How to Use This Page
Solve every exercise question on your own first, then compare your answers below. Focus on understanding why each step works — that builds exam confidence. Show all working, double-check arithmetic in statistics tables, and verify that every probability answer lies between 0 and 1. For statistics, always create a neat table with all required columns before calculating.
Key Concepts (Quick Recap)
Statistics:
- Range = Maximum value - Minimum value
- Coefficient of Range = (X_max - X_min) / (X_max + X_min)
- Variance (ungrouped) = sum(x - x_bar)^2 / n
- Standard Deviation (SD) = sqrt(Variance)
- Variance (frequency table) = sum f(x - x_bar)^2 / sum f
- SD for grouped data using step-deviation method: SD = C * sqrt[(sum f d^2 / sum f) - (sum f d / sum f)^2] where d = (x - A)/C
- Coefficient of Variation (C.V.) = (SD / Mean) * 100
- Lower C.V. means more consistent data
- If each value is changed by a constant k: new Mean = old Mean + k, SD unchanged
- If each value is multiplied by k: new Mean = k old Mean, new SD = k old SD
Probability:
- P(A) = Number of favourable outcomes / Total number of outcomes
- 0 <= P(A) <= 1
- P(not A) = P(A_bar) = 1 - P(A)
- Addition Theorem: P(A or B) = P(A) + P(B) - P(A and B)
- Mutually Exclusive Events: P(A and B) = 0, so P(A or B) = P(A) + P(B)
- Independent Events: P(A and B) = P(A) * P(B)
- Playing Cards: 52 cards, 4 suits of 13, 12 face cards, 4 aces, 26 red, 26 black
- Dice: 6 faces (1-6), Two dice: 36 outcomes, Coin: 2 outcomes (H, T), Two coins: 4 outcomes
Exercise 8.1 — Measures of Dispersion
Q1. Find the range and coefficient of range for the data: 15, 22, 9, 35, 28, 12, 7, 40.
Arranging: 7, 9, 12, 15, 22, 28, 35, 40
Maximum (X_max) = 40
Minimum (X_min) = 7
Range = X_max - X_min = 40 - 7 = 33
Coefficient of Range = (X_max - X_min) / (X_max + X_min)
= (40 - 7) / (40 + 7)
= 33 / 47
= 0.702
Q2. Find the variance and standard deviation of the data: 6, 7, 10, 12, 13, 4, 8.
Step 1: Find the mean
Mean = (6 + 7 + 10 + 12 + 13 + 4 + 8) / 7 = 60 / 7 = 8.57
Step 2: Find deviations and squares
x | x - x_bar | (x - x_bar)^2
6 | -2.57 | 6.605
7 | -1.57 | 2.465
10 | 1.43 | 2.045
12 | 3.43 | 11.765
13 | 4.43 | 19.625
4 | -4.57 | 20.885
8 | -0.57 | 0.325
Sum of (x - x_bar)^2 = 6.605 + 2.465 + 2.045 + 11.765 + 19.625 + 20.885 + 0.325 = 63.715
Variance = 63.715 / 7 = 9.102
SD = sqrt(9.102) = 3.017
Q3. Calculate the standard deviation of the following frequency distribution:
| x | 5 | 10 | 15 | 20 | 25 |
|---|---|---|---|---|---|
| f | 2 | 4 | 6 | 4 | 2 |
Step 1: Find the mean
x | f | fx
5 | 2 | 10
10 | 4 | 40
15 | 6 | 90
20 | 4 | 80
25 | 2 | 50
sum f = 2+4+6+4+2 = 18
sum fx = 10+40+90+80+50 = 270
Mean = 270/18 = 15
Step 2: Find deviations and squared deviations
x | f | x - x_bar | (x - x_bar)^2 | f(x - x_bar)^2
5 | 2 | -10 | 100 | 200
10 | 4 | -5 | 25 | 100
15 | 6 | 0 | 0 | 0
20 | 4 | 5 | 25 | 100
25 | 2 | 10 | 100 | 200
sum f(x - x_bar)^2 = 200+100+0+100+200 = 600
Variance = 600/18 = 33.33
SD = sqrt(33.33) = 5.77
Q4. The mean of 20 observations is 45. If each observation is increased by 4, find the new mean and the new standard deviation if the original SD was 8.
When each value is increased by a constant k:
New Mean = Old Mean + k = 45 + 4 = 49
New SD = Old SD = 8 (SD does not change when a constant is added)
Q5. The mean and SD of 15 observations are 20 and 5 respectively. If each observation is multiplied by 3, find the new mean and new SD.
