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Statistics and Probability - Standard deviation

Grade 11IB_AI

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

🔑Concepts

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Standard Deviation is a measure of the amount of variation or dispersion in a set of data values. A low standard deviation indicates that the data points tend to be close to the mean μ\mu, while a high standard deviation indicates that the data points are spread out over a wider range.

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The population standard deviation is denoted by the Greek letter σ\sigma (sigma), while the sample standard deviation is often denoted as ss or sn−1s_{n-1}.

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Variance is the square of the standard deviation (σ2\sigma^2). It represents the average of the squared differences from the Mean.

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Linear transformations: If every value in a data set is increased by a constant kk (i.e., x+kx + k), the standard deviation remains unchanged. If every value is multiplied by a constant kk (i.e., kxkx), the standard deviation is multiplied by ∣k∣|k|.

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In IB AI, the Graphic Display Calculator (GDC) is the primary tool for calculating standard deviation. Ensure you distinguish between σx\sigma_x (population) and sxs_x (sample) on your device.

📐Formulae

μ=∑i=1nxin\mu = \frac{\sum_{i=1}^{n} x_i}{n}

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

σ=∑fixi2∑fi−μ2\sigma = \sqrt{\frac{\sum f_i x_i^2}{\sum f_i} - \mu^2}

sn−1=∑(xi−xˉ)2n−1s_{n-1} = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}

💡Examples

Problem 1:

A small dataset consists of the following values: 4,8,10,12,164, 8, 10, 12, 16. Calculate the population standard deviation σ\sigma to 3 significant figures.

Solution:

  1. Find the mean: μ=4+8+10+12+165=505=10\mu = \frac{4 + 8 + 10 + 12 + 16}{5} = \frac{50}{5} = 10
  2. Calculate squared deviations: (4−10)2=36(4-10)^2 = 36 (8−10)2=4(8-10)^2 = 4 (10−10)2=0(10-10)^2 = 0 (12−10)2=4(12-10)^2 = 4 (16−10)2=36(16-10)^2 = 36
  3. Find the mean of squared deviations (Variance): σ2=36+4+0+4+365=805=16\sigma^2 = \frac{36 + 4 + 0 + 4 + 36}{5} = \frac{80}{5} = 16
  4. Standard Deviation: σ=16=4\sigma = \sqrt{16} = 4

Explanation:

To find the standard deviation, we first calculate the mean, then find the average of the squared distances from that mean, and finally take the square root.

Problem 2:

A set of exam scores has a mean of 6565 and a standard deviation of 1212. If the teacher decides to scale the scores by multiplying each score by 1.11.1 and then adding 55 points, what is the new standard deviation?

Solution:

Let the original standard deviation be σold=12\sigma_{old} = 12.

  1. Effect of multiplication: The standard deviation is scaled by the factor 1.11.1. σscaled=12×1.1=13.2\sigma_{scaled} = 12 \times 1.1 = 13.2
  2. Effect of addition: Adding 55 to every score does not change the spread/dispersion. σnew=13.2\sigma_{new} = 13.2

Explanation:

Standard deviation is affected by multiplication (scaling) but is invariant under addition (translation).