
Classes
Refer to the incomplete grouped frequency below. Frequency, Cumulative Midpoint, fm
fm2
(Price)
f
($)
30.00 <
4
4
Frequency m
35.00
140
4900
40.00
40.00 <
11
15
45.00
495
22275
50.00
50.00 <
A
43
135
55.00
1540
84700
60.00
60.00<
17
60
65.00
1105
71825
70.00
Σ
60
The estimated median is:
O A. $52.16
O B. $58.20
O C. $48.95
OD. $55.36
1 73°F Sunny
Esc
4×
F1
OL
川
F2
F3
KO
Ox
F4
F5
F6

The problem presents a grouped frequency table with some incomplete data. To estimate the median using this data, we need to follow the steps for finding the median in a grouped frequency distribution:

### Grouped Frequency Distribution Data (incomplete):
- Class 1: $ 30.00 < 35.00 $, $ f = 4 $, cumulative frequency = 4
- Class 2: $ 35.00 < 40.00 $, $ f = 11 $, cumulative frequency = 15
- Class 3: $ 40.00 < 45.00 $, $ f = A $, cumulative frequency = 43
- Class 4: $ 45.00 < 50.00 $, $ f = 17 $, cumulative frequency = 60

The total frequency (Σf) is given as 60. 

### Step-by-Step Solution for the Median:

1. **Find the cumulative frequency column:**
   The cumulative frequency is the running total of the frequencies up to that class. It's partially filled out, but we'll calculate it fully:
   - $ \text{Class 1:} $ Cumulative Frequency = $ 4 $
   - $ \text{Class 2:} $ Cumulative Frequency = $ 4 + 11 = 15 $
   - $ \text{Class 3:} $ Cumulative Frequency = $ 15 + A = 43 $, so $ A = 43 - 15 = 28 $
   - $ \text{Class 4:} $ Cumulative Frequency = $ 43 + 17 = 60 $

2. **Determine the median class:**
   The median class is the class where the cumulative frequency surpasses $ \frac{N}{2} $, where $ N = 60 $, so $ \frac{N}{2} = 30 $.

   From the cumulative frequencies:
   - Class 1 cumulative frequency = 4 (less than 30)
   - Class 2 cumulative frequency = 15 (less than 30)
   - Class 3 cumulative frequency = 43 (greater than 30)

   Therefore, the **median class** is $ 40.00 < 45.00 $.

3. **Apply the median formula:**
   The formula for the median is:
   \[
   \text{Median} = L + \left( \frac{\frac{N}{2} - F}{f} \right) \times h
   \]
   Where:
   - $ L $ is the lower boundary of the median class ($ L = 40.00 $)
   - $ F $ is the cumulative frequency of the class before the median class ($ F = 15 $)
   - $ f $ is the frequency of the median class ($ f = 28 $)
   - $ h $ is the class width (in this case, $ h = 5 $, because the width between classes is 5)

4. **Substitute values into the formula:**
   \[
   \text{Median} = 40.00 + \left( \frac{30 - 15}{28} \right) \times 5
   \]
   Simplifying:
   \[
   \text{Median} = 40.00 + \left( \frac{15}{28} \right) \times 5
   \]
   \[
   \text{Median} = 40.00 + \left( 0.5357 \right) \times 5
   \]
   \[
   \text{Median} = 40.00 + 2.6785
   \]
   \[
   \text{Median} \approx 42.68
   \]

Since the calculated value of $ 42.68 $ does not match any of the given choices, it seems there might be an error in the question's given options or the assumption of class boundaries. This would require double-checking the problem's data or options.

Would you like me to explore the fm column or recheck any assumptions?

Refer to the incomplete grouped frequency below. Frequency, Cumulative Midpoint, fm, fm2: (Price) f ($) 30.00 < 4, 4 Frequency m 35.00 140, 4900, 40.00, 40.00 < 11, 15, 45.00, 495, 22275, 50.00, 50.00 < A, 43, 135, 55.00, 1540, 84700, 60.00, 60.00< 17, 60, 65.00, 1105, 71825, 70.00 Σ 60. The estimated median is: O A. $52.16 O B. $58.20 O C. $48.95 O D. $55.36

Learn how to calculate the median from an incomplete grouped frequency distribution. This problem involves key mathematical concepts like cumulative frequency, class boundaries, and the median formula. Perfect for students in Grades 10-12.

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