When each value is multiplied by a constant k:
New Mean = k * Old Mean = 3 * 20 = 60
New SD = k * Old SD = 3 * 5 = 15
Q6. Find the SD for the data: 12, 8, 10, 14, 6, 16.
Mean = (12 + 8 + 10 + 14 + 6 + 16) / 6 = 66 / 6 = 11
x | x - x_bar | (x - x_bar)^2
12 | 1 | 1
8 | -3 | 9
10 | -1 | 1
14 | 3 | 9
6 | -5 | 25
16 | 5 | 25
Sum = 70
Variance = 70/6 = 11.67
SD = sqrt(11.67) = 3.42
Exercise 8.2 — Coefficient of Variation
Q1. Team A has mean score 240 with SD = 18. Team B has mean score 300 with SD = 24. Which team is more consistent?
C.V. of Team A = (SD/Mean) * 100 = (18/240) * 100 = 7.5%
C.V. of Team B = (SD/Mean) * 100 = (24/300) * 100 = 8.0%
Since 7.5% < 8.0%, Team A is more consistent.
Q2. The mean and SD of marks of 50 students in Maths are 62 and 15 respectively. The mean and SD of marks of 50 students in Science are 58 and 12 respectively. Which subject shows more consistent performance?
C.V. of Maths = (15/62) * 100 = 24.19%
C.V. of Science = (12/58) * 100 = 20.69%
Since 20.69% < 24.19%, Science shows more consistent performance.
Q3. If the C.V. of a data set is 25% and the mean is 60, find the standard deviation.
C.V. = (SD/Mean) * 100
25 = (SD/60) * 100
SD = (25 * 60) / 100 = 15
Q4. If the standard deviation of a distribution is 4.5 and its mean is 30, find the coefficient of variation.
C.V. = (SD/Mean) * 100 = (4.5/30) * 100 = 15%
Q5. Two factories A and B produce bulbs. Factory A: mean life = 1200 hours, SD = 60 hours. Factory B: mean life = 1500 hours, SD = 75 hours. Which factory produces more consistent bulbs?
C.V. of A = (60/1200) * 100 = 5%
C.V. of B = (75/1500) * 100 = 5%
Both factories have the same C.V. = 5%. They are equally consistent.
Q6. The C.V. of a data set is 20% and SD = 8. Find the mean.
C.V. = (SD/Mean) * 100
20 = (8/Mean) * 100
Mean = (8 * 100) / 20 = 40
Exercise 8.3 — Probability of an Event
Q1. A fair die is thrown once. Find the probability of getting (i) a prime number, (ii) a number greater than 4, (iii) a number not greater than 3.
Sample space S = {1, 2, 3, 4, 5, 6}
n(S) = 6
(i) Prime numbers: {2, 3, 5}, favourable = 3
P(prime) = 3/6 = 1/2
(ii) Numbers greater than 4: {5, 6}, favourable = 2
P(> 4) = 2/6 = 1/3
(iii) Numbers not greater than 3: {1, 2, 3}, favourable = 3
P(not > 3) = 3/6 = 1/2
Q2. Two coins are tossed simultaneously. Find the probability of getting (i) two heads, (ii) at least one head, (iii) no head.
Sample space S = {HH, HT, TH, TT}
n(S) = 4
(i) Two heads: {HH}, favourable = 1
P(two heads) = 1/4
(ii) At least one head: {HH, HT, TH}, favourable = 3
P(at least one head) = 3/4
(iii) No head: {TT}, favourable = 1
P(no head) = 1/4
Q3. A card is drawn at random from a well-shuffled deck of 52 playing cards. Find the probability that the card drawn is (i) a king, (ii) a red card, (iii) a face card.
n(S) = 52
(i) Number of kings = 4
P(king) = 4/52 = 1/13
(ii) Number of red cards = 26
P(red card) = 26/52 = 1/2
(iii) Number of face cards (J, Q, K of each suit) = 12
P(face card) = 12/52 = 3/13
Q4. If P(A) = 3/7, find P(not A).
P(not A) = 1 - P(A) = 1 - 3/7 = 4/7
Q5. Two dice are rolled simultaneously. Find the probability that the sum of the numbers on the two dice is (i) 7, (ii) greater than 10, (iii) less than 5.
