Ch 13: Statistics

Introduction

Page 1: Mean of Grouped Data (The Direct Method)

Welcome to the world of Statistics! In Class IX, you learned how to organize data and find its "center" using the mean, median, and mode for simple, ungrouped lists of numbers. But what happens when you're faced with massive amounts of data, like the exam scores of every student in a city? Listing them one by one is impractical. This is where grouped data comes in.

Imagine a cricket analyst trying to understand a batsman's performance over a year. Instead of looking at every single score (7, 45, 102, 33, ...), they group them: "How many times did he score between 0-20 runs?", "How many times between 21-40?", and so on. This gives a much clearer picture. In this chapter, we'll upgrade our skills to find the mean, median, and mode for exactly this kind of grouped data, starting with the most fundamental measure: the mean.

{{FORMULA: expr=x̄ = (Σfᵢxᵢ) / (Σfᵢ) | symbols=x̄:Mean, Σ:Summation, fᵢ:Frequency of the i-th class, xᵢ:Class mark of the i-th class}}

Definitions & Formulas

Before we dive in, let's clarify the key terms we'll be using for grouped data.

The Logic: From Ungrouped to Grouped Mean

Finding the mean seems simple: add everything up and divide. But how do you "add up" data when it's in ranges like "40-55"? We can't use the range itself. The logic behind the Direct Method involves a clever and crucial assumption.

1. Recall the Ungrouped Mean: For a simple frequency table where each value x appears f times, the mean is the sum of all (f × x) products, divided by the total number of observations (sum of all f).

   Mean x̄ = (f₁x₁ + f₂x₂ + ... + fₙxₙ) / (f₁ + f₂ + ... + fₙ) = (Σfᵢxᵢ) / (Σfᵢ)
   

2. The Grouped Data Problem: In a grouped frequency table (e.g., Table 13.2 in your textbook), we have class intervals (like 40-55) and their frequencies (e.g., 7 students). We know 7 students scored between 40 and 55, but we don't know their exact marks.

{{VISUAL: chart: A side-by-side comparison showing a bar graph for ungrouped marks (discrete bars at 10, 20, 36 etc.) and a histogram for the same marks grouped into class intervals like 10-25, 25-40 (adjacent bars).}}

3. The Mid-Point Solution: To solve this, we need a single value to represent each class interval. The most logical choice is the midpoint, which we call the class mark (xᵢ).

   Class Mark (xᵢ) = (Upper Class Limit + Lower Class Limit) / 2
   

4. The Core Assumption: We assume that the frequency of each class is centered around its class mark. For the 40-55 interval with 7 students, we pretend all 7 students scored exactly at the midpoint, which is (40+55)/2 = 47.5. This is an approximation, but it's a very effective one for large datasets.

{{VISUAL: diagram: A number line showing a class interval from 40 to 55, with its midpoint, the class mark 47.5, clearly labeled. An arrow points to 47.5 with the text "Representative value (xᵢ) for all 7 students in this class".}}

5. Applying the Formula: Now that we have a representative xᵢ (the class mark) for each frequency fᵢ, we can use the same formula as the ungrouped mean. We are simply replacing the individual observations with the class marks.

   Mean x̄ = (Σfᵢxᵢ) / (Σfᵢ)
   

This process gives us a very close estimate of the true mean and is known as the Direct Method.

{{KEY: type=concept | title=The Mid-Point Assumption | text=The Direct Method's accuracy relies on the assumption that the values within each class are evenly distributed, making the class mark (mid-point) a fair representative for all data points in that class. The exact mean can only be found from raw, ungrouped data.}}

Solved Examples

Let's apply the Direct Method to a few problems, increasing in difficulty.

Example 1: Finding Mean Pocket Money (Easy)

Given: The daily pocket money of 20 students in a class is recorded in the table below.

To Find: The mean daily pocket money.

Solution:

1. First, we need to find the class mark (xᵢ) for each interval and then calculate the product fᵢxᵢ. Let's create a calculation table.

2. Now, find the sum of frequencies (Σfᵢ) and the sum of the products (Σfᵢxᵢ).

   Σfᵢ = 4 + 5 + 8 + 3 = 20
   

   Σfᵢxᵢ = 60 + 125 + 280 + 135 = 600
   

3. Apply the formula for the mean.

   x̄ = (Σfᵢxᵢ) / (Σfᵢ) = 600 / 20
   

   x̄ = 30
   

Final Answer:

The mean daily pocket money is ₹30.

Example 2: Literacy Rate Analysis (Medium)

Given: The following table shows the literacy rate (in percentage) of 35 cities.

To Find: The mean literacy rate.

Solution:

1. Construct a table to calculate class marks (xᵢ) and the product fᵢxᵢ.

2. Sum the fᵢ and fᵢxᵢ columns. We have Σfᵢ = 35 and Σfᵢxᵢ = 2430.

3. Use the mean formula.

   x̄ = (Σfᵢxᵢ) / (Σfᵢ) = 2430 / 35
   

4. Perform the division.

   x̄ ≈ 69.43
   

Final Answer:

The mean literacy rate is approximately 69.43%.

Example 3: Handling Inclusive Classes (Hard)

Given: The weights of 50 wrestlers are recorded as follows. Notice the class intervals are inclusive.

To Find: The mean weight.

Solution:

1. The data is in an inclusive format (there's a gap between 109 and 110). We must first convert it to an exclusive format (continuous classes) to find the correct class marks. Subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class.

   • 100-109 becomes 99.5 - 109.5
   • 110-119 becomes 109.5 - 119.5
   • ...and so on.

