Statistics and Basic Probability

This chapter answers the Mathematics heading of the RRB computer-based tests that covers elementary statistics and probability: mean, median and mode for ungrouped and grouped data, range, mean deviation, variance and standard deviation, and probability of simple events. It is used by RRB Level-1, NTPC and Technician papers, where the questions are short numericals and 'which measure' items. It teaches each formula with its meaning, a worked example with exact answers, the shortcuts that save seconds, and the traps that swap two similar rules.

Data, frequency and class intervals

Raw data is a list of observations. Ungrouped data lists each value; the frequency of a value is how many times it occurs. When the values are many, they are put into class intervals such as 0 to 10, 10 to 20, giving grouped data. The class width (size) h is upper limit minus lower limit, and the class mark (midpoint) is (lower + upper) / 2. In grouped data every observation in a class is treated as if it sat at the class mark, which is why grouped answers are estimates.

Use the cumulative frequency (cf), the running total of frequencies, for the median. In the standard 'inclusive-exclusive' tables used in these papers the classes are continuous (0 to 10, 10 to 20), so the upper limit of one class is the lower limit of the next, and a value on the boundary is counted in the higher class.

Term	Meaning	Example for class 20 to 30
Class width h	upper − lower	10
Class mark	(lower + upper) / 2	25
Cumulative frequency	frequencies added up to and including the class	depends on table

Mean: the balancing point

The arithmetic mean is the sum of observations divided by their number: mean = Σx / n. With frequencies it is Σfx / Σf. For grouped data replace x by the class mark. The mean is the balancing point of the data: the deviations of the observations from the mean add up to zero.

Worked example (grouped). Classes 0 to 10, 10 to 20, 20 to 30, 30 to 40, 40 to 50 have frequencies 5, 8, 10, 12, 5. Class marks are 5, 15, 25, 35, 45. Σf = 40. Σfx = 25 + 120 + 250 + 420 + 225 = 1040. Mean = 1040 / 40 = 26.

Combined mean: (n₁m₁ + n₂m₂) / (n₁ + n₂). 20 students with mean 60 and 30 students with mean 70 give (1200 + 2100) / 50 = 66, not 65.
Changing one value: the total changes by the difference. Mean of 10 numbers is 15; a 12 is replaced by 22, the total rises by 10, so the new mean is 16.
Shift and scale: adding k to every observation adds k to the mean; multiplying every observation by k multiplies the mean by k.
Mean of the first n natural numbers is (n + 1) / 2; of the first n even numbers is n + 1.

🧠 Assumed mean saves time

Pick a convenient assumed mean a near the middle, find the average of the small deviations (x − a), then add a back. For 98, 102, 101, 99, 100 take a = 100: deviations −2, 2, 1, −1, 0 sum to 0, so the mean is 100.

Median and mode

The median is the middle value after arranging the data in order. For n observations: if n is odd it 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. Always sort first. For 9, 3, 15, 7, 11, 5 the sorted list is 3, 5, 7, 9, 11, 15 and the median is (7 + 9) / 2 = 8. The mode is the value that occurs most often; 3, 5, 5, 7, 8, 5, 9 has mode 5. A data set can have no mode or more than one.

Grouped median: median = l + ((n / 2 − cf) / f) × h, where l is the lower limit of the median class (the class containing the (n / 2)-th observation), cf the cumulative frequency of the class before it, f its frequency and h its width. For the table above (frequencies 5, 8, 10, 12, 5): n = 40, n / 2 = 20; cumulative frequencies 5, 13, 23, 35, 40, so the median class is 20 to 30 with cf = 13, f = 10. Median = 20 + (7 / 10) × 10 = 27.

Grouped mode: mode = l + ((f₁ − f₀) / (2f₁ − f₀ − f₂)) × h, where the modal class has the highest frequency f₁, with f₀ before it and f₂ after it. Classes 0 to 10, 10 to 20, 20 to 30, 30 to 40 with frequencies 3, 5, 8, 5: modal class 20 to 30, f₁ = 8, f₀ = 5, f₂ = 5. Mode = 20 + (3 / 6) × 10 = 25.

⚠️ Do not confuse the two class formulas

The median uses n / 2 and cumulative frequency; the mode uses only the three neighbouring frequencies. Quick check: the answer must lie inside the chosen class, otherwise the class was picked wrongly.

