CBSE Class 10 Maths: Statistics - Measures of Central Tendency

Understanding Measures of Central Tendency

Statistics is the branch of mathematics that deals with the collection, organization, analysis, interpretation, and presentation of data. In Class 10, our primary focus is on Measures of Central Tendency for grouped data. These measures — mean, median, and mode — provide a single value that attempts to describe a set of data by identifying the central position within that set. Think of them as ways to find the "average" or "typical" value in a dataset.

• Mean: This is the most common average, calculated by summing all values and dividing by the number of values. For grouped data, we use specific formulas to handle class intervals and frequencies.
• Median: This represents the middle value in an ordered dataset. When data is grouped, we find the class interval where the median lies and then use a formula to pinpoint its exact value.
• Mode: This is the value that appears most frequently in a dataset. For grouped data, we identify the modal class (the class with the highest frequency) and then apply a formula to find the mode within that class.

Understanding these measures helps us summarize complex data efficiently and draw quick, insightful conclusions about the characteristics of the data distribution. Mastery of these methods is crucial for your CBSE Class 10 board examinations.

Key Definitions in Statistics

Methods to Calculate the Mean for Grouped Data

1. 1. Direct Method — This method is suitable when the values of $x_i$ (class marks) and $f_i$ (frequencies) are small. It's straightforward and less prone to calculation errors if numbers are manageable. Formula: $\bar{x} = \frac{\Sigma f_i x_i}{\Sigma f_i}$ Steps: 1. Find the class mark ($x_i$) for each class interval. 2. Calculate the product $f_i x_i$ for each class. 3. Find the sum of all $f_i x_i$ (i.e., $\Sigma f_i x_i$). 4. Find the sum of all frequencies (i.e., $\Sigma f_i$). 5. Divide $\Sigma f_i x_i$ by $\Sigma f_i$ to get the mean.
2. 2. Assumed Mean Method — This method is useful when the class marks ($x_i$) are large, which would make the product $f_i x_i$ large and calculations tedious. By assuming a mean, we work with smaller deviations. Formula: $\bar{x} = A + \frac{\Sigma f_i d_i}{\Sigma f_i}$, where $d_i = x_i - A$ Steps: 1. Find the class mark ($x_i$) for each class interval. 2. Choose an 'assumed mean' (A) from the class marks, preferably the middle one. 3. Calculate the deviation ($d_i = x_i - A$) for each class. 4. Calculate the product $f_i d_i$ for each class. 5. Find the sum of all $f_i d_i$ (i.e., $\Sigma f_i d_i$). 6. Find the sum of all frequencies (i.e., $\Sigma f_i$). 7. Apply the formula to find the mean.
3. 3. Step-Deviation Method — This method is an extension of the assumed mean method and simplifies calculations even further when the class sizes are uniform. It involves dividing the deviations by the common class size. Formula: $\bar{x} = A + \left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right) \times h$, where $u_i = \frac{x_i - A}{h}$ Steps: 1. Find the class mark ($x_i$) for each class interval. 2. Choose an 'assumed mean' (A) from the class marks, preferably the middle one. 3. Calculate the deviation ($d_i = x_i - A$) for each class. 4. Calculate $u_i = \frac{d_i}{h}$, where $h$ is the class size (difference between upper and lower limits, assuming uniform class size). 5. Calculate the product $f_i u_i$ for each class. 6. Find the sum of all $f_i u_i$ (i.e., $\Sigma f_i u_i$). 7. Find the sum of all frequencies (i.e., $\Sigma f_i$). 8. Apply the formula to find the mean.

How to Calculate the Median for Grouped Data

1. 1. Prepare Cumulative Frequency Table — First, arrange the data in ascending order (if not already grouped) and then create a cumulative frequency (c.f.) column. The cumulative frequency for a class is the sum of its frequency and the frequencies of all preceding classes.
2. 2. Find n/2 — Calculate $n/2$, where $n$ is the total number of observations (sum of all frequencies, $\Sigma f_i$). This value helps us locate the median class.
3. 3. Identify the Median Class — The median class is the class interval whose cumulative frequency is just greater than or equal to $n/2$. This class contains the median.
4. 4. Apply the Median Formula — Once the median class is identified, use the following formula: Formula: $Median = L + \left[\frac{\frac{n}{2} - CF}{f}\right] \times h$ Where: $L$ = lower limit of the median class. $n$ = total frequency ($\Sigma f_i$). $CF$ = cumulative frequency of the class preceding the median class. $f$ = frequency of the median class. * $h$ = class size (assuming uniform class size).

