Statistics and probability — SSC CGL Quant Notes

Concepts and rules

Mean (arithmetic average) of n values = sum of the values ÷ n. The sum = mean × n, which is the form used in most SSC questions.

Adding a constant k to every value adds k to the mean, median and mode. Multiplying every value by k multiplies all three by k.

Combined mean of 2 groups = (n1 × m1 + n2 × m2) ÷ (n1 + n2). It lies between m1 and m2 and is closer to the mean of the larger group.

Corrected mean: if a value a was wrongly taken as b, the correct sum = wrong sum − b + a; divide by n again.

Median: arrange the values in order. If n is odd, the median is the ((n + 1)/2)th value; if n is even, it is the average of the (n/2)th and (n/2 + 1)th values.

Mode: the value that occurs most often. A data set can have 1 mode, more than 1 mode, or none if every value occurs equally often.

Grouped data: mean = Σfx ÷ Σf, where x is the class mark (mid-point) of each class and f its frequency.

Grouped median = l + ((N/2 − cf) ÷ f) × h, where the median class is the first class whose cumulative frequency reaches N/2, l its lower limit, cf the cumulative frequency before it, f its frequency and h its width.

Grouped mode = l + ((f1 − f0) ÷ (2f1 − f0 − f2)) × h, where the modal class has the highest frequency f1, and f0 and f2 are the frequencies of the classes before and after it.

Empirical relation: Mode ≈ 3 Median − 2 Mean. It holds only approximately, for moderately skewed data; use it only when the question asks for an approximate value from the other 2.

Method and shortcuts

Step 1: Convert every average into a total (mean × count) before adding, removing or correcting values.

Step 2: For the median, always sort first and count the values; check whether n is odd or even.

Step 3: For grouped data, make a cumulative frequency column; it locates the median class at once.

Shortcut: assumed mean. Take a convenient value A, find the average of the deviations (x − A), and add it to A. This keeps numbers small.

Shortcut: the mean of consecutive integers or of an arithmetic progression is the average of the first and last terms; so is its median.

Example 1: Mean, median and mode of a list

Question: Find the mean, median and mode of 7, 3, 9, 12, 5, 9, 4.

Solution: Sum = 49 and n = 7, so mean = 7. Sorted: 3, 4, 5, 7, 9, 9, 12; the 4th value is 7. The value 9 occurs twice, more than any other.

Answer: Mean 7, median 7, mode 9.

Example 2: Median of an even number of values

Question: Find the median of 12, 7, 15, 9, 20, 11.

Solution: Sorted: 7, 9, 11, 12, 15, 20. With 6 values, the median is the average of the 3rd and 4th: (11 + 12)/2 = 11.5.

Answer: 11.5

Example 3: Correcting a wrongly recorded value

Question: The mean of 40 observations was 52. It was later found that 87 had been recorded as 57. Find the correct mean.

Solution: Wrong sum = 40 × 52 = 2080. Correct sum = 2080 − 57 + 87 = 2110. Correct mean = 2110 ÷ 40 = 52.75.

Answer: 52.75

Example 4: Combined mean of 2 sections

Question: Section A has 30 students with average marks 64 and section B has 45 students with average marks 54. What is the average of all 75 students? (a) 59 (b) 58 (c) 60 (d) 57

Solution: Total = 30 × 64 + 45 × 54 = 1920 + 2430 = 4350, and 4350 ÷ 75 = 58. The simple average of 64 and 54 is 59, which ignores the larger size of B.

Answer: (b) 58

Example 5: Median of grouped data

Question: Classes and frequencies: 0–10: 5, 10–20: 8, 20–30: 12, 30–40: 10, 40–50: 5. Find the median. (a) 24.17 (b) 25 (c) 25.83 (d) 26.5

Solution: N = 40, so N/2 = 20. Cumulative frequencies: 5, 13, 25, 35, 40. The first to reach 20 is 25, so the median class is 20–30 with l = 20, cf = 13, f = 12, h = 10.

Median = 20 + ((20 − 13) ÷ 12) × 10 = 20 + 70/12 = 20 + 5.83 = 25.83.

Answer: (c) 25.83

Example 6: Grouped mode versus the empirical formula

Question: For the data of Example 5, find the mode using the grouped-data formula. (a) 26.5 (b) 25.5 (c) 27.5 (d) 26.67

Solution: The modal class is 20–30 (frequency 12), with f0 = 8 and f2 = 10. Mode = 20 + ((12 − 8) ÷ (24 − 8 − 10)) × 10 = 20 + (4/6) × 10 = 26.67.

Compare: the mean is Σfx ÷ Σf = (25 + 120 + 300 + 350 + 225) ÷ 40 = 1020 ÷ 40 = 25.5, and 3 × 25.83 − 2 × 25.5 = 26.5. The empirical value 26.5 is close but is not the formula answer.