ℹ️ Empirical relation

For a moderately skewed distribution, mode = 3 × median − 2 × mean. With mean 24 and median 25 the estimated mode is 75 − 48 = 27. It is an approximation, so questions say 'empirical' or 'approximately'.

Measures of spread: range, mean deviation, variance, SD

Two sets can share a mean and still differ in how scattered they are. Range = largest − smallest (for 12, 7, 25, 3, 18 it is 25 − 3 = 22). Mean deviation about the mean = Σ|x − mean| / n. Variance σ² = Σ(x − mean)² / n = Σx² / n − (mean)². Standard deviation σ = √variance, in the same unit as the data. The tables here treat the data as a whole population, so divide by n.

Worked example. Data 2, 4, 4, 4, 5, 5, 7, 9. n = 8, sum 40, mean 5. Deviations −3, −1, −1, −1, 0, 0, 2, 4. Mean deviation = (3 + 1 + 1 + 1 + 0 + 0 + 2 + 4) / 8 = 12 / 8 = 1.5. Squares 9, 1, 1, 1, 0, 0, 4, 16 add to 32, so variance = 32 / 8 = 4 and SD = 2. Check by the shortcut: Σx² = 4 + 48 + 50 + 49 + 81 = 232; 232 / 8 − 25 = 29 − 25 = 4.

Effect of shift and scale
Change to every observation	Mean	Range, mean deviation, SD	Variance
Add k	+ k	unchanged	unchanged
Multiply by k	× k	× \|k\|	× k²

⚠️ Variance is squared

If every value is doubled, the SD doubles but the variance becomes four times. For y = 3x + 5 with mean 20 and SD 4 for x: mean of y = 65 and SD of y = 12; the +5 shifts the mean but never the spread.

🧠 First n natural numbers

Variance of 1, 2, ..., n is (n² − 1) / 12. For 1 to 5 it is 24 / 12 = 2; for 1 to 7 it is 48 / 12 = 4, so the SD is 2.

Probability of simple events

Probability of an event E = (number of favourable outcomes) / (total equally likely outcomes). It lies from 0 (impossible) to 1 (certain). The complement has probability P(not E) = 1 − P(E). Bag with 5 red, 3 blue, 2 green balls: P(blue) = 3 / 10 and P(not red) = 1 − 5 / 10 = 1 / 2.

Standard sample spaces
Experiment	Total outcomes	Event and probability
One die	6	prime (2, 3, 5): 3/6 = 1/2
Two dice	36	sum 7: 6/36 = 1/6; sum 8: 5/36
Three coins	8	exactly two heads (HHT, HTH, THH): 3/8
Pack of 52 cards	52	king: 4/52 = 1/13; red face card: 6/52 = 3/26

A pack has 4 suits of 13 cards, 26 red and 26 black cards, and 12 face cards (king, queen, jack in each suit). Two rules combine events: for mutually exclusive events P(A or B) = P(A) + P(B); in general P(A or B) = P(A) + P(B) − P(A and B). For independent events P(A and B) = P(A) × P(B). If P(A) = 1/2 and P(B) = 1/3 are independent, P(both) = 1/6 and P(at least one) = 1 − (1/2)(2/3) = 2/3. Without replacement the second draw changes: two aces from a pack = 4/52 × 3/51 = 1/221.

🎯 Why add for 'or', multiply for 'and'

'Or' with no overlap simply collects favourable outcomes from the same sample space, so the counts add. 'And' for independent events needs a pair of outcomes, one from each experiment, so the number of pairs multiplies: 6 faces times 6 faces gives the 36 outcomes of two dice.

Choosing the measure and avoiding traps

Mean uses every value and is pulled by extreme values; use it for fair totals.
Median is not disturbed by extreme values; use it for skewed data such as salaries.
Mode is the most common value; the only average that also works for categories such as the commonest shoe size.
Range, mean deviation, variance and SD are measures of dispersion; mean, median and mode are measures of central tendency.

⚠️ Common slips

Median of unsorted data; forgetting to halve for an even count; using class limits instead of class marks for the grouped mean; taking SD as variance; and giving a probability above 1 (check that favourable outcomes never exceed total). For 'at least one' always try 1 − P(none).

