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To calculate the standard deviation of the data presented in the table, we will follow these steps:

### Step 1: Determine Midpoints for Each Class
The midpoints (\(x_i\)) for each interval are calculated by averaging the upper and lower limits:

- For \(15 < y \leq 20\): \(x_1 = \frac{15 + 20}{2} = 17.5\)
- For \(20 < y \leq 25\): \(x_2 = \frac{20 + 25}{2} = 22.5\)
- For \(25 < y \leq 30\): \(x_3 = \frac{25 + 30}{2} = 27.5\)
- For \(30 < y \leq 35\): \(x_4 = \frac{30 + 35}{2} = 32.5\)

### Step 2: Multiply Midpoints by Frequencies
We multiply each midpoint by its corresponding frequency (\(f_i\)):

- \(17.5 \times 3 = 52.5\)
- \(22.5 \times 6 = 135.0\)
- \(27.5 \times 5 = 137.5\)
- \(32.5 \times 5 = 162.5\)

### Step 3: Compute Total Frequency and Total \(f_ix_i\)
- Total frequency: \(3 + 6 + 5 + 5 = 19\)
- Total \(f_ix_i = 52.5 + 135.0 + 137.5 + 162.5 = 487.5\)

### Step 4: Compute the Mean (\(\bar{x}\))
The mean is given by:

\[
\bar{x} = \frac{\sum f_ix_i}{\sum f_i} = \frac{487.5}{19} \approx 25.66
\]

### Step 5: Calculate Squared Deviations and Multiply by Frequencies
Next, we compute \((x_i - \bar{x})^2\) for each midpoint and multiply by the corresponding frequency:

- \((17.5 - 25.66)^2 \times 3 = 66.474 \times 3 = 199.42\)
- \((22.5 - 25.66)^2 \times 6 = 9.986 \times 6 = 59.92\)
- \((27.5 - 25.66)^2 \times 5 = 3.366 \times 5 = 16.83\)
- \((32.5 - 25.66)^2 \times 5 = 46.664 \times 5 = 233.32\)

### Step 6: Compute Variance
The variance is:

\[
\text{Variance} = \frac{\sum f_i(x_i - \bar{x})^2}{\sum f_i} = \frac{199.42 + 59.92 + 16.83 + 233.32}{19} = \frac{509.49}{19} \approx 26.81
\]

### Step 7: Calculate Standard Deviation
The standard deviation is the square root of the variance:

\[
\text{Standard Deviation} = \sqrt{26.81} \approx 5.2
\]

### Final Answer: 
The standard deviation of the data is approximately **5.2 cm**.

Would you like further clarification on any of the steps? Here are five follow-up questions related to this process:

1. What is the importance of calculating the mean in finding the standard deviation?
2. How do class intervals affect the calculation of midpoints?
3. Why is the square of deviations used in the calculation of variance?
4. What is the difference between population and sample standard deviation?
5. Can standard deviation ever be negative?

**Tip:** Always check if the class intervals are equal when calculating frequencies and midpoints, as unequal intervals require extra considerations in statistical measures.

Joana collects the lengths of some animals and records the data in the table below:

Length (y cm) | Frequency
------------------------
15 < y ≤ 20  | 3
20 < y ≤ 25  | 6
25 < y ≤ 30  | 5
30 < y ≤ 35  | 5

Find the standard deviation of the data in the table. Give your answer correct to 1 decimal place where appropriate.

Find out how to calculate the standard deviation for the lengths of animals, recorded in a frequency table. This step-by-step solution explains the use of midpoints, mean, variance, and standard deviation for grouped data. Suitable for high school students in Grades 10-12.

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