Math Problem Statement

Find the range, sample standard deviation and inter-quartile range (IQR) of the following data set.

21 35 38 40 40 42 45 47 61 85

Solution

Let's calculate the range, sample standard deviation, and inter-quartile range (IQR) for the given data set:

Given Data Set:

21, 35, 38, 40, 40, 42, 45, 47, 61, 85

1. Range

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

  • Maximum value: 85
  • Minimum value: 21

Range=8521=64\text{Range} = 85 - 21 = 64

2. Inter-Quartile Range (IQR)

The IQR is the difference between the third quartile (Q3) and the first quartile (Q1). To calculate the quartiles:

  • Sort the data (already sorted):
    21, 35, 38, 40, 40, 42, 45, 47, 61, 85

The data set has 10 numbers, so:

  • Q1Q1 is the median of the first half of the data (excluding the overall median).
    First half: 21, 35, 38, 40, 40
    Median of this set is Q1=38Q1 = 38.

  • Q3Q3 is the median of the second half of the data (excluding the overall median).
    Second half: 42, 45, 47, 61, 85
    Median of this set is Q3=47Q3 = 47.

IQR=Q3Q1=4738=9\text{IQR} = Q3 - Q1 = 47 - 38 = 9

3. Sample Standard Deviation

The sample standard deviation is calculated as:

s=1n1i=1n(xix)2s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (x_i - \overline{x})^2}

Where:

  • nn = 10 (number of data points)
  • x\overline{x} = mean of the data
  • xix_i = each data point
Step 1: Find the Mean (x\overline{x})

x=21+35+38+40+40+42+45+47+61+8510=45410=45.4\overline{x} = \frac{21 + 35 + 38 + 40 + 40 + 42 + 45 + 47 + 61 + 85}{10} = \frac{454}{10} = 45.4

Step 2: Calculate (xix)2(x_i - \overline{x})^2 for each data point:

(2145.4)2=600.25,(3545.4)2=108.16,(21 - 45.4)^2 = 600.25, \quad (35 - 45.4)^2 = 108.16, \quad \dots

The sum of squared differences is:
600.25+108.16+55.76+28.96+28.96+11.56+0.16+2.56+244.36+1576.96=2657.2600.25 + 108.16 + 55.76 + 28.96 + 28.96 + 11.56 + 0.16 + 2.56 + 244.36 + 1576.96 = 2657.2

Step 3: Calculate the Standard Deviation:

s=2657.2101=2657.29=295.2417.18s = \sqrt{\frac{2657.2}{10 - 1}} = \sqrt{\frac{2657.2}{9}} = \sqrt{295.24} \approx 17.18

Summary:

  • Range: 64
  • Inter-Quartile Range (IQR): 9
  • Sample Standard Deviation: 17.18

Would you like more details or explanations on any of these steps?


Here are 5 related questions:

  1. How does the IQR help in identifying outliers?
  2. What is the difference between sample and population standard deviation?
  3. How would the standard deviation change if you added a constant value to each data point?
  4. Can you explain why the range is not always a reliable measure of spread?
  5. How does the shape of the data distribution affect the quartiles?

Tip: The IQR is often a better measure of spread than the range in skewed distributions, as it focuses on the middle 50% of the data.

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Math Problem Analysis

Mathematical Concepts

Descriptive Statistics
Range
Inter-Quartile Range
Sample Standard Deviation

Formulas

Range = Max - Min
Inter-Quartile Range (IQR) = Q3 - Q1
Sample Standard Deviation: s = sqrt((1/(n-1)) * Σ(x_i - x̄)^2)

Theorems

Quartile Theorem
Standard Deviation Formula

Suitable Grade Level

Grades 9-12