To find the sample standard deviation $s$ for data summarized in a frequency distribution, we use the formula:

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

Where:

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

Steps:

1. Calculate the midpoints of each class interval.
2. Calculate the mean ($\bar{x}$) of the data using the midpoints and frequencies.
3. Calculate the variance by finding $f(x - \bar{x})^2$, summing those values, and dividing by $n-1$.
4. Take the square root of the variance to get the standard deviation.

Let’s break this down:

1. Midpoints of Class Intervals

The midpoint for each class is calculated as:

$\text{Midpoint} = \frac{\text{Lower bound} + \text{Upper bound}}{2}$

Interval	Frequency $f$	Midpoint $x$
30-39	2	$\frac{30+39}{2} = 34.5$
40-49	1	$\frac{40+49}{2} = 44.5$
50-59	4	$\frac{50+59}{2} = 54.5$
60-69	1	$\frac{60+69}{2} = 64.5$
70-79	12	$\frac{70+79}{2} = 74.5$
80-89	36	$\frac{80+89}{2} = 84.5$
90-99	37	$\frac{90+99}{2} = 94.5$

2. Calculate the Mean ($\bar{x}$)

$\bar{x} = \frac{\sum f \cdot x}{n}$ Where $n = 93$ (the sum of all frequencies). I'll calculate this next.

Let me calculate these steps for you.The mean ($\bar{x}$) of the data is approximately 84.18.

3. Calculate the Variance

Next, we calculate the variance using the formula:

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

Where $x$ is the midpoint, and $f$ is the frequency.

I'll calculate this now.The computed standard deviation $s$ is approximately 12.98.

Comparison with the original list of data

The standard deviation from the original data values was given as 11.1. The computed standard deviation from the frequency distribution (12.98) is higher, which might be due to the grouping of data in intervals leading to a loss of precision.

Would you like more details or have any questions? Here are some related questions to expand on this topic:

1. How does grouping data into intervals affect the accuracy of statistical measures?
2. What is the difference between population standard deviation and sample standard deviation?
3. How would the calculation change if the data were a population rather than a sample?
4. What are other ways to calculate the mean and standard deviation if given raw data?
5. How does skewness or kurtosis of data affect the interpretation of standard deviation?

Tip: When working with grouped data, the loss of precision in midpoints can sometimes cause slight discrepancies compared to calculations on the raw data.