To calculate the approximate mean and standard deviation of the grouped data, we follow these steps:

Step 1: Define the Midpoints of Each Class

For each interval (class), we calculate the midpoint $(x_i)$, which is given by the average of the lower and upper bounds of each class.

1. 10 to under 30: Midpoint = $\frac{10 + 30}{2} = 20$
2. 30 to under 50: Midpoint = $\frac{30 + 50}{2} = 40$
3. 50 to under 70: Midpoint = $\frac{50 + 70}{2} = 60$
4. 70 to under 90: Midpoint = $\frac{70 + 90}{2} = 80$

Step 2: Create a Table with Midpoints and Frequencies

Class	Midpoint $(x_i)$	Frequency $(f_i)$
10 to under 30	20	25
30 to under 50	40	31
50 to under 70	60	57
70 to under 90	80	37

Step 3: Calculate the Approximate Mean

The approximate mean $\bar{x}$ for grouped data is given by:

$\bar{x} = \frac{\sum (f_i \cdot x_i)}{\sum f_i}$

Where:

• $f_i$ is the frequency of each class.
• $x_i$ is the midpoint of each class.

We first calculate $f_i \cdot x_i$ for each class:

1. $f_1 \cdot x_1 = 25 \cdot 20 = 500$
2. $f_2 \cdot x_2 = 31 \cdot 40 = 1240$
3. $f_3 \cdot x_3 = 57 \cdot 60 = 3420$
4. $f_4 \cdot x_4 = 37 \cdot 80 = 2960$

Now sum these products:

$\sum (f_i \cdot x_i) = 500 + 1240 + 3420 + 2960 = 8120$

Next, calculate the total frequency:

$\sum f_i = 25 + 31 + 57 + 37 = 150$

Finally, the approximate mean is:

$\bar{x} = \frac{8120}{150} = 54.133$

Step 4: Calculate the Approximate Standard Deviation

The standard deviation for grouped data is given by:

$\sigma = \sqrt{\frac{\sum f_i (x_i - \bar{x})^2}{\sum f_i}}$

We need to compute $(x_i - \bar{x})^2$ and then $f_i \cdot (x_i - \bar{x})^2$:

1. $(x_1 - \bar{x})^2 = (20 - 54.133)^2 = (-34.133)^2 = 1164.925$
2. $(x_2 - \bar{x})^2 = (40 - 54.133)^2 = (-14.133)^2 = 199.735$
3. $(x_3 - \bar{x})^2 = (60 - 54.133)^2 = (5.867)^2 = 34.419$
4. $(x_4 - \bar{x})^2 = (80 - 54.133)^2 = (25.867)^2 = 669.290$

Now multiply these by the frequencies:

1. $f_1 \cdot (x_1 - \bar{x})^2 = 25 \cdot 1164.925 = 29123.125$
2. $f_2 \cdot (x_2 - \bar{x})^2 = 31 \cdot 199.735 = 6191.785$
3. $f_3 \cdot (x_3 - \bar{x})^2 = 57 \cdot 34.419 = 1961.883$
4. $f_4 \cdot (x_4 - \bar{x})^2 = 37 \cdot 669.290 = 24763.73$

Now sum these values:

$\sum f_i (x_i - \bar{x})^2 = 29123.125 + 6191.785 + 1961.883 + 24763.73 = 62040.523$

Now calculate the standard deviation:

$\sigma = \sqrt{\frac{62040.523}{150}} = \sqrt{413.603} = 20.337$

Final Answers:

• a) Approximate Mean: 54.133 (rounded to three decimal places)
• b) Approximate Standard Deviation: 20.337 (rounded to three decimal places)

Would you like further clarification or details on any part of the process?

Here are 5 related questions you can explore:

1. How would the approximate mean change if the frequencies were different?
2. Can you explain why the midpoint is used for grouped data?
3. What is the significance of the standard deviation in this context?
4. How does the number of classes affect the accuracy of the mean and standard deviation?
5. How would the calculation differ if the intervals were of unequal widths?

Tip: For grouped data, using midpoints helps in estimating values, but it's important to remember this provides an approximation.