Answer: (d) 26.67

Common mistakes

Finding the median without sorting. The middle of the list as written is not the median.

Taking the simple average of 2 group means when the groups differ in size. Weight each mean by its group size.

In the grouped median, using the cumulative frequency of the median class itself as cf. cf is the total before that class.

Using Mode = 3 Median − 2 Mean as an exact formula when the grouped data are given. Use the class formula; the relation is approximate.

Correcting a mean by adding the error to the mean instead of to the sum. The error changes the sum; the mean changes by error ÷ n.

Practice questions

Q1. The mean of 5 numbers is 18. When one number is removed, the mean of the remaining numbers is 16. Which number was removed? (a) 24 (b) 26 (c) 28 (d) 2

Q2. What is the median of the first 10 prime numbers? (a) 11 (b) 13 (c) 12 (d) 12.9

Q3. For a moderately skewed distribution the mean is 24 and the median is 26. What is the approximate mode? (a) 28 (b) 25 (c) 32 (d) 30

Answers

Q1. (b) Sum of 5 = 90 and sum of 4 = 64, so the removed number is 90 − 64 = 26.

Q2. (c) The primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29; median = (11 + 13)/2 = 12. Option (d) 12.9 is their mean.

Q3. (d) Mode ≈ 3 × 26 − 2 × 24 = 78 − 48 = 30.

Quick revision

Sum = mean × n; combined mean = (n1m1 + n2m2) ÷ (n1 + n2); corrected sum = sum − wrong + right.

Median: sort first; odd n gives the middle value, even n the average of the middle 2.

Grouped median l + ((N/2 − cf)/f)h; grouped mode l + ((f1 − f0)/(2f1 − f0 − f2))h.

Mode ≈ 3 Median − 2 Mean is approximate; adding k shifts all 3 measures by k.

Concepts and rules

Measures of dispersion show how spread out the values are. 2 data sets can have the same mean and very different spreads.

Range = largest value − smallest value. Coefficient of range = (largest − smallest) ÷ (largest + smallest), a pure number used to compare spreads.

Mean deviation about the mean = Σ|x − mean| ÷ n, the average distance of the values from the mean, ignoring signs.

Variance σ² = Σ(x − mean)² ÷ n = (Σx² ÷ n) − mean². Standard deviation σ = √variance, in the same unit as the data.

These notes divide by n (population variance), as in school textbooks and SSC questions. Some books divide by n − 1 for a sample; use that only if the question says so.

Adding a constant to every value changes the mean but not the range, mean deviation, variance or SD. Multiplying every value by k multiplies range and SD by |k| and variance by k².

Coefficient of variation CV = σ ÷ mean × 100. The data set with the smaller CV is more consistent (less variable relative to its mean).

Useful results: SD of 2 values a and b = |a − b|/2. Variance of the first n natural numbers = (n² − 1)/12. If all values are equal, SD = 0.

Combined variance of 2 groups: σ² = [n1(σ1² + d1²) + n2(σ2² + d2²)] ÷ (n1 + n2), where d1 and d2 are the differences between each group mean and the combined mean.

Method and shortcuts

Step 1: Find the mean first. Then write each deviation (x − mean) in a row; their sum must be 0, which checks your mean.

Step 2: Square the deviations, add them and divide by n for the variance; take the square root for the SD.

Step 3: When the mean is not a whole number, use variance = Σx²/n − mean² instead of deviations.

Shortcut: for a change of origin and scale y = ax + b, SD of y = |a| × SD of x; ignore b completely.

Shortcut: to compare consistency, compare CV, not SD, when the means differ.

Example 1: Range and its coefficient

Question: Find the range and the coefficient of range of 23, 41, 17, 35, 29.

Solution: Largest = 41, smallest = 17. Range = 41 − 17 = 24. Coefficient = 24 ÷ (41 + 17) = 24/58 = 12/29, about 0.41.

Answer: Range 24; coefficient 12/29.

Example 2: Mean deviation

Question: Find the mean deviation about the mean of 4, 7, 8, 9, 12.

Solution: Mean = 40 ÷ 5 = 8. Distances from 8: 4, 1, 0, 1, 4, with sum 10. Mean deviation = 10 ÷ 5 = 2.

Answer: 2

Example 3: Variance and SD

Question: Find the variance and standard deviation of 2, 4, 4, 4, 5, 5, 7, 9.

Solution: Mean = 40 ÷ 8 = 5. Squared deviations: 9, 1, 1, 1, 0, 0, 4, 16, with sum 32. Variance = 32 ÷ 8 = 4, so SD = 2.

Answer: Variance 4, SD 2.

