Math Problem Statement

standard sample deviation

Solution

The standard deviation is a statistical measure that quantifies the amount of variation or dispersion in a dataset. It tells you how much the individual data points differ from the mean (average) of the dataset. A low standard deviation indicates that the data points are close to the mean, while a high standard deviation indicates that the data points are spread out over a wider range of values.

Formula for Standard Deviation

For a dataset x1,x2,x3,,xnx_1, x_2, x_3, \ldots, x_n:

  1. Compute the mean μ\mu (for a population) or xˉ\bar{x} (for a sample): Mean xˉ=1ni=1nxi\text{Mean } \bar{x} = \frac{1}{n} \sum_{i=1}^n x_i
  2. Subtract the mean from each data point and square the result to calculate the squared deviations: (xixˉ)2(x_i - \bar{x})^2
  3. Compute the variance:
    • For a sample: s2=i=1n(xixˉ)2n1s^2 = \frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n-1}
    • For a population: σ2=i=1n(xiμ)2n\sigma^2 = \frac{\sum_{i=1}^n (x_i - \mu)^2}{n}
  4. Take the square root of the variance to find the standard deviation:
    • Sample standard deviation: s=s2s = \sqrt{s^2}
    • Population standard deviation: σ=σ2\sigma = \sqrt{\sigma^2}

Example Calculation:

For a sample dataset: [4,8,6,5,3][4, 8, 6, 5, 3]:

  1. Mean: xˉ=4+8+6+5+35=5.2\bar{x} = \frac{4 + 8 + 6 + 5 + 3}{5} = 5.2
  2. Deviations squared: (45.2)2,(85.2)2,(65.2)2,(55.2)2,(35.2)2(4-5.2)^2, (8-5.2)^2, (6-5.2)^2, (5-5.2)^2, (3-5.2)^2 (45.2)2=1.44, (85.2)2=7.84, (65.2)2=0.64, (55.2)2=0.04, (35.2)2=4.84(4-5.2)^2 = 1.44,\ (8-5.2)^2 = 7.84,\ (6-5.2)^2 = 0.64,\ (5-5.2)^2 = 0.04,\ (3-5.2)^2 = 4.84
  3. Variance: s2=1.44+7.84+0.64+0.04+4.844=3.7s^2 = \frac{1.44 + 7.84 + 0.64 + 0.04 + 4.84}{4} = 3.7
  4. Standard deviation: s=3.71.92s = \sqrt{3.7} \approx 1.92

The standard deviation of this dataset is approximately 1.921.92.


Would you like more details, have questions, or want me to solve a specific example for you?

Here are 5 related questions for further learning:

  1. What is the difference between population and sample standard deviation?
  2. How is standard deviation used in real-world data analysis?
  3. Can you calculate the standard deviation for a dataset with given values?
  4. What does a standard deviation of 0 indicate about a dataset?
  5. How is standard deviation related to variance?

Tip: Remember, standard deviation is always non-negative, as it measures the spread of data around the mean!

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Math Problem Analysis

Mathematical Concepts

Statistics
Standard Deviation
Variance

Formulas

Mean: \( \bar{x} = \frac{1}{n} \sum_{i=1}^n x_i \)
Sample variance: \( s^2 = \frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n-1} \)
Sample standard deviation: \( s = \sqrt{s^2} \)

Theorems

Properties of Variance and Standard Deviation

Suitable Grade Level

Grades 9-12