How to Calculate the Mode for Grouped Data

1. 1. Identify the Modal Class — The modal class is the class interval with the highest frequency. This class is where the mode is expected to lie.
2. 2. Note Down Relevant Frequencies — From the modal class, identify: $f_1$: 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.
3. 3. Apply the Mode Formula — Once the modal class and frequencies are identified, use the following formula: Formula: $Mode = L + \left[\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right] \times h$ Where: $L$ = lower limit of the modal class. $f_1$ = 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. * $h$ = class size (assuming uniform class size).

Exam Tips for Statistics in Class 10

To score well in Statistics, remember these key points:

• Know Your Formulas: Memorize all formulas for mean (direct, assumed mean, step-deviation), median, and mode. Write them down correctly at the beginning of your solution.
• Accurate Calculations: Be extremely careful with arithmetic, especially when dealing with negative numbers in the assumed mean method or large sums. Use a table format to organize your calculations; this reduces errors and makes your work easy to follow.
• Identify Correct Classes: Ensure you correctly identify the 'median class' and 'modal class'. A mistake here will lead to an entirely wrong answer. For median, don't confuse the cumulative frequency of the median class ($CF$) with that of the preceding class.
• Understand 'h': The class size 'h' is crucial. Make sure your class intervals are continuous (e.g., 10-20, 20-30). If they are discontinuous (e.g., 10-19, 20-29), make them continuous first (e.g., 9.5-19.5, 19.5-29.5) by adjusting limits by 0.5 before calculating $h$ and applying formulas.
• Practice with All Methods: Practice finding the mean using all three methods to understand when each is most efficient. Sometimes the question specifies a method, so be prepared.

