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Statistics and Probability - Data manipulation and misinterpretation

Grade 10IB

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

🔑Concepts

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Sampling Bias: This occurs when the sample chosen is not representative of the entire population. For example, surveying only people at a gym about national health habits creates a biased sample.

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Misleading Scales: Graphs can be manipulated by starting the yy-axis at a value other than 00. This is called a truncated axis and makes small differences between data points appear much larger.

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Data Cherry-picking: This involve selecting only the data points that support a specific conclusion while ignoring data that contradicts it.

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Mean vs. Median: In a skewed distribution with outliers, the mean xˉ\bar{x} can be misleading. The median is often a better measure of central tendency because it is not affected by extreme values.

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Correlation vs. Causation: A mathematical relationship between two variables, xx and yy, does not imply that xx causes yy. There may be a lurking variable influencing both.

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Small Sample Size: Conclusions drawn from a very small sample (nn) are less reliable and have a higher margin of error compared to larger samples.

📐Formulae

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

Percentage Increase=New Value−Original ValueOriginal Value×100%Percentage\ Increase = \frac{New\ Value - Original\ Value}{Original\ Value} \times 100\%

Range=xmax−xminRange = x_{max} - x_{min}

💡Examples

Problem 1:

A company reports the following annual salaries for its 5 employees: Rs 20,000, Rs 22,000, Rs 25,000, Rs 28,000, and Rs 150,000. The CEO claims the 'average' salary is over Rs 40,000. Is this a fair representation of the typical worker's pay?

Solution:

First, calculate the mean: xˉ=20000+22000+25000+28000+1500005=2450005=49000\bar{x} = \frac{20000 + 22000 + 25000 + 28000 + 150000}{5} = \frac{245000}{5} = 49000. Next, find the median: The middle value of the ordered set is Rs 25,000.

Explanation:

While the mean is technically Rs 49,000, it is heavily skewed by the outlier of Rs 150,000. Most employees earn significantly less than the mean. Using the median (Rs 25,000) provides a more honest representation of a 'typical' salary.

Problem 2:

A graph showing housing price increases starts the vertical axis at Rs 400,000. In Year 1, the price is Rs 410,000. In Year 2, the price is Rs 420,000. Calculate the percentage increase and explain how the graph might be misleading.

Solution:

Percentage Increase=420000−410000410000×100%≈2.44%Percentage\ Increase = \frac{420000 - 410000}{410000} \times 100\% \approx 2.44\%

Explanation:

Because the yy-axis starts at Rs 400,000, the bar for Year 2 (representing Rs 20,000 above the baseline) will look twice as tall as the bar for Year 1 (representing Rs 10,000 above the baseline). This visual suggests a 100%100\% increase, whereas the actual mathematical increase is only 2.44%2.44\%.

Problem 3:

Determine the change in the range if an outlier of 100100 is added to the data set: {10,12,15,18,20}\{10, 12, 15, 18, 20\}.

Solution:

Original Range: 20−10=1020 - 10 = 10. New data set: {10,12,15,18,20,100}\{10, 12, 15, 18, 20, 100\}. New Range: 100−10=90100 - 10 = 90.

Explanation:

The range increased from 1010 to 9090. This demonstrates how a single outlier can drastically change the spread of the data, which can be used to misinterpret the consistency of a data set.

Data manipulation and misinterpretation Grade 10 Notes & Examples