Speed plan: for mean, look for a round assumed mean; for median, sort and count before any formula; for spread, test which of range, mean deviation or SD the stem names; for probability, write favourable over total and simplify last.

Key takeaways

Mean = Σfx / Σf (class marks for grouped data); the deviations from the mean add to zero, and combined mean = (n₁m₁ + n₂m₂) / (n₁ + n₂).
Median: sort first, middle value (average of two middle values for even n); grouped median = l + ((n/2 − cf)/f) × h.
Grouped mode = l + ((f₁ − f₀)/(2f₁ − f₀ − f₂)) × h; empirical relation: mode = 3 median − 2 mean.
Variance = Σx²/n − mean²; SD = √variance. Adding k changes only the mean; multiplying by k multiplies SD by |k| and variance by k².
Probability = favourable / total, between 0 and 1; P(not E) = 1 − P(E); add for exclusive 'or', multiply for independent 'and'.
Two dice: 36 outcomes, sum 7 has 6 ways; pack of 52 has 26 red cards, 12 face cards and 4 of each rank.

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

What is the mean of 12, 15, 18, 20 and 25?
1. 17
2. 18
3. 19
4. 20
Show answer
Answer: B — 18
The sum is 12 + 15 + 18 + 20 + 25 = 90 and 90 / 5 = 18. The value 20 is the second-largest observation, not the mean, and 19 comes from a miscount of the total.
What is the median of 9, 3, 15, 7, 11 and 5?
1. 9
2. 7
3. 10
4. 8
Show answer
Answer: D — 8
Sorted the data is 3, 5, 7, 9, 11, 15; with six values the median is the average of the 3rd and 4th, (7 + 9) / 2 = 8. Taking 7 or 9 alone ignores the even count, and reading the unsorted middle gives a wrong pair.
Which value is the mode of 3, 5, 5, 7, 8, 5, 9?
1. 5
2. 7
3. 6
4. 8
Show answer
Answer: A — 5
The mode is the most frequent value, and 5 appears three times while every other value appears once. The value 6 is the mean of the list and 7 is merely the middle entry of the unsorted list; neither is the most frequent value, which tempts candidates who mix up the three averages.
The mean marks of 20 students is 60 and of another 30 students is 70. What is the mean marks of all 50 students?
1. 65
2. 67
3. 66
4. 68
Show answer
Answer: C — 66
Total marks are 20 × 60 + 30 × 70 = 1200 + 2100 = 3300 and 3300 / 50 = 66. The simple average of 60 and 70, which is 65, ignores that the larger group has more weight.
Classes 0 to 10, 10 to 20, 20 to 30, 30 to 40, 40 to 50 have frequencies 5, 8, 10, 12 and 5. What is the mean using class marks?
1. 26
2. 25
3. 27
4. 28
Show answer
Answer: A — 26
Σfx = 5 × 5 + 8 × 15 + 10 × 25 + 12 × 35 + 5 × 45 = 1040 and Σf = 40, so the mean is 26. The middle class mark 25 is not the mean, and 27 is the median of the same table, not the mean.
Classes 0 to 10, 10 to 20, 20 to 30, 30 to 40 and 40 to 50 have frequencies 5, 8, 10, 12 and 5. What is the median of the grouped data?
1. 26
2. 25
3. 32
4. 27
Show answer
Answer: D — 27
n = 40 and n / 2 = 20; the cumulative frequencies 5, 13, 23 make 20 to 30 the median class with cf = 13 and f = 10, so median = 20 + (7 / 10) × 10 = 27. The value 26 is the mean, not the median.
Classes 0 to 10, 10 to 20, 20 to 30, 30 to 40 have frequencies 3, 5, 8 and 5. What is the mode of the grouped data?
1. 24
2. 26
3. 25
4. 30
Show answer
Answer: C — 25
The modal class is 20 to 30 with f₁ = 8, f₀ = 5, f₂ = 5, so mode = 20 + (8 − 5) / (16 − 5 − 5) × 10 = 25. The neighbours are equal, so the mode sits at the class mark 25; the class mark of the next class, 35, is not the answer either.
For a moderately skewed distribution the mean is 24 and the median is 25. Using the empirical relation, what is the approximate mode?
1. 26
2. 27
3. 22
4. 25
Show answer
Answer: B — 27