Practice Questions with Solutions

• Q: The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹18. Find the missing frequency 'f'. | Daily Pocket Allowance (₹) | Number of Children (f) | |----------------------------|------------------------| | 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 and $f_i x_i$. | Class Interval | $f_i$ | $x_i$ | $f_i x_i$ | |----------------|-------|-------|---------| | 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 | Step 2: Find the sum of frequencies and sum of $f_i x_i$. $\Sigma f_i = 7 + 6 + 9 + 13 + f + 5 + 4 = 44 + f$ $\Sigma f_i x_i = 84 + 84 + 144 + 234 + 20f + 110 + 96 = 752 + 20f$ Step 3: Use the mean formula and solve for 'f'. $\text{Mean} = \frac{\Sigma f_i x_i}{\Sigma f_i}$ $18 = \frac{752 + 20f}{44 + f}$ $18(44 + f) = 752 + 20f$ $792 + 18f = 752 + 20f$ $792 - 752 = 20f - 18f$ $40 = 2f$ $f = 20$ Final answer: The missing frequency 'f' is 20.
• Q: Find the median of the following data: | Class Interval | Frequency | |----------------|-----------| | 0-10 | 5 | | 10-20 | 8 | | 20-30 | 20 | | 30-40 | 15 | | 40-50 | 7 | | 50-60 | 5 | A: Step 1: Create a cumulative frequency table. | Class Interval | Frequency (f) | Cumulative Frequency (CF) | |----------------|---------------|---------------------------| | 0-10 | 5 | 5 | | 10-20 | 8 | 5 + 8 = 13 | | 20-30 | 20 | 13 + 20 = 33 | | 30-40 | 15 | 33 + 15 = 48 | | 40-50 | 7 | 48 + 7 = 55 | | 50-60 | 5 | 55 + 5 = 60 | Step 2: Find $n/2$. Total frequency $n = \Sigma f_i = 60$ $n/2 = 60/2 = 30$ Step 3: Identify the median class. The class whose cumulative frequency is just greater than or equal to 30 is 20-30 (c.f. = 33). So, the median class is 20-30. Step 4: Extract values for the median formula. $L = 20$ (lower limit of median class) $CF = 13$ (c.f. of class preceding median class) $f = 20$ (frequency of median class) $h = 10$ (class size) Step 5: Apply the median formula. $Median = L + \left[\frac{\frac{n}{2} - CF}{f}\right] \times h$ $Median = 20 + \left[\frac{30 - 13}{20}\right] \times 10$ $Median = 20 + \left[\frac{17}{20}\right] \times 10$ $Median = 20 + \frac{17}{2} = 20 + 8.5 = 28.5$ Final answer: The median of the data is 28.5.
• Q: Find the mode of the following data: | Class Interval | Frequency | |----------------|-----------| | 0-10 | 7 | | 10-20 | 9 | | 20-30 | 15 | | 30-40 | 12 | | 40-50 | 8 | | 50-60 | 4 | A: Step 1: Identify the modal class. The highest frequency is 15, which corresponds to the class interval 20-30. So, the modal class is 20-30. Step 2: Extract values for the mode formula. $L = 20$ (lower limit of modal class) $f_1 = 15$ (frequency of modal class) $f_0 = 9$ (frequency of class preceding modal class) $f_2 = 12$ (frequency of class succeeding modal class) $h = 10$ (class size) Step 3: Apply the mode formula. $Mode = L + \left[\frac{f_1 - f_0}{2f_1 - f_0 - f_2}\right] \times h$ $Mode = 20 + \left[\frac{15 - 9}{2(15) - 9 - 12}\right] \times 10$ $Mode = 20 + \left[\frac{6}{30 - 9 - 12}\right] \times 10$ $Mode = 20 + \left[\frac{6}{9}\right] \times 10$ $Mode = 20 + \frac{2}{3} \times 10$ $Mode = 20 + \frac{20}{3} = 20 + 6.67 \text{ (approx.)} = 26.67$ Final answer: The mode of the data is approximately 26.67.
• Q: Calculate the mean for the following distribution using the assumed mean method: | Class Interval | Frequency (f) | |----------------|---------------| | 100-120 | 10 | | 120-140 | 15 | | 140-160 | 20 | | 160-180 | 25 | | 180-200 | 10 | A: Step 1: Calculate class marks ($x_i$) and choose an assumed mean (A). Let A = 150 (mid-value of 140-160). | Class Interval | $f_i$ | $x_i$ | $d_i = x_i - A$ (where A=150) | $f_i d_i$ | |----------------|-------|-------|-------------------------------|-----------| | 100-120 | 10 | 110 | 110 - 150 = -40 | -400 | | 120-140 | 15 | 130 | 130 - 150 = -20 | -300 | | 140-160 | 20 | 150 | 150 - 150 = 0 | 0 | | 160-180 | 25 | 170 | 170 - 150 = 20 | 500 | | 180-200 | 10 | 190 | 190 - 150 = 40 | 400 | Step 2: Find the sum of frequencies and sum of $f_i d_i$. $\Sigma f_i = 10 + 15 + 20 + 25 + 10 = 80$ $\Sigma f_i d_i = -400 - 300 + 0 + 500 + 400 = 200$ Step 3: Apply the assumed mean formula. $\bar{x} = A + \frac{\Sigma f_i d_i}{\Sigma f_i}$ $\bar{x} = 150 + \frac{200}{80}$ $\bar{x} = 150 + 2.5$ $\bar{x} = 152.5$ Final answer: The mean of the data is 152.5.
• Q: The following table gives the lifetime of 400 neon lamps: | Lifetime (in hours) | Number of Lamps (Frequency) | |---------------------|-----------------------------| | 1500-2000 | 14 | | 2000-2500 | 56 | | 2500-3000 | 60 | | 3000-3500 | 86 | | 3500-4000 | 74 | | 4000-4500 | 62 | | 4500-5000 | 48 | Find the median lifetime of a lamp. A: Step 1: Create a cumulative frequency table. | Lifetime (hours) | Frequency (f) | Cumulative Frequency (CF) | |------------------|---------------|---------------------------| | 1500-2000 | 14 | 14 | | 2000-2500 | 56 | 14 + 56 = 70 | | 2500-3000 | 60 | 70 + 60 = 130 | | 3000-3500 | 86 | 130 + 86 = 216 | | 3500-4000 | 74 | 216 + 74 = 290 | | 4000-4500 | 62 | 290 + 62 = 352 | | 4500-5000 | 48 | 352 + 48 = 400 | Step 2: Find $n/2$. Total frequency $n = 400$ $n/2 = 400/2 = 200$ Step 3: Identify the median class. The class whose cumulative frequency is just greater than or equal to 200 is 3000-3500 (c.f. = 216). So, the median class is 3000-3500. Step 4: Extract values for the median formula. $L = 3000$ (lower limit of median class) $CF = 130$ (c.f. of class preceding median class) $f = 86$ (frequency of median class) $h = 500$ (class size) Step 5: Apply the median formula. $Median = L + \left[\frac{\frac{n}{2} - CF}{f}\right] \times h$ $Median = 3000 + \left[\frac{200 - 130}{86}\right] \times 500$ $Median = 3000 + \left[\frac{70}{86}\right] \times 500$ $Median = 3000 + \frac{35000}{86}$ $Median = 3000 + 406.97 \text{ (approx.)}$ $Median = 3406.97$ Final answer: The median lifetime of a lamp is approximately 3406.97 hours.

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