When two dice are rolled, n(S) = 6 * 6 = 36
(i) Sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → 6 outcomes
P(sum = 7) = 6/36 = 1/6
(ii) Sum > 10: Sum = 11 → (5,6), (6,5); Sum = 12 → (6,6) → 3 outcomes
P(sum > 10) = 3/36 = 1/12
(iii) Sum < 5: Sum = 2 → (1,1); Sum = 3 → (1,2), (2,1); Sum = 4 → (1,3), (2,2), (3,1) → 6 outcomes
P(sum < 5) = 6/36 = 1/6
Q6. A bag contains 3 red, 4 white, and 5 blue balls. One ball is drawn at random. Find the probability that the ball drawn is (i) red, (ii) white, (iii) not blue.
Total balls = 3 + 4 + 5 = 12
n(S) = 12
(i) Red balls = 3
P(red) = 3/12 = 1/4
(ii) White balls = 4
P(white) = 4/12 = 1/3
(iii) Not blue = Red + White = 3 + 4 = 7
P(not blue) = 7/12
Q7. A number is selected at random from the numbers 1 to 30. Find the probability that it is a multiple of 7.
n(S) = 30
Multiples of 7 from 1 to 30: 7, 14, 21, 28 → 4 numbers
P(multiple of 7) = 4/30 = 2/15
Exercise 8.4 — Addition Theorem of Probability
Q1. If P(A) = 0.5, P(B) = 0.3, and P(A and B) = 0.1, find P(A or B).
By Addition Theorem:
P(A or B) = P(A) + P(B) - P(A and B)
= 0.5 + 0.3 - 0.1
= 0.7
Q2. If P(A) = 3/8, P(B) = 1/4, and P(A or B) = 1/2, find P(A and B).
P(A or B) = P(A) + P(B) - P(A and B)
1/2 = 3/8 + 1/4 - P(A and B)
1/2 = 3/8 + 2/8 - P(A and B)
1/2 = 5/8 - P(A and B)
P(A and B) = 5/8 - 1/2 = 5/8 - 4/8 = 1/8
Q3. A and B are mutually exclusive events. If P(A) = 2/5 and P(B) = 1/3, find P(A or B).
Since A and B are mutually exclusive, P(A and B) = 0
P(A or B) = P(A) + P(B) = 2/5 + 1/3 = 6/15 + 5/15 = 11/15
Q4. If P(A) = 0.6, P(B) = 0.4, and P(A and B) = 0.2, find (i) P(not A and not B), (ii) P(A and not B).
(i) P(not A and not B) = P(neither A nor B)
= 1 - P(A or B)
= 1 - [P(A) + P(B) - P(A and B)]
= 1 - [0.6 + 0.4 - 0.2]
= 1 - 0.8
= 0.2
(ii) P(A and not B) = P(A) - P(A and B) = 0.6 - 0.2 = 0.4
Q5. A card is drawn from a well-shuffled pack of 52 cards. Find the probability that the card is either a king or a queen.
Drawing a king and drawing a queen are mutually exclusive events.
P(king) = 4/52, P(queen) = 4/52
P(king or queen) = P(king) + P(queen) = 4/52 + 4/52 = 8/52 = 2/13
Q6. In a class of 50 students, 30 like Maths, 25 like Science, and 10 like both. A student is selected at random. Find the probability that the student likes Maths or Science.
n(S) = 50
n(Maths) = 30, n(Science) = 25, n(both) = 10
P(Maths) = 30/50 = 3/5
P(Science) = 25/50 = 1/2
P(both) = 10/50 = 1/5
P(Maths or Science) = P(Maths) + P(Science) - P(both)
= 3/5 + 1/2 - 1/5
= 6/10 + 5/10 - 2/10
= 9/10
Q7. If P(A) = 1/2, P(B) = 1/3, and P(A or B) = 5/6, are A and B mutually exclusive events? Justify.
P(A and B) = P(A) + P(B) - P(A or B)
= 1/2 + 1/3 - 5/6
= 3/6 + 2/6 - 5/6
= 0
Since P(A and B) = 0, A and B are mutually exclusive events.
Q8. If A and B are events such that P(A) = 0.4, P(B) = 0.5, and P(A and B) = 0.2, find P(neither A nor B).
