deviation lies in the interval:
(a) [0, 9)
(b) (0, 9]
(c) [0, 3]
(d) (0, 3]
Answer: (a) [0, 9)

Solution: np = 81; npq < np so SD = √(npq) < 9. 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:
(a) 1/10
(b) 1/5
(c) 1/4
(d) 1/2
✓ Correct Answer: (a) 1/10
Solution:
k(1+2+3+4) = 1 → 10k = 1 → k = 1/10
Question 20
The mean of a binomial distribution with parameters n = 10 and p = 0.4 is:
(a) 2.4
(b) 4.0
(c) 6.0
(d) 10
✓ Correct Answer: (b) 4.0
Solution:
Mean = np = 10 × 0.4 = 4.0
Question 21
If X follows Poisson distribution with parameter λ = 3, then P(X = 2) is:
(a) 9e⁻³/2
(b) 3e⁻³/2
(c) 9e⁻³
(d) 3e⁻³
✓ Correct Answer: (a) 9e⁻³/2
Solution:
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:
(a) 4/3
(b) 2
(c) 3
(d) 4
✓ Correct Answer: (a) 4/3
Solution:
Var = npq = 6 × (1/3) × (2/3) = 4/3
Question 23
If E(X) = 5 and E(X²) = 30, then the variance of X is:
(a) 25
(b) 5
(c) 10
(d) 15
✓ Correct Answer: (b) 5
Solution:
Var(X) = E(X²) − [E(X)]² = 30 − 25 = 5
Question 24
In a Poisson distribution, if mean = variance = 4, then P(X = 0) is:
(a) e⁻⁴
(b) 4e⁻⁴
(c) e⁴
(d) 1/4
✓ Correct Answer: (a) e⁻⁴
Solution:
P(X=0) = (e⁻⁴ × 4⁰)/0! = e⁻⁴
Question 25
A fair coin is tossed 5 times. The probability of getting exactly 3 heads is:
(a) 5/16
(b) 5/32
(c) 10/32
(d) 15/32
✓ Correct Answer: (a) 5/16
Solution:
P(X=3) = ⁵C₃ × (1/2)⁵ = 10/32 = 5/16
Question 26
If X ~ B(n, p) and E(X) = 6, Var(X) = 4.2, then n is:
(a) 10
(b) 12
(c) 15
(d) 18
✓ Correct Answer: (a) 10
Solution:
q = 4.2/6 = 0.7, p = 0.3; n = 6/0.3 = 10
Question 27
The standard deviation of a Poisson distribution with parameter λ = 9 is:
(a) 3
(b) 9
(c) 81
(d) 27
✓ Correct Answer: (a) 3
Solution:
SD = √λ = √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:
(a) Binomial distribution with n = 5, p = 1/6
(b) Binomial distribution with n = 5, p = 5/6
(c) Poisson distribution with λ = 5
(d) Normal distribution
✓ Correct Answer: (a) Binomial distribution with n = 5, p = 1/6
Solution:
Matches P(X=x) = ⁿCₓ pˣ q⁽ⁿ⁻ˣ⁾ with n=5, p=1/6, q=5/6
Question 29
The mean of a random variable X is 10. Then E(3X + 5) is:
(a) 30
(b) 35
(c) 15
(d) 45
✓ Correct Answer: (b) 35
Solution:
E(3X+5) = 3E(X)+5 = 30+5 = 35
Question 30
If Var(X) = 4, then Var(2X + 3) is:
(a) 8
(b) 11
(c) 16
(d) 19
✓ Correct Answer: (c) 16
Solution:
Var(aX+b) = a²Var(X) = 4 × 4 = 16
Question 31
For a standard normal distribution, the mean and variance are:
(a) Mean = 0, Variance = 0
(b) Mean = 0, Variance = 1
(c) Mean = 1, Variance = 0
(d) Mean = 1, Variance = 1
✓ Correct Answer: (b) Mean = 0, Variance = 1
Solution:
Z ~ N(0,1): μ = 0, σ² = 1
Question 32
If X ~ N(50, 25), then the standard normal variable Z is given by:
(a) Z = (X - 50)/5
(b) Z = (X - 50)/25
(c) Z = (X - 25)/50
(d) Z = X - 50
✓ Correct Answer: (a) Z = (X - 50)/5
Solution:
μ = 50, σ² = 25, σ = 5. 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.
(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
✓ Correct Answer: (d) A is False but R is True
Solution:
Random variables can be discrete (integers) or continuous (real numbers) — A is false. R is correct.
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.
(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
✓ Correct Answer: (a) Both A and R are True and R is the correct explanation of A
Solution:
Using np − npq = 1 and (np)² − (npq)² = 5, solving gives p = 1/3. R correctly explains A.
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)
(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
✓ Correct Answer: (d) A is False but R is True
Solution:
p = 1/2; P(X=4) = 1/16; P(X ≤ 3) = 1 − 1/16 = 15/16, not 1/16. A is false. R is correct.