2. Now, create the calculation table with the new continuous class intervals.

{{VISUAL: chart: A visually annotated table for this example, showing the original inclusive classes (100-109), the conversion to exclusive classes (99.5-109.5), the calculation of class marks (xᵢ = 104.5), and the fᵢxᵢ product column, with Σfᵢ and Σfᵢxᵢ totals highlighted.}}

3. Apply the mean formula with the calculated sums.

   x̄ = (Σfᵢxᵢ) / (Σfᵢ) = 6145 / 50
   

4. Calculate the final value.

   x̄ = 122.9
   

Final Answer:

The mean weight of the wrestlers is 122.9 kg.

Example 4: Finding a Missing Frequency (Tricky)

Given: The mean of the following distribution is 50.

To Find: The value of the missing frequency p.

Solution:

1. Set up the calculation table as usual, but this time we will have the variable p in our sums.

2. Calculate Σfᵢ and Σfᵢxᵢ in terms of p.

   Σfᵢ = 17 + p + 32 + 24 + 19 = 92 + p
   

   Σfᵢxᵢ = 170 + 30p + 1600 + 1680 + 1710 = 5160 + 30p
   

3. Now, use the mean formula. We are given that x̄ = 50.

   x̄ = (Σfᵢxᵢ) / (Σfᵢ)
   

   50 = (5160 + 30p) / (92 + p)
   

4. Solve the linear equation for p.

   50 × (92 + p) = 5160 + 30p
   

   4600 + 50p = 5160 + 30p
   

   50p - 30p = 5160 - 4600
   

   20p = 560
   

   p = 560 / 20 = 28
   

Final Answer:

The value of the missing frequency p is 28.

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

1. The mean age of a group of 30 students is 15 years. If the age of their teacher, 46 years, is also included, what will be the new mean age of the group?

   💡 Answer: Initial total age = 30 × 15 = 450 years. New total age (with teacher) = 450 + 46 = 496 years. New number of people = 30 + 1 = 31. New mean = 496 / 31 = 16 years.

2. The class marks of a continuous frequency distribution are given as: 22, 28, 34, 40, 46. What is the class size and the class interval corresponding to the class mark 34?

   💡 Answer: The class size (h) is the difference between consecutive class marks: 28 - 22 = 6. For the class mark 34, the interval extends h/2 on both sides. h/2 = 6/2 = 3. Lower Limit = 34 - 3 = 31. Upper Limit = 34 + 3 = 37. The class interval is 31 - 37.

3. If the mean of a distribution is 62 (calculated using the Direct Method from grouped data) and the exact mean from the original ungrouped data was 59.3, what could be a possible reason for this difference?

   💡 Answer: The difference arises from the "mid-point assumption". The Direct Method assumes all values in a class interval are concentrated at the midpoint (class mark). In reality, the actual data points within each interval might have been skewed towards the lower or upper end, causing the approximate mean (62) to differ from the exact mean (59.3).

Mini Cheatsheet

Mean of Grouped Data — Direct Method

Mean of Grouped Data — Direct Method

{{FORMULA: expr=x̄ = Σ(fᵢ xᵢ) / Σfᵢ | symbols=x̄:Mean, fᵢ:Frequency of the i-th class, xᵢ:Class mark of the i-th class, Σ:Summation}}

Concept Introduction

Imagine you're the manager of a popular pizza place. At the end of a busy week, you want to know the average amount a customer spends. You have hundreds of bills, but looking at each one is too slow. Instead, you group them into spending ranges: ₹100-200, ₹200-300, ₹300-400, and so on.

Now, how do you find the average? You don't have the exact bill amounts anymore, just the groups. This is where we need a method for finding the mean of grouped data. We make a smart assumption: we treat everyone in a group as if they spent the mid-point amount. For the ₹100-200 group, we assume everyone spent ₹150. This clever shortcut allows us to calculate a very close estimate of the average spending, giving you valuable business insights quickly.

Definitions & Formulas

Before we jump into calculations, let's be crystal clear on the terms we'll be using. These are the building blocks for finding the mean of any grouped data.

The two key formulas for this method are:

1. Finding the Class Mark (xᵢ):

   Class Mark = (Upper Class Limit + Lower Class Limit) / 2
   

2. Finding the Mean (x̄) using the Direct Method:

   x̄ = (Σfᵢ xᵢ) / (Σfᵢ)
   

The Logic Behind the Method

Why does this method work? It's all based on one central assumption. When we group data, we lose the original individual values. To overcome this, we assume that all data points within a class are centered around its mid-point.

Here's the step-by-step logic:

1. The Problem: We have data in ranges (class intervals), not as individual points. We can't add up the original values because we don't know them anymore.

2. The Assumption: We need a single value to represent each class. The most logical choice is the middle value, the class mark (xᵢ). We assume that the frequency (fᵢ) for each class is centered around this class mark.

   {{VISUAL: diagram: A number line from 10 to 25. A large dot is placed exactly in the middle at 17.5, with a label "Class Mark (xᵢ)". Text below reads: "All values in the 10-25 interval are represented by 17.5."}}

3. The Transformation: By calculating the class mark for each interval, we transform the grouped data into a format that looks like ungrouped frequency data. The class marks (xᵢ) become our new "observations".

4. Applying the Old Formula: Now we can use the familiar formula for the mean of a frequency distribution: Mean = (Sum of all observations) / (Total number of observations).

5. The sum of all observations is estimated by f₁x₁ + f₂x₂ + ... + fₙxₙ, which we write as Σfᵢxᵢ.

6. The total number of observations is simply the sum of all frequencies: f₁ + f₂ + ... + fₙ, which we write as Σfᵢ.

7. The Result: Combining these gives us the Direct Method formula for the mean of grouped data: x̄ = Σ(fᵢ xᵢ) / Σfᵢ.

{{KEY: type=concept | title=The Mid-Point Assumption | text=The Direct Method's accuracy depends on the assumption that frequencies are centered around the class mark. This is why the calculated mean is an approximation of the true mean, but it's a very effective and widely used one.}}

Solved Examples

Let's put this into practice. We'll start easy and gradually increase the difficulty.

