To calculate the standard deviation $s$ of the sample data using the frequency distribution, we will use the formula:

$s = \sqrt{ \frac{ n \left( \sum f \cdot x^2 \right) - \left( \sum (f \cdot x) \right)^2 }{n(n-1)} }$ Where:

• $x$ is the class midpoint,
• $f$ is the frequency of the class,
• $n$ is the total number of sample values (sum of all frequencies).

Step-by-step solution:

1. Find the class midpoints $x$:

• For each class interval, the midpoint $x$ is calculated as: $x = \frac{\text{Lower boundary} + \text{Upper boundary}}{2}$
  • For the interval $30-36$, $x = \frac{30 + 36}{2} = 33$
  • For the interval $37-43$, $x = \frac{37 + 43}{2} = 40$
  • For the interval $44-50$, $x = \frac{44 + 50}{2} = 47$
  • For the interval $51-57$, $x = \frac{51 + 57}{2} = 54$
  • For the interval $58-64$, $x = \frac{58 + 64}{2} = 61$
  • For the interval $65-71$, $x = \frac{65 + 71}{2} = 68$

2. Create a table with $x$, $f$, $f \cdot x$, and $f \cdot x^2$:

\hline \text{Class Interval} & f & x & f \cdot x & f \cdot x^2 \\ \hline 30-36 & 4 & 33 & 132 & 4356 \\ 37-43 & 25 & 40 & 1000 & 40000 \\ 44-50 & 38 & 47 & 1786 & 83942 \\ 51-57 & 25 & 54 & 1350 & 72900 \\ 58-64 & 5 & 61 & 305 & 18605 \\ 65-71 & 5 & 68 & 340 & 23120 \\ \hline \end{array}$$ #### 3. Calculate $$\sum f$$, $$\sum f \cdot x$$, and $$\sum f \cdot x^2$$: - $$\sum f = 4 + 25 + 38 + 25 + 5 + 5 = 102$$ - $$\sum (f \cdot x) = 132 + 1000 + 1786 + 1350 + 305 + 340 = 4913$$ - $$\sum (f \cdot x^2) = 4356 + 40000 + 83942 + 72900 + 18605 + 23120 = 241923$$ #### 4. Apply the standard deviation formula: $$s = \sqrt{ \frac{ 102 \cdot 241923 - (4913)^2 }{ 102 \cdot (102 - 1) } }$$ Now let me compute the exact value for the standard deviation.The computed standard deviation of the sample data is approximately $$s = 7.23$$. Now, let's compare this value to the original standard deviation given as 9.0. The computed value $$7.23$$ is slightly lower than the original standard deviation of $$9.0$$, but it is close, indicating the calculation aligns with the distribution fairly well. Do you want further details on this calculation or have any questions? Here are 5 questions to deepen understanding: 1. How does the standard deviation formula for grouped data differ from ungrouped data? 2. What is the significance of the class midpoint in frequency distribution calculations? 3. Why might there be a difference between the computed standard deviation and the original value? 4. How would increasing the sample size affect the standard deviation? 5. Can you explain the importance of variance and how it relates to standard deviation? **Tip:** Standard deviation is a measure of how spread out the data values are from the mean; the larger the standard deviation, the more dispersed the data.