Math Problem Statement

Find ​(a) the range and ​(b) the standard deviation of the set of data. 30, 28​, 33​, 23​,32​, 24​, 26

Solution

Let's start by calculating both the range and the standard deviation for the given data set:
Data set: 30, 28, 33, 23, 32, 24, 26

(a) Range

The range is calculated as the difference between the maximum and minimum values of the data set.

  • Maximum value: 33
  • Minimum value: 23

Range=3323=10\text{Range} = 33 - 23 = 10

(b) Standard Deviation

To calculate the standard deviation, we will follow these steps:

  1. Find the mean (average) of the dataLet's start by calculating both the range and the standard deviation for the given data set:
    Data set: 30, 28, 33, 23, 32, 24, 26

(a) Range

The range is calculated as the difference between the maximum and minimum values of the data set.

  • Maximum value: 33
  • Minimum value: 23

Range=3323=10\text{Range} = 33 - 23 = 10

(b) Standard Deviation

To calculate the standard deviation, we will follow these steps:

  1. Find the mean (average) of the Mean=30+28+33+23+32+24+267=1967=28\text{Mean} = \frac{30 + 28 + 33 + 23 + 32 + 24 + 26}{7} = \frac{196}{7} = 28

  2. Calculate each data point's deviation from the mean and square the result: (3028)2=4,(2828)2=0,(3328)2=25(30 - 28)^2 = 4, \quad (28 - 28)^2 = 0, \quad (33 - 28)^2 = 25 (2328)2=25,(3228)2=16,(2428)2=16,(2628)2=4(23 - 28)^2 = 25, \quad (32 - 28)^2 = 16, \quad (24 - 28)^2 = 16, \quad (26 - 28)^2 = 4

  3. Find the variance by averaging these squared deviations: Variance=4+0+25+25+16+16+47=90712.86\text{Variance} = \frac{4 + 0 + 25 + 25 + 16 + 16 + 4}{7} = \frac{90}{7} \approx 12.86

  4. Standard deviation is the square root of the variance: Standard Deviation=12.863.59\text{Standard Deviation} = \sqrt{12.86} \approx 3.59

Final Answer:

  • Range: 10
  • Standard Deviation: 3.59

Would you like more details on this, or do you have any questions?


Here are some related questions you might find interesting:

  1. How does the standard deviation change if a data point is increased?
  2. How is the variance different from the standard deviation?
  3. What are some common uses of the range in data analysis?
  4. How would removing an outlier affect the standard deviation?
  5. Can the standard deviation ever be negative?

Tip: The range only tells you about the spread of the data, while the standard deviation tells you about the consistency of the data points around the mean.

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Math Problem Analysis

Mathematical Concepts

Statistics
Descriptive Statistics

Formulas

Range = max(data) - min(data)
Standard Deviation = sqrt((Σ(x - mean)^2) / n)

Theorems

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Suitable Grade Level

Grades 7-9