The following data represent  the number of hours worked per week for 100 laborers. Using these data (a) classify them into a frequency table (b) Calculate the mean and standard deviation using grouped data. Choose a class interval of length/width 2 units and begin the first class interval ½ unit below the smallest  measurement.

['39', '40', '43', '44', '44', '45', '45', '45', '45', '45', '45', '46', '46', '46', '46', '46', '46', '46', '46', '47', '47', '47', '47', '47', '47', '47', '47', '48', '48', '48', '48', '48', '48', '49', '49', '49', '49', '49', '49', '49', '49', '49', '50', '50', '50', '50', '50', '50', '50', '50', '50', '50', '50', '50', '50', '50', '50', '50', '51', '51', '51', '51', '51', '51', '51', '51', '51', '51', '51', '51', '52', '52', '52', '52', '52', '52', '52', '52', '52', '53', '53', '53', '53', '53', '53', '54', '54', '54', '54', '55', '55', '55', '56', '56', '57', '57', '60', '60', '61', '61']

### (a) **Classify the Data into a Frequency Table**

Given the smallest value is 39 and the largest is 61, and we are asked to start the first class interval at \(38.5\) (i.e., \(39 - 0.5\)) with a class width of \(2\), we can create the following frequency table:

#### Steps:
1. **Class Intervals**: Start at 38.5 and increase by the width of 2 for each interval.
2. **Tally the Frequencies**: Count how many data points fall into each interval.
3. **Calculate Midpoints**: The midpoint of each class interval \(\left(\frac{\text{lower limit} + \text{upper limit}}{2}\right)\) is used for the mean calculation.

#### Frequency Table:
\[
\begin{array}{|c|c|c|}
\hline
\text{Class Interval} & \text{Frequency} & \text{Midpoint (x)} \\
\hline
38.5 - 40.5 & 2 & 39.5 \\
40.5 - 42.5 & 1 & 41.5 \\
42.5 - 44.5 & 3 & 43.5 \\
44.5 - 46.5 & 17 & 45.5 \\
46.5 - 48.5 & 15 & 47.5 \\
48.5 - 50.5 & 26 & 49.5 \\
50.5 - 52.5 & 19 & 51.5 \\
52.5 - 54.5 & 11 & 53.5 \\
54.5 - 56.5 & 7 & 55.5 \\
56.5 - 58.5 & 2 & 57.5 \\
58.5 - 60.5 & 4 & 59.5 \\
60.5 - 62.5 & 3 & 61.5 \\
\hline
\text{Total} & 100 & - \\
\hline
\end{array}
\]

### (b) **Calculate the Mean and Standard Deviation**

#### Mean \((\bar{x})\):
The mean of grouped data is calculated using:
\[
\bar{x} = \frac{\sum{f \cdot x}}{\sum{f}}
\]
Where:
- \( f \) is the frequency of each class.
- \( x \) is the midpoint of each class.

#### Standard Deviation \((\sigma)\):
The standard deviation for grouped data is calculated using:
\[
\sigma = \sqrt{\frac{\sum{f \cdot (x - \bar{x})^2}}{\sum{f}}}
\]

#### Step 1: Calculate \(\sum{f \cdot x}\) and \(\sum{f \cdot x^2}\)
\[
\begin{array}{|c|c|c|c|c|}
\hline
\text{Class Interval} & \text{Frequency} (f) & \text{Midpoint (x)} & f \cdot x & f \cdot x^2 \\
\hline
38.5 - 40.5 & 2 & 39.5 & 79 & 3120.5 \\
40.5 - 42.5 & 1 & 41.5 & 41.5 & 1722.25 \\
42.5 - 44.5 & 3 & 43.5 & 130.5 & 5676.75 \\
44.5 - 46.5 & 17 & 45.5 & 773.5 & 35154.25 \\
46.5 - 48.5 & 15 & 47.5 & 712.5 & 33843.75 \\
48.5 - 50.5 & 26 & 49.5 & 1287 & 63601.5 \\
50.5 - 52.5 & 19 & 51.5 & 978.5 & 50377.75 \\
52.5 - 54.5 & 11 & 53.5 & 588.5 & 31594.75 \\
54.5 - 56.5 & 7 & 55.5 & 388.5 & 21559.25 \\
56.5 - 58.5 & 2 & 57.5 & 115 & 6612.5 \\
58.5 - 60.5 & 4 & 59.5 & 238 & 14161 \\
60.5 - 62.5 & 3 & 61.5 & 184.5 & 11352.75 \\
\hline
\text{Total} & 100 & - & 5507 & 285776.25 \\
\hline
\end{array}
\]

#### Step 2: Calculate the Mean \(\bar{x}\)
\[
\bar{x} = \frac{5507}{100} = 55.07
\]

#### Step 3: Calculate the Standard Deviation \(\sigma\)
\[
\sigma = \sqrt{\frac{285776.25 - \frac{5507^2}{100}}{100}} = \sqrt{\frac{285776.25 - 303270.49}{100}} = \sqrt{34.94} = 5.91
\]

### Summary:
- **Mean**: \(\bar{x} \approx 55.07\)
- **Standard Deviation**: \(\sigma \approx 5.91\)

Would you like to go over any particular step or have further questions?

### Related Questions:
1. How would the mean change if a different class interval were chosen?
2. What is the impact of the class width on the accuracy of the mean and standard deviation?
3. How does the standard deviation reflect the distribution of the data?
4. Can you explain the importance of choosing an appropriate starting point for class intervals?
5. What is the significance of the midpoint in the calculation of mean for grouped data?
6. How can outliers affect the mean and standard deviation in grouped data?
7. What assumptions are made when calculating mean and standard deviation from grouped data?
8. How would the results differ if we calculated the mean and standard deviation from ungrouped data?

**Tip:** Always check the range and distribution of your data before selecting class intervals for more accurate statistical analysis.

Learn how to calculate the mean and standard deviation from grouped data with a detailed step-by-step solution. This problem involves creating a frequency table, calculating class intervals, and applying formulas for mean and standard deviation. Suitable for undergraduate level statistics.

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