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Short Answer Questions with Step-by-Step Solutions

Practice 2-mark and 3-mark questions

Question 1CBSE 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
Solution:
E(X) = 0(0.3) + 1(0.5) + 2(0.2) = 0.9
E(X²) = 0(0.3) + 1(0.5) + 4(0.2) = 1.3
Var(X) = 1.3 − (0.9)² = 1.3 − 0.81 = 0.49
Mean = 0.9, Variance = 0.49
Question 2CBSE 2024
A die is thrown 6 times. If getting an odd number is a success, find the probability of getting exactly 4 successes.
Solution:
n = 6, p = 1/2, q = 1/2
P(X = 4) = ⁶C₄ × (1/2)⁴ × (1/2)² = 15 × (1/64) = 15/64
P(X = 4) = 15/64
Question 3CBSE 2023
If X follows Poisson distribution with mean 2, find P(X ≥ 1).
Solution:
P(X ≥ 1) = 1 − P(X = 0) = 1 − e⁻²
P(X ≥ 1) = 1 − e⁻²
Question 4CBSE 2025
If X ~ B(n, p) with mean 20 and variance 16, find n and p.
Solution:
np = 20, npq = 16 → q = 4/5, p = 1/5
n = 20/(1/5) = 100
n = 100, p = 1/5 = 0.2
Question 5CBSE 2023
The probability distribution of a discrete random variable X is: X: 1,2,3,4; P(X): k, 2k, 3k, 4k. Find k and P(X < 3).
Solution:
10k = 1 → k = 1/10
P(X < 3) = k + 2k = 3k = 3/10
k = 1/10, P(X < 3) = 3/10
Question 6CBSE 2024
If X ~ N(60, 16), find the z-score when X = 68.
Solution:
μ = 60, σ² = 16, σ = 4
Z = (68 − 60)/4 = 2
z-score = 2
Question 7
A random variable X has: P(X=0)=0.1, P(X=1)=0.4, P(X=2)=a, P(X=3)=b, P(X=4)=0.1. If mean = 2.8, find a and b.
Solution:
a + b = 0.4 … (1)
0.4 + 2a + 3b + 0.4 = 2.8 → 2a + 3b = 2.0 … (2)
From (1) and (2): b = 0.2, a = 0.2
a = 0.2, b = 0.2
Question 8
If the probability of success in a single trial is 0.01, how many minimum Bernoulli trials must be performed for P(at least one success) ≥ 0.95?
(Use log 20 = 1.301, log 2 = 0.301)
Solution:
1 − (0.99)ⁿ ≥ 0.95 → (0.99)ⁿ ≤ 0.05
n ≥ 1.301/|log(0.99)| ≈ 298.5
Minimum n = 299 trials
Question 9
A car hire firm has two cars hired day by day. Number of demands follows Poisson with mean 1.5. Calculate probability of days on which demand is refused. (e⁻¹·⁵ = 0.223)
Solution:
Demand refused when X > 2. P(X ≤ 2) = 0.223 + 0.3345 + 0.2509 = 0.8084
P(X > 2) = 1 − 0.8084 = 0.1916
Probability of demand being refused ≈ 19%
Question 10
A company averages 3 defects per batch. Assuming Poisson distribution, find P(exactly 2 defects). (e⁻³ = 0.049787)
Solution:
P(X = 2) = (3² × e⁻³)/2! = 9 × 0.049787/2 = 0.2241
P(exactly 2 defects) ≈ 0.224 (22.4%)
Question 11
A river floods twice every 10 years on average (Poisson). (a) Mean expectation? (b) P(3 or fewer overflows in 10 years)?
Solution:
(a) Mean λ = 2 floods per 10 years
(b) P(X ≤ 3) = e⁻²(1 + 2 + 2 + 4/3) = e⁻²(19/3) ≈ 0.857
(a) Mean = 2; (b) P(X ≤ 3) ≈ 85.7%
Question 12
IQ test has mean 100 and SD 10. Find P(score between 90 and 110). [P(Z < 1) = 0.8413, P(Z < −1) = 0.1587]
Solution:
Z₁ = −1, Z₂ = 1. P(−1 < Z < 1) = 0.8413 − 0.1587 = 0.6826
Probability = 0.6826 (68.26%)

