Mean, Median, Mode and Dispersion
Mean, Median, Mode and Dispersion
Elementary statistics in NTPC means three measures of the centre (mean, median, mode) and, increasingly in recent papers, the spread of data: range, variance and standard deviation. Each has one clean formula; most mistakes come from forgetting to sort the data before finding the median.
1. Formula Box
| Measure | Ungrouped data (n values) | Frequency table |
|---|---|---|
| Mean | Sum of values ÷ n | Σfx ÷ Σf (x = value or class mid-point) |
| Median | Sort first; odd n: the ((n + 1)/2)th value; even n: average of the (n/2)th and (n/2 + 1)th | Value where the cumulative frequency first passes n/2 |
| Mode | Most frequent value | Value or class with the highest frequency |
| Range | Largest − smallest | |
| Variance | Σ(x − mean)² ÷ n = (Σx² ÷ n) − mean² | Σf(x − mean)² ÷ Σf |
| Standard deviation (SD) | √variance | √variance |
Empirical relation (moderately skewed data): Mode = 3 Median − 2 Mean
Adding k to every value: mean, median and mode rise by k; range, variance and SD do not change
Multiplying every value by k: mean and SD are multiplied by k (SD by |k|), variance by k²
First n natural numbers: mean (n + 1)/2, variance (n² − 1)/12. Any n consecutive integers have the same variance
✗ Median of 23, 11, 45, 37, 19 is 45, the middle value as written | ✓ Sort first: 11, 19, 23, 37, 45, so the median is 23
✗ Adding 5 to every value raises the SD by 5 | ✓ Shifting does not change spread; SD stays the same
हिंदी नोट: माध्य = योग ÷ संख्या, माध्यिका के लिए आँकड़ों को पहले क्रम में लगाइए, बहुलक सबसे अधिक बार आने वाला मान है। बहुलक = 3 माध्यिका − 2 माध्य। हर मान में k जोड़ने पर मानक विचलन नहीं बदलता।
Exam Pointer: Verified NTPC patterns: median of an unsorted list (two June 2025 Graduate CBT-1 shifts) and standard deviation of consecutive integers (March 2026 Graduate CBT-1). Mean of a frequency table, the empirical relation and missing-value questions had no verified NTPC shift and are tagged syllabus-based.
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: Mean of a list or a frequency table
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Marks 10, 20, 30 and 40 were scored by 3, 5, 8 and 4 students. Find the mean mark.
Σfx = 30 + 100 + 240 + 160 = 530 and Σf = 20, so the mean = 26.5.
Answer: 26.5
Data for the next question:
| Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
|---|---|---|---|---|---|
| Frequency | 5 | 8 | 12 | 10 | 5 |
EXAMATLAS LEVEL
Q. Find the mean of the grouped distribution above.
Use the mid-points 5, 15, 25, 35, 45. Σfx = 25 + 120 + 300 + 350 + 225 = 1,020 and Σf = 40, so the mean = 25.5.
Answer: 25.5
Pattern 2: Median of an unsorted list
[PYQ: NTPC Graduate CBT-1 12-Jun-2025 Shift-3 | NTPC Graduate CBT-1 16-Jun-2025 Shift-1]
EXAM LEVEL
Q. Find the median of 23, 11, 45, 37, 19, 52, 30.
Sorted: 11, 19, 23, 30, 37, 45, 52. There are 7 values, so the median is the 4th: 30.
Answer: 30
EXAMATLAS LEVEL
Q. Find the median of 18, 42, 27, 35, 50, 12, 29, 44. What is the median if 60 is added to the list?
Sorted: 12, 18, 27, 29, 35, 42, 44, 50. With 8 values the median is the average of the 4th and 5th: (29 + 35)/2 = 32. With 60 added there are 9 values, and the 5th, 35, is the median.
Answer: 32; 35
Pattern 3: Mode and the empirical relation
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Find the mode of 4, 7, 4, 9, 7, 4, 10, 7, 4.
4 appears four times and 7 three times, so the mode is 4.
Answer: 4
EXAMATLAS LEVEL
Q. In a moderately skewed distribution the mean is 34.5 and the median is 33. Find the mode. For the list 3, 5, 5, 7, 8, 5, 9, find the mean of its median and mode.
Mode = 3 × 33 − 2 × 34.5 = 99 − 69 = 30. For the list, sorted 3, 5, 5, 5, 7, 8, 9, the median is 5 and the mode is 5, so their mean is 5.
Answer: 30; 5
Pattern 4: Mean given, find a missing value or frequency
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. The mean of 6, 9, x, 14 and 16 is 12. Find x.
The total must be 60, and the known values add to 45, so x = 15.
Answer: 15
Data for the next question:
| x | 5 | 10 | 15 | 20 |
|---|---|---|---|---|
| f | 4 | k | 6 | 5 |
EXAMATLAS LEVEL
Q. The mean of the distribution above is 13. Find k.
(20 + 10k + 90 + 100)/(15 + k) = 13 gives 210 + 10k = 195 + 13k, so k = 5. Check: 260/20 = 13.
Answer: k = 5
CBT-2 LEVEL
Pattern 5: Range, variance and standard deviation
[PYQ: NTPC Graduate CBT-1 22-Mar-2026 Shift-1]
EXAM LEVEL
Q. Find the range, variance and standard deviation of 2, 4, 6, 8, 10.
Range = 8. Mean = 6; squared deviations 16, 4, 0, 4, 16 add to 40, so variance = 8 and SD = √8 = 2√2 ≈ 2.83.
Answer: 8; 8; 2√2
EXAMATLAS LEVEL
Q. Find the variance of the first 9 natural numbers. If each number is multiplied by 3 and then 5 is added, find the new mean and SD.
Variance = (81 − 1)/12 = 20/3 and the mean is 5. New mean = 3 × 5 + 5 = 20. Adding 5 does not affect spread, so the new SD = 3 × √(20/3) = √60 = 2√15 ≈ 7.75.
Answer: 20/3; mean 20, SD 2√15
60-Second Revision
- Mean = Σfx/Σf; use class mid-points for grouped data.
- Sort before taking the median; even count means average of the middle two.
- Mode = 3 Median − 2 Mean.
- Shifting changes the mean, not the SD; scaling by k multiplies both.