To calculate the standard deviation using the direct method, we follow these steps:

1. **Step 1: Set up the table with given dataTo calculate the standard deviation using the direct method, we follow these steps:

2. **Step 1: Set up the table with given **

   • X (Size): 4, 5, 6, 7, 8, 9, 10
   • f (Frequency): 6, 12, 15, 28, 20, 14, 5

3. Step 2: Compute the following quantities for each $X_i$:

   • $f_i X_i$ (Frequency times Size)
   • $f_i X_i^2$ (Frequency times Size squared)

$X$	$f$	$f \cdot X$	$f \cdot X^2$
4	6	$6 \cdot 4 = 24$	$6 \cdot 4^2 = 6 \cdot 16 = 96$
5	12	$12 \cdot 5 = 60$	$12 \cdot 5^2 = 12 \cdot 25 = 300$
6	15	$15 \cdot 6 = 90$	$15 \cdot 6^2 = 15 \cdot 36 = 540$
7	28	$28 \cdot 7 = 196$	$28 \cdot 7^2 = 28 \cdot 49 = 1372$
8	20	$20 \cdot 8 = 160$	$20 \cdot 8^2 = 20 \cdot 64 = 1280$
9	14	$14 \cdot 9 = 126$	$14 \cdot 9^2 = 14 \cdot 81 = 1134$
10	5	$5 \cdot 10 = 50$	$5 \cdot 10^2 = 5 \cdot 100 = 500$

3. Step 3: Sum up the values:

   • $\sum f = 6 + 12 + 15 + 28 + 20 + 14 + 5 = 100$
   • $\sum fX = 24 + 60 + 90 + 196 + 160 + 126 + 50 = 706$
   • $\sum fX^2 = 96 + 300 + 540 + 1372 + 1280 + 1134 + 500 = 5222$

4. Step 4: Calculate the mean ($\mu$): $\mu = \frac{\sum fX}{\sum f} = \frac{706}{100} = 7.06$

5. Step 5: Calculate the variance ($\sigma^2$): $\sigma^2 = \frac{\sum fX^2}{\sum f} - \mu^2 = \frac{5222}{100} - (7.06)^2 = 52.22 - 49.8436 = 2.3764$

6. Step 6: Calculate the standard deviation ($\sigma$): $\sigma = \sqrt{2.3764} \approx 1.541$

Thus, the standard deviation of the distribution is approximately 1.541.

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Would you like further clarification or details? Here are 5 related questions:

1. How would the standard deviation change if the frequency of one of the values increased?
2. What is the relationship between variance and standard deviation?
3. How would we calculate the variance if given only raw data without frequencies?
4. What are the practical uses of standard deviation in analyzing data sets?
5. How does changing the size values affect the overall mean and variance?

Tip: Standard deviation is a key measure of the spread of data. The smaller it is, the closer the data points are to the mean.