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Long Answer Questions with Complete Solutions

Practice 5-mark questions

Question 1
A random variable X has 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)
Complete Solution:
(i) 10k² + 9k − 1 = 0 → (10k−1)(k+1) = 0 → k = 1/10
(ii) P(X < 3) = 0 + k + 2k = 3k = 3/10
(iii) P(X ≥ 3) = 1 − 3/10 = 7/10
(iv) P(0 < X < 5) = k + 2k + 2k + 3k = 8k = 4/5
(i) k = 1/10; (ii) 3/10; (iii) 7/10; (iv) 4/5
Question 2
Ten coins are tossed simultaneously. Find P(exactly 6 heads), P(at least 6 heads), P(at most 6 heads).
Complete Solution:
(i) P(X=6) = ¹⁰C₆ × (1/2)¹⁰ = 210/1024 = 105/512
(ii) P(X≥6) = (210+120+45+10+1)/1024 = 386/1024 = 193/512
(iii) P(X≤6) = 1 − (120+45+10+1)/1024 = 848/1024 = 53/64
(i) 105/512; (ii) 193/512; (iii) 53/64
Question 3CBSE 2023
Probability a bulb fuses after 150 days = 0.05. Out of 5 bulbs find: (i) none fuses, (ii) at least one fuses, (iii) not more than one fuses.
Complete Solution:
n = 5, p = 0.05, q = 0.95
(i) P(X=0) = (0.95)⁵ ≈ 0.7738
(ii) P(X≥1) = 1 − (0.95)⁵ ≈ 0.2262
(iii) P(X=1) = 5(0.05)(0.95)⁴ ≈ 0.2036; P(X≤1) ≈ 0.9774
(i) ≈ 0.7738; (ii) ≈ 0.2262; (iii) ≈ 0.9774
Question 4
Defective chips follow Poisson with mean 3. Find: (i) exactly 2 defective, (ii) at most 2, (iii) more than 2.
Complete Solution:
(i) P(X=2) = 9e⁻³/2
(ii) P(X≤2) = e⁻³(1+3+9/2) = 17e⁻³/2
(iii) P(X>2) = 1 − 17e⁻³/2
(i) 9e⁻³/2; (ii) 17e⁻³/2; (iii) 1 − 17e⁻³/2
Question 5CBSE 2023
In a binomial distribution (n = 5), P(X=1) = 0.4096 and P(X=2) = 0.2048. Find p.
Complete Solution:
P(X=2)/P(X=1) = 2p/q = 0.5 → 2p = 0.5(1−p) → p = 0.2
p = 0.2 (or 1/5)
Question 6
Marks of 500 students ~ N(65, 100). Find approx. how many scored (i) between 55 and 75, (ii) more than 75 marks. [P(0 < Z < 1) = 0.3413]
Complete Solution:
(i) Z₁=−1, Z₂=1; P = 0.6826; 500 × 0.6826 ≈ 341 students
(ii) Z=1; P(Z>1) = 0.1587; 500 × 0.1587 ≈ 79 students
(i) ≈ 341 students; (ii) ≈ 79 students
Question 7
Probability distribution: X: 1,2,3,4,5; P(X): A/2, A/3, A/5, A/5, 2A/15. (i) Find A if E(X) = 2.94. (ii) Find Variance.
Complete Solution:
(i) 41A/30 = 1 → A = 30/41
(ii) E(X²) = 305A/30 = 305/41; Var = 305/41 − (2.94)² ≈ 1.53
(i) A = 30/41; (ii) Variance ≈ 1.53
Question 8
