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Statistics and Probability - Standard deviation-extended

Grade 9IB

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

🔑Concepts

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Standard Deviation (σ\sigma) is a measure of the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.

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Variance (σ2\sigma^2) is the average of the squared differences from the Mean. The Standard Deviation is the positive square root of the Variance.

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For discrete data, the mean is represented as xˉ\bar{x}, where xˉ=∑xn\bar{x} = \frac{\sum x}{n}.

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In grouped data, we use the midpoint of each class interval as the xx value to calculate the mean and standard deviation.

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Effect of constant operations: If a constant kk is added to every data point, the standard deviation remains unchanged. If every data point is multiplied by a constant kk, the new standard deviation becomes ∣k∣×σ|k| \times \sigma.

📐Formulae

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

σ2=∑(xi−xˉ)2n\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}

σ=∑(xi−xˉ)2n\sigma = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n}}

σ=∑fi(xi−xˉ)2∑fi\sigma = \sqrt{\frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}}

σ=∑x2n−(xˉ)2\sigma = \sqrt{\frac{\sum x^2}{n} - (\bar{x})^2}

💡Examples

Problem 1:

Calculate the standard deviation for the data set: 3,7,8,10,123, 7, 8, 10, 12.

Solution:

  1. Calculate the mean (xˉ\bar{x}): xˉ=3+7+8+10+125=405=8\bar{x} = \frac{3 + 7 + 8 + 10 + 12}{5} = \frac{40}{5} = 8

  2. Calculate squared deviations from the mean (x−xˉ)2(x - \bar{x})^2:

  • (3−8)2=(−5)2=25(3 - 8)^2 = (-5)^2 = 25
  • (7−8)2=(−1)2=1(7 - 8)^2 = (-1)^2 = 1
  • (8−8)2=(0)2=0(8 - 8)^2 = (0)^2 = 0
  • (10−8)2=(2)2=4(10 - 8)^2 = (2)^2 = 4
  • (12−8)2=(4)2=16(12 - 8)^2 = (4)^2 = 16

  1. Find the sum of squared deviations: 25+1+0+4+16=4625 + 1 + 0 + 4 + 16 = 46

  2. Calculate Variance (σ2\sigma^2): σ2=465=9.2\sigma^2 = \frac{46}{5} = 9.2

  3. Calculate Standard Deviation (σ\sigma): σ=9.2≈3.03\sigma = \sqrt{9.2} \approx 3.03

Explanation:

First, find the arithmetic mean. Then, find how much each point deviates from that mean, square those values to eliminate negatives, average them to find the variance, and finally take the square root.

Problem 2:

A data set has a mean of 1515 and a standard deviation of 44. If every value in the set is multiplied by 33 and then 55 is added to each, find the new mean and the new standard deviation.

Solution:

  1. New Mean: xˉnew=(15×3)+5=45+5=50\bar{x}_{new} = (15 \times 3) + 5 = 45 + 5 = 50

  2. New Standard Deviation:

  • Adding a constant does not change σ\sigma.
  • Multiplying by a constant k=3k=3 scales σ\sigma. σnew=4×3=12\sigma_{new} = 4 \times 3 = 12

Explanation:

The mean is affected by both addition and multiplication. However, the standard deviation (measure of spread) is only affected by multiplication, as shifting the entire data set by adding a constant does not change how spread out the numbers are relative to each other.