Example 1: Basic Calculation (Easy)

Given: The distribution of marks obtained by 40 students in a science test.

To Find: The mean marks of the students using the direct method.

Solution:

1. First, we need to construct a table with columns for Class Interval, Frequency (fᵢ), Class Mark (xᵢ), and the product fᵢxᵢ.

2. Calculate the Class Mark (xᵢ) for each interval. For 0-10, x₁ = (0+10)/2 = 5. For 10-20, x₂ = (10+20)/2 = 15, and so on.

3. Calculate the product fᵢxᵢ for each row. For the first row, f₁x₁ = 5 × 5 = 25.

4. Complete the table and find the sums Σfᵢ and Σfᵢxᵢ.

   Marks (Class Interval)	No. of Students (fᵢ)	Class Mark (xᵢ)	fᵢxᵢ
   0 - 10	5	5	25
   10 - 20	8	15	120
   20 - 30	15	25	375
   30 - 40	7	35	245
   40 - 50	5	45	225
   Total	Σfᵢ = 40		Σfᵢxᵢ = 990

5. Apply the mean formula.

   x̄ = Σfᵢxᵢ / Σfᵢ
   

   x̄ = 990 / 40
   

   x̄ = 24.75
   

Final Answer:

The mean marks obtained by the students is 24.75.

Example 2: Data with Decimal Class Marks (Medium)

Given: The daily wages of 50 workers in a factory.

To Find: The mean daily wages of the workers.

Solution:

1. Set up the calculation table. The class marks will be whole numbers here, but the calculation involves larger numbers.

2. Calculate the class mark for each interval. For 100-120, x₁ = (100+120)/2 = 110.

3. Calculate the product fᵢxᵢ for each row. For the first row, f₁x₁ = 12 × 110 = 1320.

4. Fill the table and find the sums.

   Daily Wages (₹)	Number of Workers (fᵢ)	Class Mark (xᵢ)	fᵢxᵢ
   100 - 120	12	110	1320
   120 - 140	14	130	1820
   140 - 160	8	150	1200
   160 - 180	6	170	1020
   180 - 200	10	190	1900
   Total	Σfᵢ = 50		Σfᵢxᵢ = 7260

5. Use the direct method formula.

   x̄ = Σfᵢxᵢ / Σfᵢ
   

   x̄ = 7260 / 50
   

   x̄ = 145.2
   

Final Answer:

The mean daily wage of the workers is ₹145.20.

Example 3: Finding a Missing Frequency (Hard)

Given: The mean of the following distribution is 18. The data shows the daily pocket allowance of children in a locality.

To Find: The value of the missing frequency, f.

Solution:

1. In this problem, the mean x̄ is given as 18. We need to work backwards to find f. First, let's set up our table as usual.

2. Calculate class marks (xᵢ) and the products fᵢxᵢ.

   Daily Allowance (₹)	Frequency (fᵢ)	Class Mark (xᵢ)	fᵢxᵢ
   11 - 13	7	12	84
   13 - 15	6	14	84
   15 - 17	9	16	144
   17 - 19	13	18	234
   19 - 21	f	20	20f
   21 - 23	5	22	110
   23 - 25	4	24	96

3. Now, find the sums Σfᵢ and Σfᵢxᵢ in terms of f.

   Σfᵢ = 7 + 6 + 9 + 13 + f + 5 + 4 = 44 + f
   

   Σfᵢxᵢ = 84 + 84 + 144 + 234 + 20f + 110 + 96 = 752 + 20f
   

4. Substitute these sums and the given mean into the formula.

   x̄ = Σfᵢxᵢ / Σfᵢ
   

   18 = (752 + 20f) / (44 + f)
   

5. Now, solve the linear equation for f.

   18 × (44 + f) = 752 + 20f
   

   792 + 18f = 752 + 20f
   

   792 - 752 = 20f - 18f
   

   40 = 2f
   

   f = 20
   

Final Answer:

The value of the missing frequency (f) is 20.

Example 4: Unequal Class Sizes (Tricky)

Given: The age distribution of patients treated in a hospital on a particular day.

To Find: The mean age of the patients.

Solution:

1. Notice that the class sizes are not equal. The first four classes have a size of 10 (15-5=10), but the last class 45-65 has a size of 20. The Direct Method works perfectly fine even with unequal class sizes.

2. We follow the exact same procedure. Set up the table.

3. Calculate the class marks (xᵢ) and products (fᵢxᵢ).

   Age (years)	No. of Patients (fᵢ)	Class Mark (xᵢ)	fᵢxᵢ
   5 - 15	6	10	60
   15 - 25	11	20	220
   25 - 35	21	30	630
   35 - 45	23	40	920
   45 - 65	14	55	770
   Total	Σfᵢ = 75		Σfᵢxᵢ = 2600

4. Calculate the mean using the sums.

   x̄ = Σfᵢxᵢ / Σfᵢ
   

   x̄ = 2600 / 75
   

   x̄ = 34.666...
   

Final Answer:

The mean age of the patients is approximately 34.67 years.

Tips & Tricks

Working smart is just as important as working hard. Here are a few tricks to make your calculations faster and more accurate.

Common Mistakes

Many students make similar small mistakes that can cost them marks. Here’s what to watch out for.

Brain-Teaser Questions

Test your understanding with these slightly challenging problems.

1. The mean weight of a class of 30 students is 45 kg. If the weight of the teacher, who weighs 75 kg, is included, what is the new mean weight of the group?

