NCERT Solutions for Class 10 Maths Statistics Exercise 14.1: Mean of Grouped Data

Welcome to the world of statistics! In this chapter, we move from raw, individual data points to working with large, organized datasets. Exercise 14.1 is your first and most crucial step, focusing on one of the most important measures of central tendency: the mean. You've calculated the average before, but what happens when data is grouped into class intervals? How do you find the 'average' then? This page will guide you through the methods to calculate the mean of grouped data. We will explore the Direct Method for simpler data and the Assumed Mean Method for more complex calculations. By the end of this lesson from your YoLearn AI Tutor, you will not only understand the formulas but also know exactly when and how to apply each method to solve any problem from Statistics Ex 14.1 with confidence.

What is the Mean of Grouped Data?

When we have a large amount of data, it's practical to group it into class intervals. For example, instead of listing the marks of 50 students individually, we can group them: 0-10 marks, 10-20 marks, and so on. The 'mean' still represents the average value, but we can't just add up all the individual values anymore because we don't have them!

This is where we introduce the Class Mark (xᵢ). The class mark is the midpoint of a class interval. We assume that all the data points within an interval are centered around this midpoint. It's calculated as:

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

Once we have the class mark for each interval and the frequency (fᵢ - how many data points fall in that interval), we can calculate the mean. Exercise 14.1 introduces systematic methods to do this, making our calculations organized and less prone to error.

Methods for Calculating Mean

1. Direct Method
This is the most straightforward method. We multiply each class mark (xᵢ) by its corresponding frequency (fᵢ), sum up these products, and then divide by the total number of observations (total frequency). Formula: Mean (x̄) = Σ(fᵢxᵢ) / Σfᵢ When to use: This method is best when the numerical values of xᵢ and fᵢ are small, making the multiplication fᵢxᵢ easy to handle without large numbers.
2. Assumed Mean Method
This is a shortcut to the Direct Method, especially useful when xᵢ and fᵢ values are large. We 'assume' a mean (a), which is usually one of the middle xᵢ values. We then calculate the deviation of each class mark from this assumed mean (dᵢ = xᵢ - a). Formula: Mean (x̄) = a + (Σ(fᵢdᵢ) / Σfᵢ) When to use: Use this when the direct method would result in large, difficult calculations. It simplifies the multiplication step.

Worked Example: Finding Mean using Assumed Mean Method

  1. Step 1: Understand the Problem & Create a Table — Let's find the mean for the following data: | Class Interval | Frequency (fᵢ) | | :---: | :---: | | 100-120 | 12 | | 120-140 | 14 | | 140-160 | 8 | | 160-180 | 6 | | 180-200 | 10 | We will create a table with columns for Class Interval, Frequency (fᵢ), Class Mark (xᵢ), Deviation (dᵢ = xᵢ - a), and fᵢdᵢ.
  2. Step 2: Calculate Class Marks (xᵢ) — For each interval, xᵢ = (Lower Limit + Upper Limit) / 2. - 100-120: (100 + 120) / 2 = 110 - 120-140: (120 + 140) / 2 = 130 - 140-160: (140 + 160) / 2 = 150 - 160-180: (160 + 180) / 2 = 170 - 180-200: (180 + 200) / 2 = 190
  3. Step 3: Choose an Assumed Mean (a) — Let's choose a middle value from the xᵢ column to be our assumed mean. Let a = 150.
  4. Step 4: Calculate Deviations (dᵢ = xᵢ - a) — Now, we find the deviation for each xᵢ from a = 150. - 110 - 150 = -40 - 130 - 150 = -20 - 150 - 150 = 0 - 170 - 150 = 20 - 190 - 150 = 40
  5. Step 5: Calculate the Product (fᵢdᵢ) — Multiply the frequency (fᵢ) by the deviation (dᵢ) for each row. - 12 (-40) = -480 - 14 (-20) = -280 - 8 0 = 0 - 6 20 = 120 - 10 * 40 = 400
  6. Step 6: Find the Sums (Σfᵢ and Σfᵢdᵢ) — Sum the 'fᵢ' and 'fᵢdᵢ' columns. - Σfᵢ = 12 + 14 + 8 + 6 + 10 = 50 - Σfᵢdᵢ = -480 - 280 + 0 + 120 + 400 = -760 + 520 = -240
  7. Step 7: Apply the Assumed Mean Formula — Use the formula: Mean (x̄) = a + (Σ(fᵢdᵢ) / Σfᵢ) - x̄ = 150 + (-240 / 50) - x̄ = 150 - 4.8 - x̄ = 145.2 So, the mean of the data is 145.2.

