The problem presents a frequency distribution table with intervals and frequencies, and it asks to calculate the standard deviation ($s$) using the corresponding formulas.

Here's the step-by-step approach to solve it:

Step 1: Extract Midpoints ($x_i$) of Each Class Interval

The class intervals are:

• 10 – 19
• 20 – 29
• 30 – 39
• 40 – 49
• 50 – 59
• 60 – 69
• 70 – 79
• 80 – 89
• 90 – 99

To calculate the midpoint for each class interval, use the formula: $x_i = \frac{\text{Lower Limit} + \text{Upper Limit}}{2}$

Midpoints for each interval:

• For 10 – 19: $x_1 = \frac{10 + 19}{2} = 14.5$
• For 20 – 29: $x_2 = \frac{20 + 29}{2} = 24.5$
• For 30 – 39: $x_3 = \frac{30 + 39}{2} = 34.5$
• For 40 – 49: $x_4 = \frac{40 + 49}{2} = 44.5$
• For 50 – 59: $x_5 = \frac{50 + 59}{2} = 54.5$
• For 60 – 69: $x_6 = \frac{60 + 69}{2} = 64.5$
• For 70 – 79: $x_7 = \frac{70 + 79}{2} = 74.5$
• For 80 – 89: $x_8 = \frac{80 + 89}{2} = 84.5$
• For 90 – 99: $x_9 = \frac{90 + 99}{2} = 94.5$

Step 2: Compute Mean ($\bar{x}$)

The mean of the data is given by: $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$ Where $f_i$ is the frequency of each class, and $x_i$ is the midpoint of each class.

The frequency data is: $\{2, 3, 5, 8, 10, 9, 7, 4, 2\}$ So, calculate: $\sum f_i x_i = (2 \times 14.5) + (3 \times 24.5) + (5 \times 34.5) + (8 \times 44.5) + (10 \times 54.5) + (9 \times 64.5) + (7 \times 74.5) + (4 \times 84.5) + (2 \times 94.5)$

Perform the calculations:

• $2 \times 14.5 = 29$
• $3 \times 24.5 = 73.5$
• $5 \times 34.5 = 172.5$
• $8 \times 44.5 = 356$
• $10 \times 54.5 = 545$
• $9 \times 64.5 = 580.5$
• $7 \times 74.5 = 521.5$
• $4 \times 84.5 = 338$
• $2 \times 94.5 = 189$

Summing these products: $\sum f_i x_i = 29 + 73.5 + 172.5 + 356 + 545 + 580.5 + 521.5 + 338 + 189 = 2805$

Now, divide by the total frequency (50): $\bar{x} = \frac{2805}{50} = 56.1$

Step 3: Compute Variance and Standard Deviation

The variance formula is: $\sigma^2 = \frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}$

For each class, calculate $(x_i - \bar{x})^2$ and multiply by the corresponding frequency $f_i$:

• $(14.5 - 56.1)^2 \times 2$
• $(24.5 - 56.1)^2 \times 3$
• $(34.5 - 56.1)^2 \times 5$
• $(44.5 - 56.1)^2 \times 8$
• $(54.5 - 56.1)^2 \times 10$
• $(64.5 - 56.1)^2 \times 9$
• $(74.5 - 56.1)^2 \times 7$
• $(84.5 - 56.1)^2 \times 4$
• $(94.5 - 56.1)^2 \times 2$

I will now calculate these values and give the final result for standard deviation.

The calculations yield the following results:

• Mean ($\bar{x}$): $56.1$
• Variance ($\sigma^2$): $381.44$
• Standard Deviation ($s$): $19.53$

These values reflect the statistical properties of the data based on the given frequency distribution.

Let me know if you'd like more details on any step or have questions!

Here are some follow-up questions:

1. How do we interpret the standard deviation in the context of this data?
2. Can we compute the coefficient of variation using the mean and standard deviation?
3. What would happen to the standard deviation if the frequencies were doubled?
4. How does the spread of midpoints affect the variance?
5. What other statistical measures can be derived from this frequency distribution?

Tip: Standard deviation helps measure how spread out the data points are relative to the mean, with higher values indicating more variability.