   💡 Answer: Total weight of students = 30 × 45 = 1350 kg. New total weight (with teacher) = 1350 + 75 = 1425 kg. New total number of people = 30 + 1 = 31. New mean = 1425 / 31 ≈ 45.97 kg.

2. The class marks of a continuous distribution are: 22, 28, 34, 40, 46. What is the class size and what are the class intervals?

   💡 Answer: The class size (h) is the difference between consecutive class marks: h = 28 - 22 = 6. The class limits are halfway between the class marks. For the first class mark 22, the lower limit is 22 - h/2 = 22 - 3 = 19 and the upper limit is 22 + h/2 = 22 + 3 = 25. The class intervals are: 19-25, 25-31, 31-37, 37-43, 43-49.

3. If the mean of a frequency distribution is 8.1 and Σfᵢxᵢ = 132 + 5k, and Σfᵢ = 20, find the value of k.

   💡 Answer: We use the formula x̄ = Σfᵢxᵢ / Σfᵢ. 8.1 = (132 + 5k) / 20 8.1 × 20 = 132 + 5k 162 = 132 + 5k 162 - 132 = 5k 30 = 5k k = 6.

Mini Cheatsheet

Here's a quick summary of everything on this page. Screenshot this for your last-minute revision!

Mean of Grouped Data — Assumed Mean Method

{{FORMULA: expr=x̄ = a + (Σfᵢdᵢ / Σfᵢ) | symbols=x̄:Mean, a:Assumed Mean, fᵢ:Frequency, dᵢ:Deviation (xᵢ - a)}}

Mean of Grouped Data — Assumed Mean Method

Concept Introduction

In the previous section, we learned the Direct Method to find the mean of grouped data. It works perfectly, but what happens when the data involves very large numbers? Imagine calculating the average monthly salary for 5,000 employees in a factory, where salaries range from ₹18,575 to ₹85,250.

Calculating the product of class marks (xᵢ) and frequencies (fᵢ) would be tedious, time-consuming, and prone to errors. 150 × 45750.50 is not a calculation you want to do by hand!

To simplify this, we use the Assumed Mean Method. The core idea is simple: we assume a convenient value as the mean (usually from the middle of the data), calculate how far off our data points are from this assumption, find the average of these "errors" (deviations), and then adjust our original assumption. It's a clever shortcut to tame large numbers.

Definitions & Formulas

This method introduces a few new terms, primarily the concept of deviation.

The Logic Behind the Formula

Why does this method work? It's a simple algebraic rearrangement of the Direct Method formula. Let's see how.

1. We know the formula for the mean by the Direct Method is:

   x̄ = (Σfᵢxᵢ) / (Σfᵢ)
   

2. We define a new term, deviation dᵢ, for each class mark xᵢ from an assumed mean a.

   dᵢ = xᵢ - a
   

3. This means we can express any class mark xᵢ in terms of the assumed mean and its deviation.

   xᵢ = a + dᵢ
   

4. Now, let's substitute this expression for xᵢ back into the Direct Method formula.

   x̄ = (Σfᵢ(a + dᵢ)) / (Σfᵢ)
   

5. Using the distributive property of summation, we can split the numerator.

   x̄ = (Σ(fᵢa) + Σ(fᵢdᵢ)) / (Σfᵢ)
   

6. Since a is a constant, Σ(fᵢa) is the same as a × (Σfᵢ).

   x̄ = (a × (Σfᵢ) + Σfᵢdᵢ) / (Σfᵢ)
   

7. Finally, we can separate this into two fractions.

   x̄ = (a × Σfᵢ / Σfᵢ) + (Σfᵢdᵢ / Σfᵢ)
   

8. The Σfᵢ terms cancel out in the first fraction, leaving us with the final formula for the Assumed Mean Method.

   x̄ = a + (Σfᵢdᵢ / Σfᵢ)
   

{{KEY: type=concept | title=Mean of Deviations | text=The term Σfᵢdᵢ / Σfᵢ is the mean of the deviations. So, the formula is simply: Actual Mean = Assumed Mean + Mean of Deviations. This shows how we are "correcting" our initial assumption.}}

Solved Examples

Example 1: Easy Start

Given: The distribution of daily wages of 50 workers in a factory.

To Find: The mean daily wages using the Assumed Mean method.

Solution:

1. First, we create a table to find the class mark (xᵢ) for each interval. The class mark is the midpoint: (Lower Limit + Upper Limit) / 2.

2. Let's choose an assumed mean (a). A good choice is one of the middle xᵢ values. Let's take a = 150.

3. Next, we calculate the deviation dᵢ = xᵢ - a for each class.

4. Then, we find the product fᵢdᵢ for each class.

5. Finally, we sum the fᵢ and fᵢdᵢ columns.

6. Now, we apply the Assumed Mean formula.

   x̄ = a + (Σfᵢdᵢ / Σfᵢ)
   

   x̄ = 150 + (-240 / 50)
   

   x̄ = 150 - 4.8
   

   x̄ = 145.2
   

Final Answer:

The mean daily wages are ₹145.20.

Example 2: Medium (NCERT Context)

Given: The marks obtained by 30 students of Class X, condensed into grouped data.

To Find: The mean marks using the Assumed Mean method.

Solution:

1. We prepare the calculation table. Let's follow the NCERT example and choose a = 47.5 as the assumed mean.

2. Now, we use the formula x̄ = a + (Σfᵢdᵢ / Σfᵢ).

   x̄ = 47.5 + (435 / 30)
   

3. Calculate the mean of deviations.

   435 / 30 = 14.5
   

4. Add this to the assumed mean.

   x̄ = 47.5 + 14.5
   

   x̄ = 62
   

Final Answer:

The mean marks obtained by the students is 62.

