Statistics (Class 10): Notes और MCQ Practice

आसान भाषा में

Statistics में हम grouped data का औसत यानी Mean, सबसे ज़्यादा repeat होने वाली value यानी Mode, और ठीक बीच की central value यानी Median निकालना सीखते हैं। इस chapter में direct formulas हैं, इसलिए calculations ध्यान से करें तो पूरे marks पक्के हैं!

Important Formulas & Definitions

1. Class Mark (xᵢ) = (Upper class limit + Lower class limit) ÷ 2
2. Mean (x̄):

• Direct Method: x̄ = Σfᵢxᵢ ÷ Σfᵢ
• Assumed Mean Method: x̄ = a + (Σfᵢdᵢ ÷ Σfᵢ) [जहाँ dᵢ = xᵢ − a]

1. Mode = l + [(f₁ − f₀) ÷ (2f₁ − f₀ − f₂)] × h

(l = modal class की lower limit, f₁ = modal class frequency, f₀ = preceding frequency, f₂ = succeeding frequency, h = class size)

1. Median = l + [((n ÷ 2) − cf) ÷ f] × h

(l = median class की lower limit, cf = preceding class की cumulative frequency, f = median class frequency)

1. Empirical Relationship: 3 Median = Mode + 2 Mean

याद रखने के Tricks (Memory Tricks)

1. 3 Median = Mode + 2 Mean याद रखने की ट्रिक: शब्दों की length देखें! 'Median' सबसे लंबा शब्द है, इसलिए इसके साथ बड़ा नंबर 3 आएगा; 'Mean' छोटा है तो 2 आएगा; 'Mode' अकेला 1 रहेगा (3 = 1 + 2).
2. Mode Formula (10 / 102 Trick): Formula के bracket में ऊपर आता है 10 (f₁ − f₀) और नीचे आता है 102 (2f₁ − f₀ − f₂).

Common Board-Exam Mistakes

1. Median में गलत cf चुनना: Median formula में cf हमेशा median class से पहले (preceding) वाली class की ली जाती है, median class की नहीं!
2. Non-continuous Data: अगर class intervals continuous नहीं हैं (जैसे 1-5, 6-10), तो formula लगाने से पहले lower limit में से 0.5 घटाकर और upper limit में 0.5 जोड़कर continuous बनाना ज़रूरी है (0.5-5.5, 5.5-10.5)।

Chapter 14: Statistics

1. Chapter in Short

In Class 9, you learned to represent ungrouped data using bar graphs, histograms, and frequency polygons, as well as finding basic measures of central tendency. In Class 10, Statistics takes this a step further by focusing on grouped data.

A central tendency is a single representative value around which the data clusters. In this chapter, you will learn how to calculate three major measures of central tendency for grouped frequency distributions:

1. Mean (Average value)
2. Mode (Most frequently occurring value)
3. Median (Middle-most value)

You will also learn the empirical relationship connecting all three measures.

2. Key Concepts Explained Simply

A. Mean of Grouped Data

The mean is the numerical average of all observations. For grouped data, we use the midpoint of each class interval, called the class mark ($x_i$).

• Class mark ($x_i$) = (Upper class limit + Lower class limit) ÷ 2

There are three methods to calculate the mean:

1. Direct Method: Multiply each frequency ($f_i$) by its class mark ($x_i$), sum them up, and divide by the total frequency.
2. Assumed Mean Method: Pick a central value from the class marks as an assumed mean ($a$), calculate deviations ($d_i = x_i - a$), and use them to simplify large calculations.
3. Step-Deviation Method: Divide deviations by the class size ($h$) so numbers become very small ($u_i = (x_i - a) ÷ h$), making calculations quick and error-free.

Everyday Example: If a batsman scores 20, 40, 60 runs in three matches, the mean score is (20 + 40 + 60) ÷ 3 = 40 runs.

B. Mode of Grouped Data

The mode represents the value that appears most often.

