Math Problem Statement

Find teh standard deviation of the following by direct method class interval and frequency are given as follows. 0-10, 10-20, 20-30, 30-40, 40/50 and frequency are 10,3,2,1,4 respectively.

Solution

To find the standard deviation using the direct method, we need to follow these steps:

Step 1: Find the Midpoints (xx) of Each Class Interval

The midpoint of each class interval is calculated as: x=Lower Limit+Upper Limit2x = \frac{\text{Lower Limit} + \text{Upper Limit}}{2} For the given intervals:

  • 010:0+102=50-10: \frac{0 + 10}{2} = 5
  • 1020:10+202=1510-20: \frac{10 + 20}{2} = 15
  • 2030:20+302=2520-30: \frac{20 + 30}{2} = 25
  • 3040:30+402=3530-40: \frac{30 + 40}{2} = 35
  • 4050:40+502=4540-50: \frac{40 + 50}{2} = 45

Step 2: Set Up the Table

We create a table for xx, ff, fxfx, and (xxˉ)2f(x - \bar{x})^2f, where:

  • ff: Frequency
  • fx=f×xfx = f \times x
  • (xxˉ)2f(x - \bar{x})^2f: Squared deviation multiplied by the frequency
Class Intervalxxfffxfx(xxˉ)2f(x - \bar{x})^2f
0100-105510105050
102010-201515334545
203020-302525225050
304030-403535113535
405040-50454544180180

Step 3: Find xˉ\bar{x} (Mean)

xˉ=ΣfxΣf\bar{x} = \frac{\Sigma fx}{\Sigma f} Σf=10+3+2+1+4=20\Sigma f = 10 + 3 + 2 + 1 + 4 = 20 Σfx=50+45+50+35+180=360\Sigma fx = 50 + 45 + 50 + 35 + 180 = 360 xˉ=36020=18\bar{x} = \frac{360}{20} = 18

Step 4: Calculate (xxˉ)2f(x - \bar{x})^2f

Now compute (xxˉ)2(x - \bar{x})^2 and (xxˉ)2f(x - \bar{x})^2f for each class.

Class Intervalxxfffxfxxxˉx - \bar{x}(xxˉ)2(x - \bar{x})^2(xxˉ)2f(x - \bar{x})^2f
0100-10551010505013-1316916916901690
102010-2015153345453-3992727
203020-3025252250507749499898
304030-4035351135351717289289289289
405040-50454544180180272772972929162916

Σ((xxˉ)2f)=1690+27+98+289+2916=5020\Sigma \left((x - \bar{x})^2f\right) = 1690 + 27 + 98 + 289 + 2916 = 5020

Step 5: Calculate Standard Deviation

The formula for standard deviation is: σ=Σ(xxˉ)2fΣf\sigma = \sqrt{\frac{\Sigma (x - \bar{x})^2f}{\Sigma f}} σ=502020\sigma = \sqrt{\frac{5020}{20}} σ=25115.84\sigma = \sqrt{251} \approx 15.84

Final Answer:

The standard deviation is approximately 15.84.


Would you like detailed explanations or assistance with related concepts?

Relative Questions:

  1. How is the direct method for standard deviation different from the assumed mean method?
  2. What are the advantages of calculating midpoints in grouped data?
  3. How do changes in class intervals affect standard deviation?
  4. Can the direct method be applied to continuous data as well?
  5. How does the frequency distribution shape affect the standard deviation?

Tip:

Always ensure to use consistent class intervals to avoid bias in statistical calculations.

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Math Problem Analysis

Mathematical Concepts

Statistics
Standard Deviation
Grouped Data Analysis

Formulas

Midpoint formula: x = (Lower Limit + Upper Limit) / 2
Mean formula: x̄ = Σ(fx) / Σf
Standard deviation formula: σ = √(Σ((x - x̄)^2f) / Σf)

Theorems

Measures of Dispersion
Direct Method for Standard Deviation

Suitable Grade Level

Grades 10-12