Example 3: Hard (Larger Numbers)

Given: The distribution of the monthly expenditure of 200 families in a locality.

To Find: The mean monthly expenditure.

Solution:

1. The numbers are large, so the Assumed Mean method is ideal. Let's create our table. The xᵢ values are 1250, 1750, 2250, 2750, 3250, 3750, 4250, 4750.
2. Let's choose a middle value for a. We have 8 classes, so either x₄=2750 or x₅=3250 are good choices. Let's pick a = 2750.

3. Apply the formula.

   x̄ = a + (Σfᵢdᵢ / Σfᵢ)
   

   x̄ = 2750 + (-17500 / 200)
   

4. Simplify the fraction.

   x̄ = 2750 - (175 / 2)
   

   x̄ = 2750 - 87.5
   

   x̄ = 2662.5
   

Final Answer:

The mean monthly expenditure is ₹2662.50.

Example 4: Tricky (Missing Frequency)

Given: The mean of the following distribution is 18.

To Find: The value of the missing frequency, p.

Solution:

1. This problem requires us to work backward. We are given x̄ = 18. We'll use the assumed mean method to build an equation involving p.
2. Let's calculate the class marks (xᵢ): 12, 14, 16, 18, 20, 22, 24.
3. A perfect choice for the assumed mean a is 18, as it's one of the xᵢ values and is also the given mean. This will simplify calculations. Let a = 18.

4. Now, substitute these sums into the Assumed Mean formula.

   x̄ = a + (Σfᵢdᵢ / Σfᵢ)
   

   18 = 18 + ((2p - 40) / (44 + p))
   

5. Subtract 18 from both sides.

   0 = (2p - 40) / (44 + p)
   

6. For a fraction to be zero, its numerator must be zero.

   2p - 40 = 0
   

   2p = 40
   

   p = 20
   

Final Answer:

The value of the missing frequency p is 20.

Tips & Tricks

Common Mistakes to Avoid

Brain-Teaser Questions

1. In the 'Hard' example above, we chose a = 2750. What would happen to the final answer if you chose the first class mark, a = 1250, instead? Would the calculation be easier or harder?

   💡 Answer: The final answer for the mean x̄ would be exactly the same (₹2662.50). However, the calculations for dᵢ and fᵢdᵢ would become much larger because all dᵢ values (xᵢ - 1250) would be positive and large, leading to a larger Σfᵢdᵢ. This makes the calculation more difficult and error-prone.

2. If, for a given dataset, you calculate Σfᵢdᵢ = 0, what can you conclude about your choice of the assumed mean a?

   💡 Answer: If Σfᵢdᵢ = 0, then the term (Σfᵢdᵢ / Σfᵢ) becomes 0. The formula simplifies to x̄ = a + 0, which means x̄ = a. This implies that your assumed mean was, by chance, the exact actual mean of the data.

3. Suppose you calculate the mean height of a group of students using a = 160 cm. If every student then grows by exactly 2 cm, and you re-calculate the mean using a new assumed mean of a = 162 cm, how will your new dᵢ values compare to the old dᵢ values?

   💡 Answer: The dᵢ values will remain exactly the same. The new class marks will be xᵢ' = xᵢ + 2. The new assumed mean is a' = a + 2 = 162. The new deviation is dᵢ' = xᵢ' - a' = (xᵢ + 2) - (a + 2) = xᵢ - a = dᵢ. Because both the data and the assumption shifted by the same amount, their difference (deviation) remains constant.

Mini Cheatsheet

Mean of Grouped Data — Step-deviation Method

Mean of Grouped Data — Step-deviation Method

Welcome back! In the previous section, we learned the Assumed Mean Method to simplify our calculations. We subtracted an assumed mean a from each class mark xᵢ to get a smaller deviation dᵢ. But what if the deviations dᵢ are still large numbers? What if they all share a common factor?

This is where the Step-deviation Method comes in. It's a fantastic upgrade to the assumed mean method, designed to make your calculations even simpler and faster, especially when dealing with large data sets with uniform class sizes. Imagine you're a city planner analyzing traffic data. The number of cars passing a junction per hour is huge. Instead of working with these large numbers, you could simplify them into smaller, manageable steps. That's precisely what this method helps us do.

{{FORMULA: expr=x̄ = a + (Σfᵢuᵢ / Σfᵢ) × h | symbols=x̄:Mean, a:Assumed Mean, fᵢ:Frequency, uᵢ:Step Deviation, h:Class Size}}

The core idea is to "step down" the deviations dᵢ by dividing them by a common factor, which is usually the class size h. This gives us a new, even smaller variable uᵢ. We then find the mean of these uᵢ values and adjust it back at the end to find the actual mean of the original data.

Definitions & Formulas

Let's formally define the components of the Step-deviation Method.

Derivation: How the Formula Works

Understanding the logic behind a formula is key to mastering it. Let's see how the step-deviation formula is derived directly from the assumed mean method.

1. We start with the relationship from the assumed mean method, where the mean x̄ is the assumed mean a plus the mean of the deviations d̄.

   x̄ = a + d̄
   

2. The mean of the deviations, d̄, is calculated as:

   d̄ = Σfᵢdᵢ / Σfᵢ
   

3. Now, we introduce our new variable, the step deviation uᵢ. We define it by dividing the deviation dᵢ by the constant class size h.

   uᵢ = dᵢ / h
   

4. By rearranging this, we can express dᵢ in terms of uᵢ and h.

   dᵢ = uᵢ × h
   

5. Let's substitute this new expression for dᵢ back into the formula for d̄ (from Step 2).

   d̄ = Σfᵢ(uᵢ × h) / Σfᵢ
   

6. Since h is a constant for all classes, we can take it outside the summation (Σ) sign.

   d̄ = h × (Σfᵢuᵢ / Σfᵢ)
   

7. Finally, we substitute this new, complete expression for d̄ back into our starting equation from Step 1. This gives us the final Step-deviation Method formula.

   x̄ = a + (Σfᵢuᵢ / Σfᵢ) × h
   

This elegant derivation shows that the step-deviation method is not a new concept, but a clever simplification of the assumed mean method.