Example 4: Change of scale and origin

Question: The SD of a data set is 3. Each value is multiplied by 4 and then 5 is added. What is the new variance? (a) 12 (b) 17 (c) 144 (d) 149

Solution: Adding 5 has no effect. Multiplying by 4 makes the SD 4 × 3 = 12, so the variance is 12² = 144. Option (a) is the new SD, not the variance.

Answer: (c) 144

Example 5: Which player is more consistent

Question: Batsman A has mean score 48 with SD 12. Batsman B scored 30, 42, 30, 42 in 4 innings. Who is more consistent, and what is that batsman's CV? (a) A, CV 25% (b) B, CV 16.67% (c) A, CV 16.67% (d) B, CV 25%

Solution: B's mean = 144 ÷ 4 = 36; each score is 6 away from 36, so B's variance = 36 and SD = 6.

CV of A = 12 ÷ 48 × 100 = 25%; CV of B = 6 ÷ 36 × 100 = 16.67%. B has the smaller CV, so B is more consistent.

Answer: (b) B, CV 16.67%

Example 6: Combined variance

Question: Group 1 has 20 values with mean 50 and SD 4; group 2 has 30 values with mean 60 and SD 6. What is the variance of all 50 values together? (a) 40 (b) 26 (c) 49 (d) 52

Solution: Combined mean = (20 × 50 + 30 × 60) ÷ 50 = 2800 ÷ 50 = 56, so d1 = 50 − 56 = −6 and d2 = 60 − 56 = 4.

Variance = [20(16 + 36) + 30(36 + 16)] ÷ 50 = (1040 + 1560) ÷ 50 = 2600 ÷ 50 = 52.

Averaging the 2 variances (16 and 36, weighted to 28) ignores the gap between the group means; that gap adds the d² terms.

Answer: (d) 52

Common mistakes

Reporting the SD when the variance is asked, or the reverse. Variance is in squared units; SD is in the original units.

Thinking that adding a constant increases the SD. A shift moves every value equally, so the spread is unchanged.

Adding the signed deviations for mean deviation; they always add to 0. Use absolute values.

Dividing by n − 1 when the question expects n, which turns a variance of 8.25 for 1 to 10 into about 9.17.

Comparing consistency by SD alone when the means are different. Use CV.

Practice questions

Q1. What is the variance of the first 10 natural numbers? (a) 8.25 (b) 9.17 (c) 5.5 (d) 2.87

Q2. What is the standard deviation of the 2 values 6 and 14? (a) 8 (b) 2 (c) 4 (d) 16

Q3. A data set has mean 40 and coefficient of variation 15%. What is its variance? (a) 6 (b) 36 (c) 2.25 (d) 60

Answers

Q1. (a) (n² − 1)/12 with n = 10 gives 99/12 = 8.25.

Q2. (c) Mean = 10 and each value is 4 away, so SD = |14 − 6|/2 = 4.

Q3. (b) SD = 15% of 40 = 6, so variance = 36.

Quick revision

Range = max − min; coefficient = (max − min)/(max + min); mean deviation = Σ|x − mean|/n.

Variance = Σ(x − mean)²/n = Σx²/n − mean²; SD = √variance.

y = ax + b: SD multiplies by |a|, variance by a², b has no effect.

CV = SD/mean × 100; smaller CV means more consistent.

Concepts and rules

An experiment has a set of possible outcomes called the sample space. If all outcomes are equally likely, P(E) = number of outcomes in E ÷ total number of outcomes.

0 ≤ P(E) ≤ 1. An impossible event has probability 0 and a certain event has probability 1. P(not E) = 1 − P(E).

Coins: n coins (or 1 coin tossed n times) give 2ⁿ equally likely outcomes. Dice: 1 die gives 6 outcomes, 2 dice give 6 × 6 = 36 ordered outcomes.

Cards: a pack has 52 cards in 4 suits of 13 (spades and clubs black, hearts and diamonds red), so 26 red and 26 black. There are 4 aces and 12 face cards (jack, queen, king of each suit).

Addition rule: P(A or B) = P(A) + P(B) − P(A and B). If A and B cannot happen together (mutually exclusive), P(A or B) = P(A) + P(B).

Independent events: one does not affect the other, and P(A and B) = P(A) × P(B). Drawing with replacement keeps draws independent; without replacement it does not.

At least one: P(at least one) = 1 − P(none). For independent events with probabilities p1, p2, p3, P(at least one happens) = 1 − (1 − p1)(1 − p2)(1 − p3).

Selections: the number of ways to choose r items from n is C(n, r) = n!/(r!(n − r)!). For example C(5, 2) = 10 and C(12, 2) = 66.

Odds in favour of E are a : b means P(E) = a/(a + b); odds against E are b : a.

Method and shortcuts

Step 1: Write the total number of equally likely outcomes (2ⁿ, 36, 52, C(n, r)).

Step 2: Count the favourable outcomes by listing them when they are few (sums on 2 dice) or by C(n, r) when choosing.

