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Statistics - Compute and interpret mean, median, and mode for decision-making

Grade 9CBSE

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

🔑Concepts

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Mean (Arithmetic Average): It is the sum of the values of all observations divided by the total number of observations. It is denoted by xˉ\bar{x}.

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Median: It is the value of the middle-most observation when the data is arranged in ascending or descending order. It effectively splits the data into two equal halves.

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Mode: The value that appears most frequently in a data set is called the mode. A data set can have more than one mode (bimodal/multimodal) or no mode at all.

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Decision-Making Criteria: The Mean is best for data without extreme variations. The Median is preferred when the data contains 'outliers' (extremely high or low values) because it is not affected by them. The Mode is ideal for qualitative data or identifying the most popular item (e.g., most sold shoe size).

📐Formulae

xˉ=∑i=1nxin\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}

xˉ=∑fixi∑fi (For frequency distributions)\bar{x} = \frac{\sum f_i x_i}{\sum f_i} \text{ (For frequency distributions)}

Median=(n+12)th observation (if n is odd)\text{Median} = \left( \frac{n+1}{2} \right)^{th} \text{ observation (if } n \text{ is odd)}

Median=(n2)th obs+(n2+1)th obs2 (if n is even)\text{Median} = \frac{\left( \frac{n}{2} \right)^{th} \text{ obs} + \left( \frac{n}{2} + 1 \right)^{th} \text{ obs}}{2} \text{ (if } n \text{ is even)}

💡Examples

Problem 1:

Find the mean, median, and mode for the following marks obtained by 9 students: 15,12,15,20,18,15,14,10,2215, 12, 15, 20, 18, 15, 14, 10, 22.

Solution:

  1. Mean: Sum =15+12+15+20+18+15+14+10+22=141= 15+12+15+20+18+15+14+10+22 = 141. Total n=9n = 9. xˉ=1419=15.67\bar{x} = \frac{141}{9} = 15.67
  2. Median: Arrange in ascending order: 10,12,14,15,15,15,18,20,2210, 12, 14, 15, 15, 15, 18, 20, 22. Since n=9n=9 (odd), Median =(9+12)th term=5th term=15= \left( \frac{9+1}{2} \right)^{th} \text{ term} = 5^{th} \text{ term} = 15.
  3. Mode: 1515 occurs most frequently (3 times). Mode =15= 15.

Explanation:

In this case, all three measures are quite close, suggesting a relatively balanced distribution of marks.

Problem 2:

A company pays 5 employees monthly salaries of ₹10,000₹ 10,000, ₹11,000₹ 11,000, ₹10,000₹ 10,000, ₹12,000₹ 12,000, and ₹80,000₹ 80,000. Which measure of central tendency best represents the typical salary?

Solution:

  1. Mean Calculation: 10000110001000012000+80000123000\begin{array}{r} 10000 \\ 11000 \\ 10000 \\ 12000 \\ + 80000 \\ \hline 123000 \end{array} Mean =1230005=₹24,600= \frac{123000}{5} = ₹ 24,600.
  2. Median Calculation: Ordered salaries: 10000,10000,11000,12000,8000010000, 10000, 11000, 12000, 80000. Median (3rd3^{rd} term) =₹11,000= ₹ 11,000.

Explanation:

The Mean (₹24,600₹ 24,600) is much higher than what 80% of employees earn due to the outlier (₹80,000₹ 80,000). Therefore, the Median (₹11,000₹ 11,000) is a better representative for decision-making regarding 'typical' pay.