{{KEY: type=concept | title=When to Use Step-deviation | text=The Step-deviation Method is most effective when the class size h is the same for all class intervals and the deviations dᵢ = xᵢ - a all share h as a common factor. This significantly reduces calculation complexity.}}

Solved Examples

Let's apply our knowledge to solve some problems, starting from easy and moving to more challenging ones.

Example 1: Basic Application (Easy)

Given: The distribution of daily wages of 50 workers in a factory.

To Find: The mean daily wages using the step-deviation method.

Solution:

1. First, let's create our calculation table. The class size h is 120 - 100 = 20. Let's choose the middle class mark xᵢ = 150 as our assumed mean a.

2. We have the necessary sums: Σfᵢ = 50 and Σfᵢuᵢ = -12. We also know a = 150 and h = 20.

3. Now, apply the step-deviation formula.

   x̄ = a + (Σfᵢuᵢ / Σfᵢ) × h
   

4. Substitute the values into the formula.

   x̄ = 150 + (-12 / 50) × 20
   

5. Simplify the expression.

   x̄ = 150 - (12 × 20) / 50
   

   x̄ = 150 - 240 / 50
   

   x̄ = 150 - 4.8
   

   x̄ = 145.2
   

Final Answer: The mean daily wages are Rs 145.20.

Example 2: Data with Decimal Class Marks (Medium)

Given: The concentration of SO₂ in the air (in parts per million, ppm) was measured for 30 localities.

To Find: The mean concentration of SO₂ in the air.

Solution:

1. The class size h is 0.04 - 0.00 = 0.04. Let's pick a = 0.10 as the assumed mean.

2. We have Σfᵢ = 30, Σfᵢuᵢ = -1, a = 0.10, and h = 0.04.

3. Apply the step-deviation formula.

   x̄ = a + (Σfᵢuᵢ / Σfᵢ) × h
   

4. Substitute the values.

   x̄ = 0.10 + (-1 / 30) × 0.04
   

5. Calculate the result.

   x̄ = 0.10 - 0.04 / 30
   

   x̄ = 0.10 - 0.00133...
   

   x̄ ≈ 0.0987
   

Rounding to a reasonable number of decimal places for this context (e.g., 4):

x̄ ≈ 0.0987

Final Answer: The mean concentration of SO₂ is approximately 0.099 ppm.

Example 3: Finding a Missing Frequency (Hard)

Given: The mean pocket allowance of children is Rs 18. The frequency p for the class interval 19-21 is missing.

To Find: The value of the missing frequency p.

Solution:

1. Here, we are given the mean x̄ = 18. The class size h = 13 - 11 = 2. Let's choose a = 18.

2. Let's sum the fᵢ and fᵢuᵢ columns. Σfᵢ = 7 + 6 + 9 + 13 + p + 5 + 4 = 44 + p Σfᵢuᵢ = (-21) + (-12) + (-9) + 0 + p + 10 + 12 = p - 20

3. Now, we use the step-deviation formula with the known mean x̄ = 18.

   x̄ = a + (Σfᵢuᵢ / Σfᵢ) × h
   

4. Substitute all known values into the equation.

   18 = 18 + ((p - 20) / (44 + p)) × 2
   

5. Solve the equation for p. First, subtract 18 from both sides.

   0 = ((p - 20) / (44 + p)) × 2
   

6. Since (expression) × 2 = 0, the expression itself must be zero.

   (p - 20) / (44 + p) = 0
   

7. For a fraction to be zero, the numerator must be zero.

   p - 20 = 0
   

   p = 20
   

Final Answer: The value of the missing frequency p is 20.

Example 4: Non-Continuous Class Intervals (Tricky)

Given: The distribution of the lengths of 40 leaves of a plant.

To Find: The mean length of the leaves.

Solution:

1. First, notice the class intervals are not continuous (e.g., 126 is followed by 127). We must make them continuous. The gap is 127 - 126 = 1. We add 1/2 = 0.5 to each upper limit and subtract 0.5 from each lower limit.

2. Now we can proceed. The class size h = 126.5 - 117.5 = 9. Let's choose a = 149 as the assumed mean (the class mark of 144.5-153.5).

3. We have Σfᵢ = 40, Σfᵢuᵢ = -9, a = 149, and h = 9.

4. Apply the step-deviation formula.

   x̄ = a + (Σfᵢuᵢ / Σfᵢ) × h
   

5. Substitute the values.

   x̄ = 149 + (-9 / 40) × 9
   

   x̄ = 149 - 81 / 40
   

   x̄ = 149 - 2.025
   

   x̄ = 146.975
   

Final Answer: The mean length of the leaves is 146.975 mm.

Tips & Tricks

Common Mistakes

Brain-Teaser Questions

Test your understanding with these higher-order thinking problems.

1. If the mean of a distribution calculated by the step-deviation method is 45, what will be the new mean if the assumed mean a was increased by 5, but the original data remains unchanged?

   💡 Answer: The final mean will still be 45. The choice of assumed mean a is just a calculation tool. A different a will change the intermediate Σfᵢuᵢ value, but the final x̄ will always be the same for the same data set. If a increases by 5, Σfᵢuᵢ / Σfᵢ will decrease by an amount that, when multiplied by h, exactly cancels out the increase, leaving x̄ unchanged.

