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Statistics - Measures of Central Tendency (Mean, Median, Mode for grouped and ungrouped data)

Grade 10ICSE

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

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Mean (Arithmetic Average): This is the sum of all observations divided by the total number of observations. Visually, the mean can be thought of as the 'balance point' of a data set; if you placed the data points on a seesaw, the mean is where the fulcrum would be to keep it perfectly level.

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Median: The median is the middle-most value when data is arranged in ascending or descending order. If you visualize a line of students ordered by height, the student standing exactly in the center is the median height. For grouped data, the median is often located using an 'Ogive' or cumulative frequency curve, which is an S-shaped graph where the median corresponds to the xx-value at the N2\frac{N}{2} position on the yy-axis.

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Mode: The mode is the value that occurs most frequently in a data set. In a histogram (a bar graph where the area of bars represents frequency), the mode is found within the tallest bar, known as the modal class. A distribution can be unimodal (one mode), bimodal (two modes), or multimodal.

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Grouped vs. Ungrouped Data: Ungrouped data is a simple list of numbers, while grouped data is organized into class intervals (e.g., 10βˆ’20,20βˆ’3010-20, 20-30) with corresponding frequencies. To calculate the mean for grouped data, we use the 'class mark' (the midpoint of each interval) as the representative value xx for that group.

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Assumed Mean Method: This is a technique to simplify mean calculations for large numbers. You pick a central value from the data (the 'assumed mean' AA) and calculate deviations (d=xβˆ’Ad = x - A) from it. This shifts the entire data set toward zero on the number line, making the arithmetic easier.

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Cumulative Frequency and the Ogive: The cumulative frequency is the running total of frequencies. When plotted against the upper class limits, it forms a 'Less Than Ogive'. To find the median visually, you draw a horizontal line from N2\frac{N}{2} on the cumulative frequency axis to the curve, then drop a vertical line to the xx-axis; that xx-intercept is the median.

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Empirical Relationship: In a moderately asymmetrical distribution, there is a fixed numerical relationship between the three measures. This can be visualized as a rule of thumb where the distance between the mean and the mode is roughly three times the distance between the mean and the median: Mode=3Γ—Medianβˆ’2Γ—Mean\text{Mode} = 3 \times \text{Median} - 2 \times \text{Mean}.

πŸ“Formulae

Mean for ungrouped data: xΛ‰=βˆ‘xn\bar{x} = \frac{\sum x}{n}

Mean for grouped data (Direct Method): xΛ‰=βˆ‘fixiβˆ‘fi\bar{x} = \frac{\sum f_i x_i}{\sum f_i}

Mean for grouped data (Assumed Mean Method): xΛ‰=A+βˆ‘fidiβˆ‘fi\bar{x} = A + \frac{\sum f_i d_i}{\sum f_i}, where di=xiβˆ’Ad_i = x_i - A

Mean for grouped data (Step-deviation Method): xΛ‰=A+(βˆ‘fiuiβˆ‘fiΓ—h)\bar{x} = A + (\frac{\sum f_i u_i}{\sum f_i} \times h), where ui=xiβˆ’Ahu_i = \frac{x_i - A}{h} and hh is class size

Median for ungrouped data (nn is odd): Median=(n+12)thΒ term\text{Median} = (\frac{n+1}{2})^{\text{th}} \text{ term}

Median for ungrouped data (nn is even): Median=12[(n2)thΒ term+(n2+1)thΒ term]\text{Median} = \frac{1}{2} [ (\frac{n}{2})^{\text{th}} \text{ term} + (\frac{n}{2} + 1)^{\text{th}} \text{ term} ]

Empirical Formula: Mode=3Medianβˆ’2Mean\text{Mode} = 3 \text{Median} - 2 \text{Mean}

πŸ’‘Examples

Problem 1:

Find the mean of the following frequency distribution using the Assumed Mean Method: Class intervals 0βˆ’10,10βˆ’20,20βˆ’30,30βˆ’40,40βˆ’500-10, 10-20, 20-30, 30-40, 40-50 with frequencies 7,8,12,13,107, 8, 12, 13, 10 respectively.

Solution:

  1. Find Class Marks (xx): 5,15,25,35,455, 15, 25, 35, 45.
  2. Choose Assumed Mean A=25A = 25.
  3. Calculate deviations di=xiβˆ’Ad_i = x_i - A: βˆ’20,βˆ’10,0,10,20-20, -10, 0, 10, 20.
  4. Calculate fidif_i d_i: 7(βˆ’20)=βˆ’140,8(βˆ’10)=βˆ’80,12(0)=0,13(10)=130,10(20)=2007(-20)=-140, 8(-10)=-80, 12(0)=0, 13(10)=130, 10(20)=200.
  5. Find sums: βˆ‘fi=50\sum f_i = 50, βˆ‘fidi=βˆ’140βˆ’80+0+130+200=110\sum f_i d_i = -140 - 80 + 0 + 130 + 200 = 110.
  6. Apply formula: xΛ‰=A+βˆ‘fidiβˆ‘fi=25+11050=25+2.2=27.2\bar{x} = A + \frac{\sum f_i d_i}{\sum f_i} = 25 + \frac{110}{50} = 25 + 2.2 = 27.2.

Explanation:

We use the Assumed Mean Method to reduce the size of the numbers being multiplied. By subtracting 25 from every class mark, we work with smaller integers, then add the average deviation back at the end.

Problem 2:

The marks obtained by 7 students in a test are: 12,18,11,25,30,15,2112, 18, 11, 25, 30, 15, 21. Find the median marks.

Solution:

  1. Arrange the data in ascending order: 11,12,15,18,21,25,3011, 12, 15, 18, 21, 25, 30.
  2. Count the number of observations: n=7n = 7 (which is odd).
  3. Use the formula for odd nn: Median=(n+12)thΒ term\text{Median} = (\frac{n+1}{2})^{\text{th}} \text{ term}.
  4. Calculate position: 7+12=4thΒ term\frac{7+1}{2} = 4^{\text{th}} \text{ term}.
  5. Identify the 4th term: 11,12,15,18,21,25,3011, 12, 15, \mathbf{18}, 21, 25, 30.
  6. The Median is 1818.

Explanation:

Since the number of observations is odd, the median is the single value located exactly in the middle of the ordered list. There are 3 values smaller than 18 and 3 values larger than 18.