• In grouped data, we first identify the modal class (the class interval with the highest frequency).
• The actual mode lies inside this class and is calculated using a specific formula.

Everyday Example: A shoe store owner wants to know which shoe size to stock the most. The size sold most frequently is the mode.

C. Median of Grouped Data

The median is the value of the middle-most observation when data is arranged in order.

• To find the median, we first find the cumulative frequency (cf) by continuously adding frequencies.
• Find $n ÷ 2$ (where $n$ is total frequency).
• The class interval whose cumulative frequency is just greater than or equal to $n ÷ 2$ is called the median class.

Everyday Example: If 9 students stand in a line according to height, the height of the 5th student (exact middle) is the median.

3. Important Definitions, Formulas, and Facts

• Class Mark ($x_i$):

$x_i = \frac{\text{Upper Limit} + \text{Lower Limit}}{2}$

• Class Size ($h$):

$h = \text{Upper Limit} - \text{Lower Limit}$

• Mean Formulas:

1. Direct Method:

$\bar{x} = \frac{\Sigma (f_i \times x_i)}{\Sigma f_i}$

1. Assumed Mean Method:

$\bar{x} = a + \frac{\Sigma (f_i \times d_i)}{\Sigma f_i} \quad \text{where } d_i = x_i - a$

1. Step-Deviation Method:

$\bar{x} = a + \left(\frac{\Sigma (f_i \times u_i)}{\Sigma f_i}\right) \times h \quad \text{where } u_i = \frac{x_i - a}{h}$

• Mode Formula:

$\text{Mode} = l + \left[\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right] \times h$

• $l$ = lower limit of modal class
• $h$ = class size
• $f_1$ = frequency of the modal class
• $f_0$ = frequency of the class preceding the modal class
• $f_2$ = frequency of the class succeeding the modal class

• Median Formula:

$\text{Median} = l + \left[\frac{\frac{n}{2} - cf}{f}\right] \times h$

• $l$ = lower limit of median class
• $n$ = total number of observations ($\Sigma f_i$)
• $cf$ = cumulative frequency of the class preceding the median class
• $f$ = frequency of the median class
• $h$ = class size

• Empirical Relationship (Golden Rule):

$3 \times \text{Median} = \text{Mode} + 2 \times \text{Mean}$

$\text{or } \text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}$

4. Solved Examples (Board-Exam Style)

Example 1: Finding the Mean (Direct Method)

Question: Find the mean of the following distribution:

Class Interval — Frequency ($f_i$)

10 - 25 — 2

25 - 40 — 3

40 - 55 — 7

55 - 70 — 6

70 - 85 — 6

85 - 100 — 6

Solution:

1. Calculate class marks $x_i = (\text{Upper} + \text{Lower}) ÷ 2$ and products $f_i \times x_i$:

• 10 - 25: $x_i = 17.5$; $f_i \times x_i = 2 \times 17.5 = 35.0$
• 25 - 40: $x_i = 32.5$; $f_i \times x_i = 3 \times 32.5 = 97.5$
• 40 - 55: $x_i = 47.5$; $f_i \times x_i = 7 \times 47.5 = 332.5$
• 55 - 70: $x_i = 62.5$; $f_i \times x_i = 6 \times 62.5 = 375.0$
• 70 - 85: $x_i = 77.5$; $f_i \times x_i = 6 \times 77.5 = 465.0$
• 85 - 100: $x_i = 92.5$; $f_i \times x_i = 6 \times 92.5 = 555.0$

1. Sum the frequencies: $\Sigma f_i = 2 + 3 + 7 + 6 + 6 + 6 = 30$
2. Sum $f_i \times x_i$: $\Sigma(f_i \times x_i) = 35 + 97.5 + 332.5 + 375 + 465 + 555 = 1860$
3. Mean = $\Sigma(f_i \times x_i) ÷ \Sigma f_i = 1860 ÷ 30 = 62$

