Question 1

The mean of a dataset is 10 and the sum of squared deviations from the mean is 80. If there are 8 observations, the standard deviation is:

Accepted Answer

Step 1: Variance = (Sum of squared deviations) ÷ n = 80 ÷ 8 = 10. Step 2: Standard Deviation = √Variance = √10 ≈ 3.16.

Question 2

The variance of {1, 2, 3} is 2/3. What is the variance of {2, 4, 6} (each value doubled)?

Accepted Answer

Step 1: Recall the property: if each data point is multiplied by k, variance is multiplied by k². Step 2: Here k = 2, so new variance = (2/3) × 2² = (2/3) × 4 = 8/3.

Question 3

The standard deviation of the dataset {5, 5, 5, 5, 5} is:

Accepted Answer

Step 1: All values are identical (5). Step 2: Mean = 5. Step 3: Each deviation from mean = 5 − 5 = 0. Step 4: Sum of squared deviations = 0. Step 5: Variance = 0 ÷ 5 = 0. Step 6: Standard Deviation = √0 = 0.

Question 4

If the standard deviation of a dataset is 5, what is the variance?

Accepted Answer

Step 1: Recall that Variance = (Standard Deviation)². Step 2: Variance = 5² = 25.

Question 5

The variance of the dataset {2, 4, 6, 8, 10} is:

Accepted Answer

Step 1: Find the mean = (2 + 4 + 6 + 8 + 10) ÷ 5 = 30 ÷ 5 = 6. Step 2: Find squared deviations from mean: (2−6)² = 16, (4−6)² = 4, (6−6)² = 0, (8−6)² = 4, (10−6)² = 16. Step 3: Sum of squared deviations = 16 + 4 + 0 + 4 + 16 = 40. Step 4: Variance = 40 ÷ 5 = 8.

Question 6

A bag contains 5 red balls, 3 blue balls, and 2 green balls. If one ball is drawn at random, what is the probability that it is either red or green?

Accepted Answer

Step 1: Total number of balls = 5 + 3 + 2 = 10. Step 2: Number of red or green balls = 5 + 2 = 7. Step 3: Probability = Favourable outcomes / Total outcomes = 7/10 = 0.7. Therefore, the answer is 7/10.

Question 7

How many 2-digit numbers can be formed using the digits 3, 5, 7, and 9 without repetition?

Accepted Answer

Step 1: We need to form 2-digit numbers using 4 distinct digits without repetition. Step 2: For the first position (tens place), we have 4 choices. Step 3: For the second position (units place), we have 3 remaining choices (since repetition is not allowed). Step 4: Total numbers = 4 × 3 = 12. Alternatively, this is P(4,2) = 4!/(4-2)! = 4!/2! = 12. Therefore, the answer is 12.

Question 8

In how many ways can 4 red balls and 3 blue balls be arranged in a line?

Accepted Answer

Step 1: Total balls = 4 + 3 = 7. Step 2: We have 4 identical red balls and 3 identical blue balls. Step 3: Number of arrangements = 7!/(4! × 3!) = (7 × 6 × 5 × 4!)/(4! × 3 × 2 × 1) = (7 × 6 × 5)/6 = 210/6 = 35. Therefore, the answer is 35.

Question 9

A student must answer 5 questions out of 8 questions in an exam. In how many ways can the student choose the questions?

Accepted Answer

Step 1: The student must select 5 questions from 8 available questions. Step 2: The order of selection does not matter, so this is a combination problem. Step 3: Number of ways = C(8,5) = 8!/(5! × 3!) = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56. Therefore, the answer is 56.

Question 10

If the standard deviation of a dataset increases from 4 to 8, by what factor has the variance increased?

Accepted Answer

Step 1: Original variance = 4² = 16. Step 2: New variance = 8² = 64. Step 3: Factor of increase = 64 ÷ 16 = 4.

Question 11

In how many ways can 5 different books be arranged on a shelf?

Accepted Answer

Step 1: We need to arrange 5 different books in a line. Step 2: This is a permutation problem where all 5 objects are used. Step 3: Number of arrangements = 5! = 5 × 4 × 3 × 2 × 1 = 120. Therefore, the answer is 120.

Question 12

How many ways can a committee of 3 people be selected from a group of 8 people?

Accepted Answer

Step 1: We need to select 3 people from 8 where order does not matter. Step 2: This is a combination problem. Step 3: Number of ways = C(8,3) = 8!/(3! × 5!) = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56. Therefore, the answer is 56.