2. A student calculates the mean using the step-deviation method. They correctly find Σfᵢ = 50, Σfᵢuᵢ = 20, and choose a = 25. They then realize they used h = 10 but the actual class size was h = 5. What is the correct mean?

   💡 Answer: The values of uᵢ were calculated as (xᵢ - a) / 10. The correct uᵢ values should be (xᵢ - a) / 5, which are exactly double the student's values. Therefore, the correct Σfᵢuᵢ would be 2 × 20 = 40. Now, let's calculate the correct mean: x̄ = a + (correct Σfᵢuᵢ / Σfᵢ) × (correct h) x̄ = 25 + (40 / 50) × 5 x̄ = 25 + (4/5) × 5 x̄ = 25 + 4 = 29. The correct mean is 29.

3. For a given frequency distribution, the sum of deviations from the assumed mean a = 50 is Σfᵢdᵢ = -200. If the total frequency Σfᵢ = 40 and the class size h = 10, what is the value of Σfᵢuᵢ?

   💡 Answer: We know that uᵢ = dᵢ / h. Therefore, fᵢuᵢ = fᵢdᵢ / h. Summing both sides gives us Σfᵢuᵢ = Σ(fᵢdᵢ / h). Since h is a constant, we can write Σfᵢuᵢ = (1/h) × Σfᵢdᵢ. Σfᵢuᵢ = (1/10) × (-200) = -20.

Mini Cheatsheet

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Mean of Grouped Data — Choosing a Method and Examples

Mean of Grouped Data: Choosing the Right Method

Welcome back! So far, we've learned how to organize data into groups and find a representative value for each group, the class mark (xᵢ). We also looked at the Direct Method for calculating the mean. But what happens when the numbers get really big? Calculating fᵢ × xᵢ can become tedious and prone to errors.

Imagine you're an economist analyzing the average monthly income for 10,000 households in a city. The incomes might range from ₹8,500 to ₹1,50,000. Using the Direct Method here would involve multiplying very large numbers, making the calculation a nightmare! This is where statisticians use clever shortcuts to simplify their work without losing accuracy. Today, we'll master two such powerful techniques: the Assumed Mean Method and the Step-Deviation Method. We'll also learn the most crucial skill: when to use which method.

{{FORMULA: expr=x̄ = a + h(Σfᵢuᵢ / Σfᵢ) | symbols=x̄:mean, a:assumed mean, h:class size, fᵢ:frequency, uᵢ:step deviation}}

Definitions & Formulas

Let's organize our toolkit. These are the three methods we'll be working with to find the mean of grouped data.

The Logic Behind the Shortcuts

Why do these methods work? It's all about shifting the data to make it simpler, calculating the average of the simplified data, and then shifting the result back.

1. The Assumed Mean Method

This method is based on shifting the origin. We "assume" a mean (a), find out how far off our assumption is on average, and then add this "correction factor" back. The NCERT text started this derivation, let's complete it.

1. We define the deviation of each class mark from our assumed mean a as dᵢ = xᵢ - a.

2. Rearranging this, we get xᵢ = a + dᵢ. This is the key relationship.

3. To find the mean x̄, we use the fundamental formula x̄ = (Σfᵢxᵢ) / (Σfᵢ). Let's substitute our expression for xᵢ:

   x̄ = (Σfᵢ(a + dᵢ)) / (Σfᵢ)
   

4. Distributing fᵢ inside the summation, we get:

   x̄ = (Σ(fᵢa + fᵢdᵢ)) / (Σfᵢ)
   

5. The sum of two terms is the sum of their sums:

   x̄ = (Σfᵢa + Σfᵢdᵢ) / (Σfᵢ)
   

6. Since a is a constant, Σfᵢa is just a × Σfᵢ. So we can split the fraction:

   x̄ = (aΣfᵢ / Σfᵢ) + (Σfᵢdᵢ / Σfᵢ)
   

7. The Σfᵢ terms cancel in the first part, leaving us with the final formula:

   x̄ = a + (Σfᵢdᵢ / Σfᵢ)
   

2. The Step-Deviation Method

This method goes one step further. After shifting the origin (by subtracting a), it also scales the data down (by dividing by h). This is useful when all the deviations dᵢ are multiples of the class size h.

1. We start from the Assumed Mean formula: x̄ = a + (Σfᵢdᵢ / Σfᵢ).

2. We define a new, simpler variable uᵢ = dᵢ / h, which means dᵢ = h × uᵢ.

3. Substitute this expression for dᵢ into the Assumed Mean formula:

   x̄ = a + (Σfᵢ(h × uᵢ)) / (Σfᵢ)
   

4. Since the class size h is a constant for all terms, we can factor it out of the summation:

   x̄ = a + (h × Σfᵢuᵢ) / (Σfᵢ)
   

5. This gives us the final, and often simplest, formula for calculation:

   x̄ = a + h(Σfᵢuᵢ / Σfᵢ)
   

   This method is incredibly efficient as you often end up multiplying fᵢ with small integers like -2, -1, 0, 1, 2.

Solved Examples

Let's apply these methods to see them in action. We'll start simple and move to more complex problems.

Example 1: Direct Method (Easy)

Given: The number of wickets taken by a bowler in 10 cricket matches are: 2, 6, 4, 5, 0, 2, 1, 3, 2, 3.

To Find: The mean number of wickets using the direct method for ungrouped data (a quick recap).