Tips for Acing Mean Problems

Choosing the Right Method:

  • If xᵢ and fᵢ values are small (e.g., single or small double-digit numbers), use the Direct Method. It's faster and has fewer steps.
  • If xᵢ or fᵢ values are large, always use the Assumed Mean Method. It will save you from complex multiplication and reduce the chances of calculation errors.

Common Mistakes to Avoid:

  1. Forgetting 'a': A very common mistake is calculating (Σfᵢdᵢ / Σfᵢ) and forgetting to add the assumed mean a back at the end. Always write the full formula before substituting values.
  2. Sign Errors: Be extremely careful with positive and negative signs when calculating fᵢdᵢ. Sum the positive values and negative values separately before finding the final Σfᵢdᵢ.
  3. Class Mark Error: Don't accidentally use the class limit instead of the class mark. Always calculate xᵢ as your first step after writing down the frequencies.

Practice Questions with Solutions

  • Q: The following distribution shows the daily pocket allowance of children of a locality. Find the mean daily pocket allowance using the Direct Method. Class Interval | Frequency (f_i) ------------------|------------------ 100-120 | 12 120-140 | 14 140-160 | 8 160-180 | 6 180-200 | 10 A: Step 1: Calculate the class mark (x_i) for each interval. x_i = (Lower limit + Upper limit) / 2. For 100-120, x_i = 110. For 120-140, x_i = 130. For 140-160, x_i = 150. For 160-180, x_i = 170. For 180-200, x_i = 190. Step 2: Calculate the product f_i x_i for each class. 12 110 = 1320. 14 130 = 1820. 8 150 = 1200. 6 170 = 1020. 10 190 = 1900. Step 3: Calculate the sum of frequencies (Σf_i) and the sum of products (Σf_i x_i). Σf_i = 12 + 14 + 8 + 6 + 10 = 50. Σf_i x_i = 1320 + 1820 + 1200 + 1020 + 1900 = 7260. Step 4: Apply the direct method formula for mean: Mean (x̄) = Σf_i * x_i / Σf_i. x̄ = 7260 / 50 = 145.2. Final answer: The mean daily pocket allowance is Rs. 145.2.
  • Q: The marks obtained by 30 students of Class X in a mathematics paper are given below. Find the mean marks using the Assumed Mean Method. Class Interval | Frequency (f_i) ------------------|------------------ 10-25 | 2 25-40 | 3 40-55 | 7 55-70 | 6 70-85 | 6 85-100 | 6 A: Step 1: Calculate the class mark (x_i) for each interval. 10-25 -> 17.5. 25-40 -> 32.5. 40-55 -> 47.5. 55-70 -> 62.5. 70-85 -> 77.5. 85-100 -> 92.5. Step 2: Choose an assumed mean (A). Let A = 62.5. Calculate the deviation d_i = x_i - A. d_i values: -45, -30, -15, 0, 15, 30. Step 3: Calculate the product f_i d_i for each class. 2 (-45) = -90. 3 (-30) = -90. 7 (-15) = -105. 6 0 = 0. 6 15 = 90. 6 30 = 180. Step 4: Calculate the sum of frequencies (Σf_i) and the sum of products (Σf_i d_i). Σf_i = 2 + 3 + 7 + 6 + 6 + 6 = 30. Σf_i d_i = -90 - 90 - 105 + 0 + 90 + 180 = -15. Step 5: Apply the assumed mean method formula: Mean (x̄) = A + (Σf_i d_i / Σf_i). x̄ = 62.5 + (-15 / 30) = 62.5 - 0.5 = 62. Final answer: The mean marks obtained is 62.
