What is the standard deviation of the following data ?
`{:("Measurement",0-10,10-20,20-30,30-40),("Frequency",1,3,4,2):}` (a) 81 (b) 7.6 (c) 9 (d) 2.26

Options

A. 81
B. 7.6
C. 9
D. 2.26

Explanation

To find the standard deviation of the given data, we will follow these steps:

### Step 1: Create a Frequency Table
We have the following data:

| Measurement (Interval) | Frequency (f) |
|------------------------|---------------|
| 0 - 10 | 1 |
| 10 - 20 | 3 |
| 20 - 30 | 4 |
| 30 - 40 | 2 |

### Step 2: Calculate the Midpoint (y_i) for Each Interval
The midpoint for each interval is calculated as follows:

- For 0 - 10: \( y_1 = \frac{0 + 10}{2} = 5 \)
- For 10 - 20: \( y_2 = \frac{10 + 20}{2} = 15 \)
- For 20 - 30: \( y_3 = \frac{20 + 30}{2} = 25 \)
- For 30 - 40: \( y_4 = \frac{30 + 40}{2} = 35 \)

### Step 3: Calculate \( u_i \)
We will choose a value for \( A \) (the assumed mean) as the midpoint of the third interval, which is 25. The formula for \( u_i \) is:

\[ u_i = \frac{y_i - A}{h} \]

where \( h \) is the width of the interval (10).

Calculating \( u_i \):

- \( u_1 = \frac{5 - 25}{10} = -2 \)
- \( u_2 = \frac{15 - 25}{10} = -1 \)
- \( u_3 = \frac{25 - 25}{10} = 0 \)
- \( u_4 = \frac{35 - 25}{10} = 1 \)

### Step 4: Calculate \( f_i u_i \) and \( f_i u_i^2 \)
Now we will calculate \( f_i u_i \) and \( f_i u_i^2 \):

| Frequency (f) | Midpoint (y_i) | \( u_i \) | \( f_i u_i \) | \( u_i^2 \) | \( f_i u_i^2 \) |
|---------------|----------------|-----------|----------------|--------------|------------------|
| 1 | 5 | -2 | -2 | 4 | 4 |
| 3 | 15 | -1 | -3 | 1 | 3 |
| 4 | 25 | 0 | 0 | 0 | 0 |
| 2 | 35 | 1 | 2 | 1 | 2 |

### Step 5: Calculate Summations
Now, we will calculate the necessary summations:

- \( \sum f_i = 1 + 3 + 4 + 2 = 10 \)
- \( \sum f_i u_i = -2 - 3 + 0 + 2 = -3 \)
- \( \sum f_i u_i^2 = 4 + 3 + 0 + 2 = 9 \)

### Step 6: Calculate the Standard Deviation
The formula for the standard deviation \( \sigma \) is:

\[ \sigma = h \sqrt{\frac{\sum f_i u_i^2}{n} - \left(\frac{\sum f_i u_i}{n}\right)^2} \]

Substituting the values we have:

- \( n = \sum f_i = 10 \)
- \( h = 10 \)

Now substituting these into the formula:

\[ \sigma = 10 \sqrt{\frac{9}{10} - \left(\frac{-3}{10}\right)^2} \]

Calculating the terms:

\[ \sigma = 10 \sqrt{\frac{9}{10} - \frac{9}{100}} \]

Finding a common denominator:

\[ \sigma = 10 \sqrt{\frac{90}{100} - \frac{9}{100}} = 10 \sqrt{\frac{81}{100}} = 10 \cdot \frac{9}{10} = 9 \]

Thus, the standard deviation of the data is:

\[ \boxed{9} \]