Question 13

In how many ways can the letters of the word 'APPLE' be arranged?

Accepted Answer

Step 1: The word 'APPLE' has 5 letters total. Step 2: The letter 'P' appears 2 times (repeated), while A, L, E appear once each. Step 3: Number of arrangements = 5!/(2!) = (5 × 4 × 3 × 2 × 1)/2 = 120/2 = 60. Therefore, the answer is 60.

Question 14

How many 4-digit numbers can be formed using the digits 2, 3, 5, 7, 8, 9 without repetition, such that the number is even?

Accepted Answer

Step 1: For a number to be even, it must end in an even digit. From {2, 3, 5, 7, 8, 9}, the even digits are 2 and 8 (2 choices). Step 2: Fix one even digit at the units place. Step 3: Now we need to fill the remaining 3 positions (thousands, hundreds, tens) with 5 remaining digits. Step 4: This can be done in P(5,3) = 5!/(5-3)! = 5!/2! = 60 ways. Step 5: Total = 2 × 60 = 120.

Question 15

In how many ways can 5 different books be arranged on a shelf such that two specific books (A and B) are always together?

Accepted Answer

Step 1: Treat the two specific books (A and B) as a single unit. Now we have 4 units to arrange (the combined unit + 3 other books). Step 2: These 4 units can be arranged in 4! = 24 ways. Step 3: Within the combined unit, books A and B can be arranged in 2! = 2 ways. Step 4: Total arrangements = 4! × 2! = 24 × 2 = 48.

Question 16

A committee of 4 people is to be selected from 6 men and 5 women. In how many ways can this be done if the committee must have at least 2 women?

Accepted Answer

Step 1: 'At least 2 women' means 2 women or 3 women or 4 women. Step 2: Case 1 (2 women, 2 men): C(5,2) × C(6,2) = 10 × 15 = 150. Step 3: Case 2 (3 women, 1 man): C(5,3) × C(6,1) = 10 × 6 = 60. Step 4: Case 3 (4 women, 0 men): C(5,4) × C(6,0) = 5 × 1 = 5. Step 5: Total = 150 + 60 + 5 = 215.

Question 17

In how many ways can the letters of the word 'MISSISSIPPI' be arranged?

Accepted Answer

Step 1: Count the total letters in MISSISSIPPI: M(1), I(4), S(4), P(2) = 11 letters total. Step 2: When letters repeat, the formula is n! / (n₁! × n₂! × ... × nₖ!), where n is total letters and n₁, n₂, etc. are frequencies of each repeated letter. Step 3: Arrangements = 11! / (1! × 4! × 4! × 2!) = 39,916,800 / (1 × 24 × 24 × 2) = 39,916,800 / 1,152 = 34,650.

Question 18

From a group of 8 students, in how many ways can a President, Vice-President, and Secretary be selected such that no student holds more than one position?

Accepted Answer

Step 1: This is a permutation problem because the positions are distinct (order matters). Step 2: We need to select 3 students from 8 for 3 different positions. Step 3: President can be chosen in 8 ways, Vice-President in 7 ways (from remaining), Secretary in 6 ways (from remaining). Step 4: Total = 8 × 7 × 6 = 336. Alternatively, P(8,3) = 8!/(8-3)! = 8!/5! = 336.

Question 19

A password consists of 2 letters followed by 3 digits. If letters can be repeated but digits cannot, how many different passwords are possible? (Consider 26 letters and 10 digits: 0–9)

Accepted Answer

Step 1: The password has 2 letter positions and 3 digit positions. Step 2: For the 2 letters (repetition allowed): First letter has 26 choices, second letter has 26 choices = 26 × 26 = 676. Step 3: For the 3 digits (no repetition): First digit has 10 choices, second digit has 9 choices (one used), third digit has 8 choices (two used) = 10 × 9 × 8 = 720. Step 4: Total passwords = 676 × 720 = 486,720.

Question 20

A student has a 70% chance of passing Mathematics and a 60% chance of passing English. Assuming independence, what is the probability that the student passes at least one subject?

Accepted Answer

Step 1: P(Math) = 0.7, P(English) = 0.6. Step 2: P(at least one) = 1 - P(neither). Step 3: P(fail Math) = 1 - 0.7 = 0.3, P(fail English) = 1 - 0.6 = 0.4. Step 4: P(neither) = 0.3 × 0.4 = 0.12. Step 5: P(at least one) = 1 - 0.12 = 0.88 = 22/25.

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