P(neither A nor B) = 1 - P(A or B)
P(A or B) = P(A) + P(B) - P(A and B) = 0.4 + 0.5 - 0.2 = 0.7
P(neither A nor B) = 1 - 0.7 = 0.3
Exercise 8.5 — Multiple Choice Questions
Q1. If the mean of a data set is 25 and SD is 5, then the coefficient of variation is
(a) 5% (b) 10% (c) 20% (d) 25%
C.V. = (SD/Mean) * 100 = (5/25) * 100 = 20%
Q2. The range of the data 8, 12, 6, 15, 10, 3 is
(a) 9 (b) 12 (c) 15 (d) 6
Range = Max - Min = 15 - 3 = 12
Q3. If P(A) = 0.35, then P(not A) is
(a) 0.35 (b) 0.65 (c) 1.35 (d) 0.55
P(not A) = 1 - P(A) = 1 - 0.35 = 0.65
Q4. Two events A and B are mutually exclusive if
(a) P(A and B) = 1 (b) P(A and B) = 0 (c) P(A or B) = 0 (d) P(A) = P(B)
Q5. When a die is rolled, the probability of getting a number less than 7 is
(a) 0 (b) 1/6 (c) 1/2 (d) 1
All outcomes {1,2,3,4,5,6} are less than 7. Favourable = 6, Total = 6.
P(< 7) = 6/6 = 1
Q6. If SD = 0, then the data is
(a) Very scattered (b) Perfectly consistent (c) Negatively skewed (d) Bimodal
Q7. If P(A) = 3/5 and A and B are independent events with P(B) = 1/2, then P(A and B) is
(a) 3/10 (b) 1/5 (c) 4/5 (d) 6/5
For independent events: P(A and B) = P(A) * P(B) = (3/5) * (1/2) = 3/10
Q8. The addition theorem of probability states that P(A or B) =
(a) P(A) + P(B) (b) P(A) + P(B) + P(A and B) (c) P(A) + P(B) - P(A and B) (d) P(A) * P(B)
Unit Exercise 8 — Advanced Mixed Problems
Q1. Find the mean, variance, and standard deviation of the following data: 12, 16, 20, 24, 28.
Mean = (12 + 16 + 20 + 24 + 28) / 5 = 100/5 = 20
x | x - x_bar | (x - x_bar)^2
12 | -8 | 64
16 | -4 | 16
20 | 0 | 0
24 | 4 | 16
28 | 8 | 64
Sum of (x - x_bar)^2 = 160
Variance = 160/5 = 32
SD = sqrt(32) = 4 * sqrt(2) = 5.66
Q2. Two dice are thrown together. Find the probability that the product of the numbers on the dice is (i) 12, (ii) a perfect square.
n(S) = 36
(i) Product = 12: (2,6), (3,4), (4,3), (6,2) → 4 outcomes
P(product = 12) = 4/36 = 1/9
(ii) Perfect square products: 1, 4, 9, 16, 25, 36
Product = 1: (1,1) → 1
Product = 4: (1,4), (2,2), (4,1) → 3
Product = 9: (1,9 not possible), (3,3) → 1
Product = 16: (2,8 no), (4,4) → 1
Product = 25: (5,5) → 1
Product = 36: (6,6) → 1
Total favourable = 1 + 3 + 1 + 1 + 1 + 1 = 8
P(perfect square) = 8/36 = 2/9
Q3. If A and B are two events such that P(A) = 1/3, P(B) = 1/4, and P(A and B) = 1/12, find (i) P(A or B), (ii) P(A and not B), (iii) P(neither A nor B).
(i) P(A or B) = P(A) + P(B) - P(A and B)
= 1/3 + 1/4 - 1/12
= 4/12 + 3/12 - 1/12
= 6/12 = 1/2
(ii) P(A and not B) = P(A) - P(A and B) = 1/3 - 1/12 = 4/12 - 1/12 = 3/12 = 1/4
(iii) P(neither A nor B) = 1 - P(A or B) = 1 - 1/2 = 1/2
Q4. A box contains 5 red balls, 3 green balls, and 4 blue balls. Two balls are drawn at random (without replacement). Find the probability that both are red.
Total balls = 5 + 3 + 4 = 12
n(S) = C(12, 2) = 66
Favourable: Drawing 2 red from 5 = C(5, 2) = 10
P(both red) = 10/66 = 5/33
Alternative method:
P(first red) = 5/12
P(second red | first red) = 4/11
P(both red) = (5/12) * (4/11) = 20/132 = 5/33
Q5. The mean and SD of 20 observations are 50 and 10 respectively. If each observation is increased by 5 and then multiplied by 3, find the new mean and new SD.
After adding 5: New Mean = 50 + 5 = 55, New SD = 10 (unchanged)
After multiplying by 3: Final Mean = 3 * 55 = 165, Final SD = 3 * 10 = 30
Q6. Three coins are tossed simultaneously. Find the probability of getting (i) exactly two heads, (ii) at least two heads, (iii) at most one head.