Solution:

1. First, list the observations (xᵢ) and their frequencies (fᵢ).

   • Wickets (xᵢ): 0, 1, 2, 3, 4, 5, 6
   • Frequency (fᵢ): 1, 1, 3, 2, 1, 1, 1

2. Calculate the total number of matches, Σfᵢ.

   Σfᵢ = 1 + 1 + 3 + 2 + 1 + 1 + 1 = 10
   

3. Calculate Σfᵢxᵢ.

   Σfᵢxᵢ = (0×1) + (1×1) + (2×3) + (3×2) + (4×1) + (5×1) + (6×1)
   Σfᵢxᵢ = 0 + 1 + 6 + 6 + 4 + 5 + 6 = 28
   

4. Apply the Direct Method formula x̄ = Σfᵢxᵢ / Σfᵢ.

   x̄ = 28 / 10 = 2.8
   

Final Answer: The mean number of wickets taken is 2.8.

Example 2: Assumed Mean Method (Medium)

Given: The daily wages of 50 workers in a factory are given below.

To Find: The mean daily wages using the Assumed Mean Method.

Solution:

1. First, create a table with class intervals, frequencies (fᵢ), and class marks (xᵢ).

   Class	fᵢ	xᵢ
   100-120	12	110
   120-140	14	130
   140-160	8	150
   160-180	6	170
   180-200	10	190
   Total	50

2. Choose an Assumed Mean (a). Let's pick the middle value of xᵢ, which is a = 150.

3. Calculate the deviations dᵢ = xᵢ - a for each class.

   xᵢ	dᵢ = xᵢ - 150
   110	-40
   130	-20
   150	0
   170	20
   190	40

4. Now calculate the product fᵢdᵢ for each class and find their sum Σfᵢdᵢ.

   fᵢ	dᵢ	fᵢdᵢ
   12	-40	-480
   14	-20	-280
   8	0	0
   6	20	120
   10	40	400
   Σfᵢ=50		Σfᵢdᵢ = -240

5. Apply the Assumed Mean formula: x̄ = a + (Σfᵢdᵢ / Σfᵢ).

   x̄ = 150 + (-240 / 50)
   

6. Simplify the calculation.

   x̄ = 150 - 4.8 = 145.2
   

Final Answer: The mean daily wage is ₹145.20.

Example 3: Step-Deviation Method (Hard)

Given: The distribution below shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

To Find: The mean runs scored using the Step-Deviation Method.

Solution:

1. Identify the class size h. Here, h = 4000 - 3000 = 1000.
2. Choose an Assumed Mean a. Let's pick a = 6500.
3. Prepare the calculation table. First, find xᵢ. Then dᵢ = xᵢ - a. Then uᵢ = dᵢ / h. Finally, fᵢuᵢ.

4. Calculate Σfᵢ and Σfᵢuᵢ from the table. We have Σfᵢ = 47 and Σfᵢuᵢ = -45.

5. Apply the Step-Deviation formula: x̄ = a + h(Σfᵢuᵢ / Σfᵢ).

   x̄ = 6500 + 1000 × (-45 / 47)
   

6. Calculate the value.

   x̄ = 6500 - (45000 / 47)
   x̄ ≈ 6500 - 957.45
   x̄ ≈ 5542.55
   

Final Answer: The mean runs scored is approximately 5542.55.

Example 4: Finding a Missing Frequency (Tricky)

Given: The mean pocket allowance of children is ₹18. The following frequency distribution is for the daily pocket allowance.

To Find: The value of the missing frequency p.

Solution:

1. Since the numbers are relatively small and the mean is given, the Direct Method is suitable. First, prepare the table with xᵢ and fᵢxᵢ.

2. Sum the frequencies to find Σfᵢ.

   Σfᵢ = 7 + 6 + 9 + 13 + p + 5 + 4 = 44 + p
   

3. Sum the fᵢxᵢ column to find Σfᵢxᵢ.

   Σfᵢxᵢ = 84 + 84 + 144 + 234 + 20p + 110 + 96 = 752 + 20p
   

4. Use the formula for the mean: x̄ = Σfᵢxᵢ / Σfᵢ. We are given x̄ = 18.

   18 = (752 + 20p) / (44 + p)
   

5. Now, solve the linear equation for p.

   18(44 + p) = 752 + 20p
   792 + 18p = 752 + 20p
   792 - 752 = 20p - 18p
   40 = 2p
   p = 20
   

Final Answer: The missing frequency p is 20.

{{KEY: type=strategy | title=How to Choose the Right Method | text=

1. Direct Method: Use when the values of class marks (xᵢ) and frequencies (fᵢ) are small, making fᵢ × xᵢ easy to calculate.
2. Assumed Mean Method: Use when xᵢ values are large. This method simplifies xᵢ into smaller dᵢ values, making multiplication easier.
3. Step-Deviation Method: This is the best choice when xᵢ values are large AND the class size (h) is uniform. It converts deviations into very small integers (uᵢ), making calculations the simplest. }}

Tips & Tricks

Common Mistakes to Avoid

Brain-Teaser Questions

1. If the mean of a distribution is 25, and every observation is increased by 3, what will be the new mean?

   💡 Answer: The new mean will be 25 + 3 = 28. The mean is affected by a change of origin. If every value xᵢ becomes xᵢ + k, the new mean becomes x̄ + k.

2. In a frequency distribution, the assumed mean a = 55, Σfᵢ = 100, h = 10, and Σfᵢuᵢ = -30. What is the actual mean x̄?

   💡 Answer: Using the Step-Deviation formula: x̄ = a + h(Σfᵢuᵢ / Σfᵢ). x̄ = 55 + 10 × (-30 / 100) = 55 - 3 = 52. The mean is 52.

3. The mean of 10 numbers is 20. If one number, 15, is removed, what is the mean of the remaining 9 numbers?

   💡 Answer: Original sum of 10 numbers = Mean × Number of items = 20 × 10 = 200. After removing 15, the new sum = 200 - 15 = 185. The new mean of 9 numbers = 185 / 9 ≈ 20.56.

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