5% of 100 students fail. Using Poisson, find: (i) none failed, (ii) 5 failed, (iii) at most 3 failed.
Complete Solution:
λ = np = 100 × 0.05 = 5
(i) P(X=0) = e⁻⁵
(ii) P(X=5) = 3125e⁻⁵/120
(iii) P(X≤3) = e⁻⁵(1+5+12.5+20.83) ≈ 39.33e⁻⁵
(i) e⁻⁵; (ii) 3125e⁻⁵/120; (iii) ≈ 39.33e⁻⁵
Question 9
1000 students, mean = 14, SD = 2.5. How many scored (i) between 12 and 15, (ii) above 18, (iii) below 8?
Complete Solution:
(i) Z₁=−0.8, Z₂=0.4; P=0.4435 → 444 students
(ii) Z=1.6; P=0.0548 → 55 students
(iii) Z=−2.4; P=0.0082 → 8 students
(i) 444; (ii) 55; (iii) 8 students
Question 10
Bicycle riders arrive at 3.2/hour (Poisson). Find: (i) P(2 or fewer/hour), (ii) P(3 or more/hour), (iii) Mean and variance.
Complete Solution:
(i) P(X≤2) = e⁻³·²(1+3.2+5.12) = 9.32e⁻³·²
(ii) P(X≥3) = 1 − 9.32e⁻³·²
(iii) Mean = Variance = λ = 3.2
(i) 9.32e⁻³·²; (ii) 1−9.32e⁻³·²; (iii) 3.2
Question 11
Airport arrival rate: 30/hour. Find: (i) Expected arrivals in 10 min, (ii) P(exactly 4 in 10 min), (iii) P(4 or fewer in 10 min), (iv) P(10+ total given 8 in first 10 min).
Complete Solution:
(i) λ₁₀ = 30 × (1/6) = 5
(ii) P(X=4) = 625e⁻⁵/24
(iii) P(X≤4) ≈ 65.37e⁻⁵
(iv) λ₅₀ = 25; P(Y≥2) ≈ 1
(i) 5; (ii) 625e⁻⁵/24; (iii) ≈65.37e⁻⁵; (iv) ≈1
Question 12
300 students, mean = 700, SD = 180. Ramesh scored 800, Sudha scored 420. Find z-scores and Abhay's score if he beat 44.83% of batchmates.
Complete Solution:
(i) Ramesh: Z = (800−700)/180 ≈ 0.556 (above average)
(ii) Sudha: Z = (420−700)/180 ≈ −1.56 (below average)
(iii) Z ≈ −0.13 → X = 700 − 23.4 ≈ 677 marks
(i) Z ≈ 0.556; (ii) Z ≈ −1.56; (iii) ≈ 677 marks
Question 13
800 students, height ~ N(165, 100). How many are (i) between 155–175 cm, (ii) taller than 180 cm? [P(0 < Z < 1) = 0.3413, P(0 < Z < 1.5) = 0.4332]
Complete Solution:
(i) Z₁=−1, Z₂=1; P=0.6826 → 546 students
(ii) Z=1.5; P(Z>1.5)=0.0668 → 53 students
(i) 546 students; (ii) 53 students
Question 14CBSE Sample 2024
Proof reading: 4 errors per 10 pages. Using Poisson, find P(no error) and P(one error) per 10 pages. [e⁻⁰·⁴ = 0.6703]
Complete Solution:
λ = 4/10 = 0.4
(i) P(X=0) = e⁻⁰·⁴ = 0.6703
(ii) P(X=1) = 0.4 × 0.6703 = 0.2681
(i) P(no error) = 0.6703; (ii) P(one error) = 0.2681
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Case Studies