Step 3: For 'at least', 'not' or 'none' questions, the complement is usually shorter: count what you do not want and subtract from 1.

Step 4: For 'or', check for overlap before adding; for 'and', check independence before multiplying.

Shortcut for 2 dice: the number of ways to get a sum s is s − 1 for s from 2 to 7 and 13 − s for s from 7 to 12.

Example 1: 2 coins

Question: 2 coins are tossed. Find the probability of getting at least 1 head.

Solution: Outcomes HH, HT, TH, TT; only TT has no head. P = 1 − 1/4 = 3/4.

Answer: 3/4

Example 2: Sum on 2 dice

Question: 2 dice are thrown. Find the probability that the sum is 8.

Solution: Favourable: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), that is 5 of the 36 outcomes. Shortcut: 13 − 8 = 5.

Answer: 5/36

Example 3: A king or a heart

Question: 1 card is drawn from a well-shuffled pack of 52. Find the probability that it is a king or a heart.

Solution: Kings 4, hearts 13, and the king of hearts is counted in both. P = 4/52 + 13/52 − 1/52 = 16/52 = 4/13.

Answer: 4/13

Example 4: 2 balls without replacement

Question: A bag has 5 red, 4 blue and 3 green balls. 2 balls are drawn at random together. What is the probability that both are red? (a) 25/144 (b) 5/33 (c) 5/12 (d) 10/33

Solution: Total ways = C(12, 2) = 66; ways to pick 2 red = C(5, 2) = 10. P = 10/66 = 5/33. Check by steps: 5/12 × 4/11 = 20/132 = 5/33.

Option (a) 25/144 would be the answer with replacement; drawing together means without replacement.

Answer: (b) 5/33

Example 5: Overlapping events on 2 dice

Question: 2 dice are thrown. What is the probability that the product of the numbers is even or their sum is 7? (a) 11/12 (b) 2/3 (c) 3/4 (d) 5/6

Solution: The product is odd only when both numbers are odd: 3 × 3 = 9 outcomes. So P(product even) = 1 − 9/36 = 27/36.

A sum of 7 always pairs an odd number with an even number, so all 6 of those outcomes already have an even product. The overlap is the whole event.

P = 27/36 + 6/36 − 6/36 = 27/36 = 3/4. Adding without the overlap gives 33/36 = 11/12, the trap in option (a).

Answer: (c) 3/4

Example 6: At least one of 3 solves it

Question: A, B and C try a problem independently. Their chances of solving it are 1/3, 1/4 and 1/5. What is the probability that the problem is solved? (a) 3/5 (b) 47/60 (c) 2/5 (d) 1/60

Solution: The problem stays unsolved only if all 3 fail: (2/3) × (3/4) × (4/5) = 24/60 = 2/5.

So P(solved) = 1 − 2/5 = 3/5. Adding 1/3 + 1/4 + 1/5 = 47/60 counts the cases where 2 or 3 of them succeed more than once; option (c) is the probability of failure.

Answer: (a) 3/5

Common mistakes

Counting (2, 6) and (6, 2) as one outcome. With 2 dice there are 36 ordered outcomes, and both orders count.

Adding probabilities of overlapping events. Subtract P(A and B) unless the events cannot happen together.

Multiplying probabilities of dependent draws as if independent: without replacement the second draw has 1 ball fewer.

Forgetting that a pack has 12 face cards, not 16; aces are not face cards.

Answering a probability greater than 1 or negative. Any such result means an error in counting.

Practice questions

Q1. 3 coins are tossed together. What is the probability of getting exactly 2 heads? (a) 1/2 (b) 3/8 (c) 1/4 (d) 7/8

Q2. A number is chosen at random from 1 to 50. What is the probability that it is divisible by 4 or 6? (a) 2/5 (b) 7/25 (c) 9/25 (d) 8/25

Q3. 2 cards are drawn together from a pack of 52. What is the probability that both are aces? (a) 1/221 (b) 1/169 (c) 2/221 (d) 1/13

Answers

Q1. (b) 8 outcomes; exactly 2 heads: HHT, HTH, THH, so 3/8.

Q2. (d) Multiples of 4: 12, of 6: 8, of both (of 12): 4. Favourable = 12 + 8 − 4 = 16, so 16/50 = 8/25.

Q3. (a) C(4, 2)/C(52, 2) = 6/1326 = 1/221.

Quick revision

P = favourable ÷ total; P(not E) = 1 − P(E); at least one = 1 − P(none).

P(A or B) = P(A) + P(B) − P(A and B); independent: P(A and B) = P(A)P(B).

2 dice: 36 outcomes; sum s has s − 1 ways up to 7 and 13 − s after 7.

Cards: 52, 4 suits of 13, 26 red, 4 aces, 12 face cards; 'together' means without replacement.