Calculate the standard deviation of the following class interval 0-20,20-40,40-60,60-80Frequency 5,16,14,5

To calculate the standard deviation of a frequency distribution, we need to follow these steps:

### Step 1: Find the midpoint (class mark) of each class interval.

\[
\text{Midpoint (}x_i\text{)} = \frac{\text{Lower limit} + \text{Upper limit}}{2}
\]

Class intervals: \(0-20, 20-40, 40-60, 60-80\)

\[
x_1 = \frac{0 + 20}{2} = 10,\quad x_2 = \frac{20 + 40}{2} = 30,\quad x_3 = \frac{40 + 60}{2} = 50,\quad x_4 = \frac{60 + 80}{2} = 70
\]

### Step 2: Multiply the midpoint by the frequency to get \(f_i \times x_i\).

Given frequencies \(f_1 = 5\), \(f_2 = 16\), \(f_3 = 14\), \(f_4 = 5\).

\[
f_1 \times x_1 = 5 \times 10 = 50,\quad f_2 \times x_2 = 16 \times 30 = 480,\quad f_3 \times x_3 = 14 \times 50 = 700,\quad f_4 \times x_4 = 5 \times 70 = 350
\]

### Step 3: Find the mean (\(\bar{x}\)).

\[
\bar{x} = \frac{\sum f_i \times x_i}{\sum f_i} = \frac{50 + 480 + 700 + 350}{5 + 16 + 14 + 5} = \frac{1580}{40} = 39.5
\]

### Step 4: Find the squared deviation from the mean \((x_i - \bar{x})^2\) and multiply by the frequency \(f_i \times (x_i - \bar{x})^2\).

| Class | \(x_i\) | \(f_i\) | \(f_i \times x_i\) | \((x_i - \bar{x})\) | \((x_i - \bar{x})^2\) | \(f_i \times (x_i - \bar{x})^2\) |
|-------|---------|---------|--------------------|---------------------|----------------------|----------------------------------|
| 0-20  | 10      | 5       | 50                 | \(10 - 39.5 = -29.5\) | 870.25               | \(5 \times 870.25 = 4351.25\)    |
| 20-40 | 30      | 16      | 480                | \(30 - 39.5 = -9.5\)  | 90.25                | \(16 \times 90.25 = 1444\)       |
| 40-60 | 50      | 14      | 700                | \(50 - 39.5 = 10.5\)  | 110.25               | \(14 \times 110.25 = 1543.5\)    |
| 60-80 | 70      | 5       | 350                | \(70 - 39.5 = 30.5\)  | 930.25               | \(5 \times 930.25 = 4651.25\)    |

### Step 5: Find the variance.

\[
\text{Variance} = \frac{\sum f_i \times (x_i - \bar{x})^2}{\sum f_i} = \frac{4351.25 + 1444 + 1543.5 + 4651.25}{40} = \frac{11990}{40} = 299.75
\]

### Step 6: Find the standard deviation.

\[
\text{Standard deviation} = \sqrt{\text{Variance}} = \sqrt{299.75} \approx 17.32
\]

The standard deviation of the given frequency distribution is approximately **17.32**.

Would you like any further details or clarifications?

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Here are 5 related questions to expand your understanding:

1. How does the class width affect the standard deviation?
2. What would happen to the standard deviation if the frequencies were doubled?
3. How can we interpret the value of the standard deviation in a real-world context?
4. How is the standard deviation different from variance?
5. Can you calculate the mean and median of this distribution?

**Tip:** Always double-check if the class intervals are equal before applying the standard deviation formula.

Calculate the standard deviation of the following class interval 0-20, 20-40, 40-60, 60-80. Frequency 5, 16, 14, 5.

Learn how to calculate the standard deviation for a frequency distribution with class intervals 0-20, 20-40, 40-60, 60-80 and frequencies 5, 16, 14, 5. This detailed step-by-step guide covers essential statistical concepts such as midpoints, variance, and standard deviation. Suitable for students in Grades 10-12.

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