Real-world application based questions

Case Study 1Applied Maths
Photocopier Machines
An office has four copying machines, and the random variable X measures how many are in use at a given moment.
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)?
(a) 0.50
(b) 0.60
(c) 0.70
(d) 0.80
✓ Correct Answer: (b) 0.60
Solution:
P(X≤2) = 0.10+0.20+0.30 = 0.60
(ii) What is P(X > 1)?
(a) 0.60
(b) 0.65
(c) 0.70
(d) 0.75
✓ Correct Answer: (c) 0.70
Solution:
P(X>1) = 0.30+0.25+0.15 = 0.70
(iii)(a) Expected number of machines in use?
(a) 2.05
(b) 2.15
(c) 2.25
(d) 2.35
✓ Correct Answer: (b) 2.15
Solution:
E(X) = 0(0.10)+1(0.20)+2(0.30)+3(0.25)+4(0.15) = 2.15
(iii)(b) Calculate the variance and standard deviation.
✓ Variance = 1.5275, SD ≈ 1.236
Solution:
E(X²) = 0+0.20+1.20+2.25+2.40 = 6.05
Var = 6.05 − (2.15)² = 1.5275; SD ≈ 1.236
Case Study 2Applied Maths
Student Scores (Normal Distribution)
Maths scores of 400 students are normally distributed with Mean (μ) = 70 and Standard Deviation (σ) = 10.
(i) What percentage of students scored below 70 marks?
(a) 25%
(b) 40%
(c) 50%
(d) 60%
✓ Correct Answer: (c) 50%
Solution:
Normal distribution is symmetric about the mean — 50% score below the mean.
(ii) Students who scored more than 80 marks? [P(0 < Z < 1) = 0.3413]
(a) 58
(b) 63
(c) 68
(d) 73
✓ Correct Answer: (b) 63
Solution:
Z=1; P(Z>1) = 0.5−0.3413 = 0.1587; 400 × 0.1587 ≈ 63 students
(iii)(a) Students scoring between 60 and 80? [P(0 < Z < 1) = 0.3413]
(a) 250
(b) 265
(c) 273
(d) 285
✓ Correct Answer: (c) 273
Solution:
P(−1<Z<1) = 0.6826; 400 × 0.6826 ≈ 273 students
(iii)(b) Top 5% receive certificates. Find minimum qualifying score (Z = 1.645).
✓ Minimum score = 86.45 marks
Solution:
1.645 = (X−70)/10 → X = 86.45
Case Study 3Applied Maths
Phone Calls at Reservation Desk
Phone calls arrive at 48 per hour at an airline reservation desk (Poisson distribution).
(i) Probability of exactly 3 calls in 5 minutes?
(a) 4e⁻⁴/3
(b) 16e⁻⁴/3
(c) 32e⁻⁴/3
(d) 64e⁻⁴/6
✓ Correct Answer: (c) 32e⁻⁴/3
Solution:
λ per min = 0.8; λ for 5 min = 4. P(X=3) = (e⁻⁴ × 64)/6 = 32e⁻⁴/3
(ii) Agent takes 5 min on a call. Expected calls waiting? P(none waiting)?
✓ Expected = 4 calls; P(none) = e⁻⁴
Solution:
λ=4; Expected=4; P(X=0) = e⁻⁴
(iii) P(agent can take 3-minute break without being interrupted)?
(a) e⁻²·⁴
(b) e⁻³
(c) e⁻³·⁶
(d) e⁻⁴
✓ Correct Answer: (a) e⁻²·⁴
Solution:
λ for 3 min = 3 × 0.8 = 2.4; P(X=0) = e⁻²·⁴

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