Statisticsसांख्यिकी: आंकड़ों का वर्गीकरण, बारम्बारता
This module covers foundational and advanced statistics for Super TET Junior Part 2C Mathematics & Science, focusing on data classification, frequency distribution, and measures of central tendency (Mean, Median, Mode) for raw, discrete, and continuous data. High-yield weightage is assigned to empirical relationships, grouped data shortcuts, and handling edge cases with strict adherence to NCERT and SCERT UP guidelines.
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Syllabus Focus & High-Yield Strategy: This module covers foundational and advanced statistics for Super TET Junior Part 2C Mathematics & Science, focusing on data classification, frequency distribution, and measures of central tendency (Mean, Median, Mode) for raw, discrete, and continuous data. High-yield weightage is assigned to empirical relationships, grouped data shortcuts, and handling edge cases with strict adherence to NCERT and SCERT UP guidelines.
1. Fundamental Axioms, Definitions & Foundational Properties
MID — Midpoint Is Divided by two: $x_i = (\text{Lower} + \text{Upper})/2$.2. Formulas, Standard Theorems & Algorithmic Procedures
| Category / Case | Formula / Mathematical Rule | Topper Calculation Shortcut |
|---|---|---|
| Arithmetic Mean (Raw) | $\bar{x} = \frac{\sum x_i}{n}$ | Assume an arbitrary mean $A$ and use $\bar{x} = A + \frac{\sum d_i}{n}$ where $d_i = x_i - A$. |
| Mean (Grouped) | $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$ | Use Step-Deviation Method: $\bar{x} = A + h \left(\frac{\sum f_i u_i}{\sum f_i}\right)$ where $u_i = \frac{x_i - A}{h}$. |
| Median (Raw) | Odd $n$: $\left(\frac{n+1}{2}\right)^{\text{th}}$ term. Even $n$: Mean of $\left(\frac{n}{2}\right)^{\text{th}}$ and $\left(\frac{n}{2}+1\right)^{\text{th}}$ terms. | Always arrange data in ascending or descending order first before locating the position! |
| Median (Grouped) | $M_d = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$ | Identify median class where cumulative frequency ($cf$) $\ge \frac{N}{2}$, then substitute directly. |
3. Word Problem Modeling & Practical Applications
4. Mathematical Edge Cases, Sign Conventions & Common Traps
SORT — Sort Observations Right Then find median.Comprehensive Mathematical Matrix & Method Comparison
| Parameter / Dimension | Arithmetic Mean | Median | Mode |
|---|---|---|---|
| Definition | Sum of all observations divided by total count. | Middle value when data is arranged in order. | Most frequently occurring observation. |
| Algebraic Treatment | Rigorous mathematical properties; based on all observations. | Positional average; unaffected by extreme values beyond the middle. | Inspects peak frequency; ignores magnitude of non-modal values. |
| Suitability | Best for symmetrical distributions without extreme outliers. | Best for skewed distributions and ordinal data. | Best for qualitative data and quick retail/manufacturing insights. |
| Exam Speed & Accuracy Takeaway | Use step-deviation method for large grouped numbers. | Always verify cumulative frequencies ($cf$) before applying formula. | Check modal class boundaries carefully (inclusive vs exclusive). |
1. If the mean of five observations $x, x+2, x+4, x+6, x+8$ is $11$, then the mean of the first three observations is:
Given observations are $x, x+2, x+4, x+6, x+8$. Total number of observations $n = 5$. Sum = $5x + 20$. Mean = $\frac{5x + 20}{5} = x + 4$. Given mean is $11$, so $x + 4 = 11 \implies x = 7$. The first three observations are $x, x+2, x+4$, which are $7, 9, 11$. Their mean is $\frac{7 + 9 + 11}{3} = \frac{27}{3} = 9$. Option A is correct.
2. In a moderately symmetrical distribution, if the arithmetic mean is $26.8$ and the median is $27.9$, then the mode of the distribution is:
Using Pearson's empirical formula: $\text{Mode} = 3(\text{Median}) - 2(\text{Mean})$. Substituting the given values: $\text{Mode} = 3(27.9) - 2(26.8) = 83.7 - 53.6 = 30.1$. Option A is correct.
