Math Problem Statement

Find the standard deviation of the following data set. Assume the data set is a sample. Round your answer to the nearest hundredth, if necessary.  

20

40

16

22

35

23

23

29

28

28

38

Solution

To find the standard deviation of a sample data set, we follow these steps:

  1. Calculate the mean (average) of the data set.
  2. Subtract the mean from each data point and square the result (these are called squared deviations).
  3. Sum all the squared deviations.
  4. Divide this sum by the number of data points minus one (this is the sample variance).
  5. Take the square root of the sample variance to get the sample standard deviation.

Here is the given data set: 20,40,16,22,35,23,23,29,28,28,3820, 40, 16, 22, 35, 23, 23, 29, 28, 28, 38

Step 1: Calculate the Mean

First, find the mean of the data set.

Mean(xˉ)=i=1nxin\text{Mean} (\bar{x}) = \frac{\sum_{i=1}^{n} x_i}{n}

Mean(xˉ)=20+40+16+22+35+23+23+29+28+28+3811\text{Mean} (\bar{x}) = \frac{20 + 40 + 16 + 22 + 35 + 23 + 23 + 29 + 28 + 28 + 38}{11}

Mean(xˉ)=3021127.45\text{Mean} (\bar{x}) = \frac{302}{11} \approx 27.45

Step 2: Calculate the Squared Deviations

Subtract the mean from each data point and square the result.

(2027.45)255.50(20 - 27.45)^2 \approx 55.50 (4027.45)2156.80(40 - 27.45)^2 \approx 156.80 (1627.45)2131.60(16 - 27.45)^2 \approx 131.60 (2227.45)229.70(22 - 27.45)^2 \approx 29.70 (3527.45)256.10(35 - 27.45)^2 \approx 56.10 (2327.45)219.80(23 - 27.45)^2 \approx 19.80 (2327.45)219.80(23 - 27.45)^2 \approx 19.80 (2927.45)22.40(29 - 27.45)^2 \approx 2.40 (2827.45)20.30(28 - 27.45)^2 \approx 0.30 (2827.45)20.30(28 - 27.45)^2 \approx 0.30 (3827.45)2111.10(38 - 27.45)^2 \approx 111.10

Step 3: Sum of Squared Deviations

(xixˉ)2=55.50+156.80+131.60+29.70+56.10+19.80+19.80+2.40+0.30+0.30+111.10=583.40\sum (x_i - \bar{x})^2 = 55.50 + 156.80 + 131.60 + 29.70 + 56.10 + 19.80 + 19.80 + 2.40 + 0.30 + 0.30 + 111.10 = 583.40

Step 4: Calculate the Sample Variance

s2=(xixˉ)2n1s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}

s2=583.40111=583.4010=58.34s^2 = \frac{583.40}{11 - 1} = \frac{583.40}{10} = 58.34

Step 5: Calculate the Sample Standard Deviation

s=s2=58.347.64s = \sqrt{s^2} = \sqrt{58.34} \approx 7.64

Summary:

  • Standard Deviation: 7.64\approx 7.64

Do you have any questions or need further details?

Relative Questions:

  1. What is the formula for calculating the standard deviation of a sample?
  2. How is the standard deviation different from the variance?
  3. Why do we subtract one from the sample size when calculating the sample variance?
  4. How does the standard deviation help in understanding the dispersion of data?
  5. What are some practical applications of standard deviation in real-world scenarios?
  6. Can the standard deviation ever be negative?
  7. How would you interpret a high standard deviation versus a low standard deviation?
  8. What are the limitations of using standard deviation as a measure of spread?

Tip:

Always ensure that you are distinguishing between calculating the standard deviation for a sample versus a population, as the formulas differ slightly by dividing by n1n-1 for a sample and by nn for a population.

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Math Problem Analysis

Mathematical Concepts

Statistics
Descriptive Statistics
Standard Deviation

Formulas

Sample Standard Deviation Formula: s = sqrt( (sum((x_i - mean)^2)) / (n - 1) )

Theorems

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Suitable Grade Level

Grades 9-12