Mode ≈ 3 × median − 2 × mean = 75 − 48 = 27. The value 22 results from swapping the multipliers (3 × mean − 2 × median = 72 − 50 = 22), which is the common slip.
Which of the following is NOT a measure of dispersion?
1. Median
2. Range
3. Variance
4. Mean deviation
Show answer
Answer: A — Median
Range, variance and mean deviation all describe how scattered the data are. The median is a measure of central tendency that locates the middle of the data.
The data 2, 4, 4, 4, 5, 5, 7, 9 are given. What is their standard deviation?
1. 4
2. 1.5
3. 5
4. 2
Show answer
Answer: D — 2
The mean is 5 and the squared deviations add to 32, so variance = 32 / 8 = 4 and SD = √4 = 2. The value 4 is the variance and 1.5 is the mean deviation, both different measures.
Observations x have mean 20 and standard deviation 4. Each is changed to y = 3x + 5. What are the mean and the SD of y?
1. Mean 65, SD 17
2. Mean 65, SD 12
3. Mean 60, SD 12
4. Mean 65, SD 144
Show answer
Answer: B — Mean 65, SD 12
The mean becomes 3 × 20 + 5 = 65 and the SD becomes 3 × 4 = 12, because adding 5 never changes the spread. Adding 5 to the SD (17) is the usual error; 144 is the variance of y (9 × 16), not its SD.
What is the variance of the first seven natural numbers 1, 2, ..., 7?
1. 4
2. 2
3. 7
4. 3.5
Show answer
Answer: A — 4
The variance of 1 to n is (n² − 1) / 12 = 48 / 12 = 4. The value 2 is the standard deviation, not the variance, and 3.5 is the mean of the numbers.
Judge these statements. P: the sum of deviations of observations from their mean is zero. Q: the median is disturbed by extreme values as much as the mean. R: adding the same constant to every observation changes the standard deviation. S: doubling every observation doubles the variance. Which are correct?
1. P and S
2. Q and R
3. P only
4. R and S
Show answer
Answer: C — P only
Only P is true. Q is false because the median ignores extreme values; R is false because a shift leaves spread unchanged; S is false because doubling quadruples the variance, which makes P and S tempting but wrong.
A fair die is thrown once. What is the probability that the number is a prime?
1. 1/3
2. 2/3
3. 1/6
4. 1/2
Show answer
Answer: D — 1/2
The primes on a die are 2, 3 and 5, so the probability is 3 / 6 = 1 / 2. The number 1 is not prime, so counting four primes (2/3) is wrong.
What is the probability of drawing a king from a well-shuffled pack of 52 playing cards?
1. 1/13
2. 1/4
3. 1/26
4. 4/13
Show answer
Answer: A — 1/13
There are 4 kings in 52 cards, so the probability is 4 / 52 = 1 / 13. The fraction 1/4 is the chance of a particular suit, not a particular rank.
Two fair dice are thrown together. What is the probability that the sum of the numbers is 8?
1. 1/6
2. 1/9
3. 5/36
4. 7/36
Show answer
Answer: C — 5/36
Sum 8 occurs as (2,6), (3,5), (4,4), (5,3), (6,2): five of the 36 outcomes. The value 1/6 belongs to sum 7, which has six ways.
A bag holds 5 red, 3 blue and 2 green balls. Two balls are drawn one after another without replacement. What is the probability that both are red?
1. 1/4
2. 2/9
3. 5/18
4. 1/5
Show answer
Answer: B — 2/9
The first is red with 5/10 and the second with 4/9, so the probability is 20 / 90 = 2 / 9. The value 1/4 results from replacing the ball (5/10 × 5/10), which does not match the stem.
Events A and B are independent with P(A) = 1/2 and P(B) = 1/3. What is the probability that at least one of them happens?
1. 2/3
2. 5/6
3. 1/6
4. 1/2
Show answer
Answer: A — 2/3
P(at least one) = 1 − P(neither) = 1 − (1/2)(2/3) = 2/3. The sum 5/6 double counts the case in which both happen, which is exactly what the subtraction of the overlap 1/6 removes.