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How Quantities Combine: Understanding Data - Average of Averages

Grade 9CBSE

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

🔑Concepts

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The arithmetic mean (or simply mean) is the sum of all observations divided by the total number of observations.

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When combining two or more groups of data, the 'Average of Averages' is the weighted mean of the individual group averages.

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The combined mean depends on both the average of each group and the size (number of observations) of each group.

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The simple average of group means, such as xˉ1+xˉ22\frac{\bar{x}_1 + \bar{x}_2}{2}, is only equal to the combined mean if the number of observations in each group is identical (n1=n2n_1 = n_2).

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To calculate the combined mean, first find the total sum of all observations by multiplying each group's mean by its size, then divide by the total number of observations.

📐Formulae

Mean (xˉ)=∑xin\text{Mean } (\bar{x}) = \frac{\sum x_i}{n}

Total Sum=Mean×Number of Observations\text{Total Sum} = \text{Mean} \times \text{Number of Observations}

Combined Mean (xˉc)=n1xˉ1+n2xˉ2+⋯+nkxˉkn1+n2+⋯+nk\text{Combined Mean } (\bar{x}_c) = \frac{n_1\bar{x}_1 + n_2\bar{x}_2 + \dots + n_k\bar{x}_k}{n_1 + n_2 + \dots + n_k}

💡Examples

Problem 1:

In a school, Section A has 3030 students with an average score of 8080 marks, and Section B has 2020 students with an average score of 9090 marks. Calculate the combined average score of both sections.

Solution:

Step 1: Find the total marks for Section A. Total marks for A = n1×xˉ1=30×80=2400n_1 \times \bar{x}_1 = 30 \times 80 = 2400

Step 2: Find the total marks for Section B. Total marks for B = n2×xˉ2=20×90=1800n_2 \times \bar{x}_2 = 20 \times 90 = 1800

Step 3: Calculate the total number of students. Total students = n1+n2=30+20=50n_1 + n_2 = 30 + 20 = 50

Step 4: Calculate the combined mean. xˉc=2400+180050\bar{x}_c = \frac{2400 + 1800}{50} xˉc=420050=84\bar{x}_c = \frac{4200}{50} = 84

The combined average score is 8484 marks.

Explanation:

We cannot simply average 8080 and 9090 (which would be 8585) because the group sizes are different. Section A has more students, so its average carries more weight.

Problem 2:

The average height of a group of 1010 girls is 150150 cm and the average height of a group of 1010 boys is 160160 cm. What is the combined average height?

Solution:

Step 1: Identify group sizes and means. n1=10n_1 = 10, xˉ1=150\bar{x}_1 = 150 n2=10n_2 = 10, xˉ2=160\bar{x}_2 = 160

Step 2: Since n1=n2n_1 = n_2, we can use the simple average of averages. xˉc=150+1602\bar{x}_c = \frac{150 + 160}{2} xˉc=3102=155 cm\bar{x}_c = \frac{310}{2} = 155 \text{ cm}

Alternatively, using the formula: xˉc=(10×150)+(10×160)10+10=1500+160020=310020=155 cm\bar{x}_c = \frac{(10 \times 150) + (10 \times 160)}{10 + 10} = \frac{1500 + 1600}{20} = \frac{3100}{20} = 155 \text{ cm}

Explanation:

When the number of observations in all groups is the same, the combined mean is exactly equal to the arithmetic mean of the individual averages.