Sample space = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
n(S) = 8
(i) Exactly two heads: {HHT, HTH, THH} → 3 outcomes
P = 3/8
(ii) At least two heads: {HHH, HHT, HTH, THH} → 4 outcomes
P = 4/8 = 1/2
(iii) At most one head: {HTT, THT, TTH, TTT} → 4 outcomes
P = 4/8 = 1/2
Q7. A card is drawn from a well-shuffled deck. Find the probability that it is either a spade or an ace.
P(spade) = 13/52
P(ace) = 4/52
P(spade and ace) = 1/52 (the ace of spades is counted in both)
P(spade or ace) = P(spade) + P(ace) - P(spade and ace)
= 13/52 + 4/52 - 1/52
= 16/52
= 4/13
Q8. Find the SD of the first 10 natural numbers.
Mean of first 10 natural numbers = (1+2+3+...+10)/10 = 55/10 = 5.5
x | x - x_bar | (x - x_bar)^2
1 | -4.5 | 20.25
2 | -3.5 | 12.25
3 | -2.5 | 6.25
4 | -1.5 | 2.25
5 | -0.5 | 0.25
6 | 0.5 | 0.25
7 | 1.5 | 2.25
8 | 2.5 | 6.25
9 | 3.5 | 12.25
10 | 4.5 | 20.25
Sum = 82.50
Variance = 82.50/10 = 8.25
SD = sqrt(8.25) = 2.87
Alternatively, SD of first n natural numbers = sqrt((n^2 - 1)/12)
= sqrt((100 - 1)/12) = sqrt(99/12) = sqrt(8.25) = 2.87
Common Mistakes to Avoid
1. Forgetting the fx column in frequency tables — Always multiply each value by its frequency before summing. The mean is sum(fx)/sum(f), not sum(x)/number of rows.
2. Dividing by number of rows instead of total frequency — In a frequency table, N = sum(f), not the number of distinct values. This is the most common statistics error.
3. Mixing up variance and standard deviation — SD is the square root of variance. If the question asks for SD, you must take the square root at the end. If it asks for variance, do not take the square root.
4. Probability answers greater than 1 — Every probability must be between 0 and 1. If your answer exceeds 1, you have made an arithmetic or counting error.
5. Not simplifying probability fractions — Always simplify to lowest terms (e.g., 3/6 = 1/2, 4/52 = 1/13). Examiners deduct marks for unsimplified fractions.
6. Confusing mutually exclusive with independent events — Mutually exclusive means P(A and B) = 0. Independent means P(A and B) = P(A) * P(B). They are completely different concepts. Mutually exclusive events are never independent.
7. Double-counting in addition theorem — When events are not mutually exclusive, you must subtract P(A and B) in P(A or B) = P(A) + P(B) - P(A and B). Forgetting to subtract gives an inflated probability.
8. Ignoring the "without replacement" condition — In probability problems with drawing multiple items, "without replacement" means the denominator changes on each draw. "With replacement" means it stays the same.
9. Not checking if SD changes with shift/scale — Adding a constant does not change SD. Multiplying by a constant multiplies SD by the same factor. Students often change SD in both cases incorrectly.
10. Wrong card counting — A standard deck has 52 cards, not 54. Face cards are J, Q, K (12 total), not including aces. Each suit has exactly 13 cards.
Exam Checklist
- [ ] Mean formula for raw data and frequency tables memorised
- [ ] SD = sqrt(Variance) — always take the square root for SD
- [ ] C.V. = (SD/Mean) * 100 — used for comparing consistency
- [ ] Range and Coefficient of Range formulas ready
- [ ] Effect of adding a constant (Mean changes, SD unchanged) vs multiplying (both change)
- [ ] P(A) = favourable/total, always between 0 and 1
- [ ] P(not A) = 1 - P(A) memorised and verified
- [ ] Addition theorem: P(A or B) = P(A) + P(B) - P(A and B)
- [ ] Mutually exclusive: P(A and B) = 0, so P(A or B) = P(A) + P(B)
- [ ] Independent: P(A and B) = P(A) * P(B)
- [ ] Card deck composition: 52 cards, 4 suits, 13 per suit, 12 face cards, 4 aces
- [ ] Two dice: 36 outcomes, Two coins: 4 outcomes, Three coins: 8 outcomes
- [ ] Always simplify probability fractions
- [ ] Write all intermediate steps in statistics table calculations
- [ ] Verify probability answers are between 0 and 1
Frequently Asked Questions
Create columns for x, f, d = x - mean, d^2, and fd^2. Sum all fd^2 values, then SD = sqrt(sum fd^2 / N) for population SD or sqrt(sum fd^2 / (N-1)) for sample SD. For frequency tables, multiply each d^2 by its frequency.