To find the standard deviation $s$ of the sample data summarized in the frequency distribution table, we will use the following formula:

$s = \sqrt{\frac{ \sum f(x - \bar{x})^2 }{ n - 1 }}$

where:

• $x$ represents the class midpoint,
• $f$ represents the class frequency,
• $n$ represents the total number of sample values,
• $\bar{x}$ is the sample mean.

Step-by-Step Calculation

1. Calculate the class midpoints ($x$):

   • 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: $x = \frac{65 + 71}{2} = 68$
   • For the interval 72-78: $x = \frac{72 + 78}{2} = 75$

2. Calculate the total number of sample values ($n$): $n = 2 + 4 + 4 + 4 + 14 + 30 + 39 = 97$

3. Calculate the mean ($\bar{x}$): $\bar{x} = \frac{ \sum f \cdot x }{ n } = \frac{ (2 \cdot 33) + (4 \cdot 40) + (4 \cdot 47) + (4 \cdot 54) + (14 \cdot 61) + (30 \cdot 68) + (39 \cdot 75) }{ 97 }$

   $\sum f \cdot x = 66 + 160 + 188 + 216 + 854 + 2040 + 2925 = 6449$

   $\bar{x} = \frac{6449}{97} \approx 66.48$

4. Calculate the squared deviations and sum them up: $\sum f(x - \bar{x})^2$

   • For the interval 30-36: $2(33 - 66.48)^2 \approx 2(1137.1904) = 2274.3808$
   • For the interval 37-43: $4(40 - 66.48)^2 \approx 4(705.2304) = 2820.9216$
   • For the interval 44-50: $4(47 - 66.48)^2 \approx 4(384.5904) = 1538.3616$
   • For the interval 51-57: $4(54 - 66.48)^2 \approx 4(156.2304) = 624.9216$
   • For the interval 58-64: $14(61 - 66.48)^2 \approx 14(30.1904) = 422.6656$
   • For the interval 65-71: $30(68 - 66.48)^2 \approx 30(2.3104) = 69.312$
   • For the interval 72-78: $39(75 - 66.48)^2 \approx 39(72.7104) = 2835.7056$

   $\sum f(x - \bar{x})^2 \approx 2274.3808 + 2820.9216 + 1538.3616 + 624.9216 + 422.6656 + 69.312 + 2835.7056 \approx 10586.2688$

5. Calculate the sample standard deviation ($s$): $s = \sqrt{ \frac{ \sum f(x - \bar{x})^2 }{ n - 1 } } = \sqrt{ \frac{ 10586.2688 }{ 96 } } \approx \sqrt{110.2757} \approx 10.50$

Comparison with Given Standard Deviation

The computed standard deviation is approximately $10.50$. This is close to the given standard deviation of $11.1$.

Do you want more details or have any questions?

Here are some related questions you might ask:

1. How is the class midpoint calculated in a frequency distribution?
2. What is the formula for the sample mean in a frequency distribution?
3. How do you interpret the standard deviation in a frequency distribution?
4. What is the difference between sample standard deviation and population standard deviation?
5. Can you explain the concept of variance in a frequency distribution?

Tip: When dealing with frequency distributions, always double-check your class intervals and frequencies to ensure accuracy in your calculations.