3. The median of a set of 9 distinct observations arranged in ascending order is $20.5$. If each of the largest 4 observations is increased by $2$, then the median of the new set will be:
For 9 distinct observations in ascending order, the median is the $\left(\frac{9+1}{2}\right)^{\text{th}} = 5^{\text{th}}$ observation. Increasing the largest 4 observations (i.e., $6^{\text{th}}, 7^{\text{th}}, 8^{\text{th}}, 9^{\text{th}}$ observations) does not affect the $5^{\text{th}}$ observation. Hence, the median remains unchanged at $20.5$. Option B is correct.
4. If the mean of frequency distribution is $25$ and $\sum f_i = 50$, while $\sum f_i x_i = 500 + k$, what is the value of $k$?
Mean $\bar{x} = \frac{\sum f_i x_i}{\sum f_i} \implies 25 = \frac{500 + k}{50}$. Cross-multiplying gives $500 + k = 25 \times 50 = 1250 \implies k = 1250 - 500 = 750$. Option A is correct.
5. The mean weight of 50 students in a class is $45\text{ kg}$. If the mean weight of boys is $50\text{ kg}$ and that of girls is $40\text{ kg}$, the number of boys in the class is:
Let the number of boys be $b$ and girls be $g$. Given $b + g = 50 \implies g = 50 - b$. Total weight = $50 \times 45 = 2250$. Sum of weights = $50b + 40g = 2250$. Substitute $g = 50 - b$: $50b + 40(50 - b) = 2250 \implies 50b + 2000 - 40b = 2250 \implies 10b = 250 \implies b = 25$. Option A is correct.
6. What is the class mark of the continuous class interval $35-45$?
Class mark = $\frac{\text{Lower Limit} + \text{Upper Limit}}{2} = \frac{35 + 45}{2} = \frac{80}{2} = 40$. Option A is correct.
7. If the difference between Mode and Median of a data is $2$, then the difference between Median and Mean is:
Using empirical relation: $\text{Mode} - \text{Median} = 2(\text{Median} - \text{Mean})$. Given $\text{Mode} - \text{Median} = 2$. Therefore, $2 = 2(\text{Median} - \text{Mean}) \implies \text{Median} - \text{Mean} = 1$. Option A is correct.
8. The mean of 10 numbers is $20$. If each number is multiplied by $3$ and then $5$ is subtracted from each, the new mean is:
If each observation $x_i$ is transformed to $a x_i \pm b$, the new mean is $a \bar{x} \pm b$. Here, $\bar{x} = 20$, multiplier $a = 3$, subtraction $b = 5$. New mean = $3(20) - 5 = 60 - 5 = 55$. Option A is correct.
9. For the following distribution: Class $0-10, 10-20, 20-30, 30-40, 40-50$ with frequencies $3, 9, 15, 30, 18$, the modal class is:
The modal class is the class interval corresponding to the highest frequency. The highest frequency in the given distribution is $30$, which corresponds to the class interval $30-40$. Option A is correct.
10. The median of the first 10 prime numbers is:
The first 10 prime numbers are $2, 3, 5, 7, 11, 13, 17, 19, 23, 29$. Here $n = 10$ (even). The median is the mean of the $5^{\text{th}}$ and $6^{\text{th}}$ observations. The $5^{\text{th}}$ term is $11$ and the $6^{\text{th}}$ term is $13$. Median = $\frac{11 + 13}{2} = \frac{24}{2} = 12$. Option A is correct.
5-Minute Exam Simulation Mode
Experience the real pressure of Super TET Junior (Part 2C Mathematics). You will face 10 randomized speed-math questions with negative marking (-1 for incorrect answers) and a strict 5-minute countdown.
High-Yield Rapid Recall Formulas & Laws (Must-Remember Pointers)
Speed Math & Memory Mnemonics Vault (Exam Retention Anchors)
Empirical Relation Order
MMM
M - Mode equals 3 Median minus 2 Mean.
Median Position Rule
ODD-EVEN
O - Over plus 1 by 2 for odd; E - Exact half and next for even.
Negative Marking Traps & Common Blunders (-1 Penalty Avoidance)
Negative Marking Trap 1: Forgetting to arrange raw data in ascending order before computing the median.
Negative Marking Trap 2: Using exclusive class interval formulas on inclusive frequency distributions without converting limits.
Negative Marking Trap 3: Confusing frequency ($f_i$) with cumulative frequency ($cf$) in the median formula denominator.
Negative Marking Trap 4: Sign errors when subtracting negative terms in the step-deviation or empirical mode formulas.