  • Q: The following table shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure using the Step-Deviation Method. Class Interval | Frequency (f_i) ------------------|------------------ 100-150 | 4 150-200 | 5 200-250 | 12 250-300 | 2 300-350 | 2 A: Step 1: Calculate the class mark (x_i) for each interval. 100-150 -> 125. 150-200 -> 175. 200-250 -> 225. 250-300 -> 275. 300-350 -> 325. Step 2: Choose an assumed mean (A) and class size (h). Let A = 225 (middle x_i). Class size h = 150 - 100 = 50. Step 3: Calculate u_i = (x_i - A) / h. u_i values: (125-225)/50 = -2. (175-225)/50 = -1. (225-225)/50 = 0. (275-225)/50 = 1. (325-225)/50 = 2. Step 4: Calculate the product f_i u_i for each class. 4 (-2) = -8. 5 (-1) = -5. 12 0 = 0. 2 1 = 2. 2 2 = 4. Step 5: Calculate the sum of frequencies (Σf_i) and the sum of products (Σf_i u_i). Σf_i = 4 + 5 + 12 + 2 + 2 = 25. Σf_i u_i = -8 - 5 + 0 + 2 + 4 = -7. Step 6: Apply the step-deviation method formula: Mean (x̄) = A + (Σf_i u_i / Σf_i) h. x̄ = 225 + (-7 / 25) 50 = 225 + (-7 2) = 225 - 14 = 211. Final answer: The mean daily expenditure on food is Rs. 211.
  • Q: The following distribution shows the daily pocket allowance of children. The mean pocket allowance is Rs. 18. Find the missing frequency 'f'. Class Interval | Frequency (f_i) ------------------|------------------ 11-13 | 7 13-15 | 6 15-17 | 9 17-19 | 13 19-21 | f 21-23 | 5 23-25 | 4 A: Step 1: Calculate the class mark (x_i) for each interval. 11-13 -> 12. 13-15 -> 14. 15-17 -> 16. 17-19 -> 18. 19-21 -> 20. 21-23 -> 22. 23-25 -> 24. Step 2: Calculate the product f_i x_i for each class. 7 12 = 84. 6 14 = 84. 9 16 = 144. 13 18 = 234. f 20 = 20f. 5 22 = 110. 4 24 = 96. Step 3: Calculate the sum of frequencies (Σf_i) and the sum of products (Σf_i x_i). Σf_i = 7 + 6 + 9 + 13 + f + 5 + 4 = 44 + f. Σf_i x_i = 84 + 84 + 144 + 234 + 20f + 110 + 96 = 752 + 20f. Step 4: Use the direct method formula for mean and the given mean (x̄ = 18). Mean (x̄) = Σf_i x_i / Σf_i. 18 = (752 + 20f) / (44 + f). Step 5: Solve the equation for 'f'. 18 (44 + f) = 752 + 20f. 792 + 18f = 752 + 20f. 792 - 752 = 20f - 18f. 40 = 2f. f = 40 / 2 = 20. Final answer: The missing frequency 'f' is 20.

Frequently Asked Questions

Which method should I use: Direct or Assumed Mean?

If the class marks (xᵢ) and frequencies (fᵢ) are small, use the Direct Method. If the numbers are large, the Assumed Mean Method is better as it simplifies calculations and reduces the risk of errors.

What is the Step-Deviation Method and why isn't it used here?

The Step-Deviation Method is another shortcut that further simplifies calculations by dividing deviations by the class size. However, as per the rationalized syllabus for recent CBSE exams, this topic has been removed, so we focus on the Direct and Assumed Mean methods.

Can I use any value for the assumed mean 'a'?

Yes, you can mathematically choose any value for 'a'. However, it is highly recommended to choose one of the middle values from the class mark (xᵢ) column. This makes the deviations (dᵢ) smaller and centers them around zero, which simplifies the final calculation.

What if the class intervals are not continuous (e.g., 1-10, 11-20)?

For non-continuous (inclusive) intervals, you must first make them continuous. Subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class. For example, 1-10 becomes 0.5-10.5 and 11-20 becomes 10.5-20.5. Then, proceed with the calculations as usual.