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Statistics and Probability - Measures of spread

Grade 11IB_AI

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

🔑Concepts

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Measures of spread describe the variability or dispersion within a data set, indicating how much the data values deviate from the center.

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The Range is the simplest measure of spread, calculated as the difference between the maximum and minimum values: Range=xmax−xminRange = x_{max} - x_{min}.

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The Interquartile Range (IQR) measures the spread of the middle 50% of the data. It is calculated as IQR=Q3−Q1IQR = Q_3 - Q_1 and is more robust than the range because it is not affected by extreme outliers.

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Variance (sigma2\\sigma^2) is the average of the squared deviations from the mean. It provides a measure of how spread out the numbers are.

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Standard Deviation (sigma\\sigma) is the square root of the variance. It is expressed in the same units as the data, representing the 'typical' distance from the mean.

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Outliers are data points that differ significantly from the rest of the set. A common boundary for outliers is any value smaller than Q1−1.5timesIQRQ_1 - 1.5 \\times IQR or larger than Q3+1.5timesIQRQ_3 + 1.5 \\times IQR.

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Effect of Transformations: If every value in a data set is multiplied by a constant aa and increased by bb (y=ax+by = ax + b), the new standard deviation becomes ∣a∣timessigmaold|a| \\times \\sigma_{old} and the new IQR becomes ∣a∣timesIQRold|a| \\times IQR_{old}. Adding bb does not affect the measures of spread.

📐Formulae

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

IQR=Q3−Q1IQR = Q_3 - Q_1

sigma=sqrtfracsumi=1n(xi−mu)2n\\sigma = \\sqrt{\\frac{\\sum_{i=1}^{n} (x_i - \\mu)^2}{n}}

textLowerFence=Q1−1.5timesIQR\\text{Lower Fence} = Q_1 - 1.5 \\times IQR

textUpperFence=Q3+1.5timesIQR\\text{Upper Fence} = Q_3 + 1.5 \\times IQR

💡Examples

Problem 1:

Given the data set: 3,7,8,5,12,14,21,15,18\\{3, 7, 8, 5, 12, 14, 21, 15, 18\\}, calculate the Interquartile Range (IQRIQR) and determine if there are any outliers.

Solution:

  1. Order the data: 3,5,7,8,12,14,15,18,21\\{3, 5, 7, 8, 12, 14, 15, 18, 21\\}.
  2. Median (Q2Q_2) = 1212 (the 5th value).
  3. Q1Q_1 is the median of the lower half 3,5,7,8rightarrowQ1=frac5+72=6\\{3, 5, 7, 8\\} \\rightarrow Q_1 = \\frac{5+7}{2} = 6.
  4. Q3Q_3 is the median of the upper half 14,15,18,21rightarrowQ3=frac15+182=16.5\\{14, 15, 18, 21\\} \\rightarrow Q_3 = \\frac{15+18}{2} = 16.5.
  5. IQR=Q3−Q1=16.5−6=10.5IQR = Q_3 - Q_1 = 16.5 - 6 = 10.5.
  6. Outlier boundaries: Lower: 6−1.5(10.5)=6−15.75=−9.756 - 1.5(10.5) = 6 - 15.75 = -9.75 Upper: 16.5+1.5(10.5)=16.5+15.75=32.2516.5 + 1.5(10.5) = 16.5 + 15.75 = 32.25. Since no values are <−9.75< -9.75 or >32.25> 32.25, there are no outliers.

Explanation:

To find spread using quartiles, first sort the data. The IQR represents the range of the central half of the observations. Outliers are then checked using the 1.5timesIQR1.5 \\times IQR rule.

Problem 2:

A data set has a mean of mu=50\\mu = 50 and a standard deviation of sigma=8\\sigma = 8. If every value in the set is multiplied by 1.51.5 and then 1010 is added to each result, find the new mean and new standard deviation.

Solution:

  1. New Mean: munew=1.5(50)+10=75+10=85\\mu_{new} = 1.5(50) + 10 = 75 + 10 = 85.
  2. New Standard Deviation: Addition of 1010 does not change the spread. Multiplication by 1.51.5 scales the spread. sigmanew=1.5times8=12\\sigma_{new} = 1.5 \\times 8 = 12.

Explanation:

Linear transformations ax+bax + b affect the mean by both aa and bb, but measures of spread (like standard deviation and IQR) are only affected by the scaling factor ∣a∣|a|.