Answer: Mean = 62

Example 2: Finding the Mode

Question: Find the mode of the following data:

Class — 5 - 15 — 15 - 25 — 25 - 35 — 35 - 45 — 45 - 55 — 55 - 65

Frequency — 6 — 11 — 21 — 23 — 14 — 5

Solution:

1. Identify the modal class: The maximum frequency is 23, corresponding to the interval 35 - 45.
2. Note down values:

• $l$ = 35
• $h$ = 10
• $f_1$ (frequency of modal class) = 23
• $f_0$ (frequency of preceding class) = 21
• $f_2$ (frequency of succeeding class) = 14

1. Apply Mode formula:

$\text{Mode} = l + \left[\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right] \times h$

$\text{Mode} = 35 + \left[\frac{23 - 21}{2(23) - 21 - 14}\right] \times 10$

$\text{Mode} = 35 + \left[\frac{2}{46 - 35}\right] \times 10$

$\text{Mode} = 35 + \left[\frac{2}{11}\right] \times 10 = 35 + \frac{20}{11} = 35 + 1.818 = 36.82$

Answer: Mode ≈ 36.82

Example 3: Finding the Median

Question: Calculate the median for the following data:

Class Interval — Frequency ($f$)

65 - 85 — 4

85 - 105 — 5

105 - 125 — 13

125 - 145 — 20

145 - 165 — 14

165 - 185 — 8

185 - 205 — 4

Solution:

1. Form the cumulative frequency table:

• 65 - 85: $f = 4 \implies cf = 4$
• 85 - 105: $f = 5 \implies cf = 4 + 5 = 9$
• 105 - 125: $f = 13 \implies cf = 9 + 13 = 22$
• 125 - 145: $f = 20 \implies cf = 22 + 20 = 42$
• 145 - 165: $f = 14 \implies cf = 42 + 14 = 56$
• 165 - 185: $f = 8 \implies cf = 56 + 8 = 64$
• 185 - 205: $f = 4 \implies cf = 64 + 4 = 68$

1. Here, $n = 68$, so $n ÷ 2 = 68 ÷ 2 = 34$.
2. The cumulative frequency just greater than 34 is 42, which belongs to class 125 - 145.

• Median class = 125 - 145
• $l = 125$
• $cf$ (preceding class) = 22
• $f$ (median class) = 20
• $h = 20$

1. Apply formula:

$\text{Median} = l + \left[\frac{\frac{n}{2} - cf}{f}\right] \times h$

$\text{Median} = 125 + \left[\frac{34 - 22}{20}\right] \times 20$

$\text{Median} = 125 + 12 = 137$

Answer: Median = 137

Example 4: Using Empirical Relationship

Question: In a frequency distribution, the mean is 24 and the median is 26. Find the mode.

Solution:

We use the empirical formula:

$\text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}$

$\text{Mode} = (3 \times 26) - (2 \times 24)$

$\text{Mode} = 78 - 48 = 30$

Answer: Mode = 30

5. Common Mistakes and Quick Revision Points

• The 'cf' Trap in Median: In the median formula, $cf$ is the cumulative frequency of the preceding (previous) class, NOT the median class itself. However, $f$ is the frequency of the median class itself.
• Continuous Classes: Check if intervals are continuous. If they are inclusive (e.g., 1-10, 11-20), convert them into continuous boundaries (0.5-10.5, 10.5-20.5) by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit before calculating mode or median.
• Order of Frequencies in Mode:
• $f_0$ = frequency before modal class (0 comes first)
• $f_1$ = frequency of modal class
• $f_2$ = frequency after modal class (2 comes next)
• Denominator Check for Mode: Ensure you calculate $2f_1 - f_0 - f_2$ accurately; do not multiply 2 with $(f_1 - f_0)$.
• Quick Sanity Check: The calculated Mode must always lie within the modal class, and the calculated Median must always lie within the median class. If your answer falls outside that interval, you have made a calculation error!