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Sagot :
To solve this problem using the Assumed Mean Method, we need to follow these steps:
### Step 1: Define the Given Data
Suppose we are given the following frequency distribution of masses:
| Mass Range (x) | Midpoint [tex]\( m_i \)[/tex] | Frequency [tex]\( f_i \)[/tex] |
|----------------|-----------------|--------------------|
| 50 - 54 | 52.0 | 8 |
| 55 - 59 | 57.0 | 12 |
| 60 - 64 | 62.0 | 20 |
| 65 - 69 | 67.0 | 25 |
| 70 - 74 | 72.0 | 10 |
| 75 - 79 | 77.0 | 5 |
### Step 2: Calculate the Midpoints [tex]\( m_i \)[/tex]
The midpoints have already been calculated and provided in the table above.
### Step 3: Choose an Assumed Mean (A)
Let's choose [tex]\( A \)[/tex] as the midpoint of the class with the highest frequency. Here, the class with the highest frequency is 65 - 69, and its midpoint [tex]\( m_i \)[/tex] is:
[tex]\[ A = 67.0 \][/tex]
### Step 4: Calculate the Deviations [tex]\( d_i = m_i - A \)[/tex]
Now, calculate [tex]\( d_i \)[/tex] for each class:
| Midpoint [tex]\( m_i \)[/tex] | Deviation [tex]\( d_i = m_i - 67.0 \)[/tex] |
|--------------------|----------------------------------|
| 52.0 | 52.0 - 67.0 = -15.0 |
| 57.0 | 57.0 - 67.0 = -10.0 |
| 62.0 | 62.0 - 67.0 = -5.0 |
| 67.0 | 67.0 - 0.0 |
| 72.0 | 72.0 - 67.0 = 5.0 |
| 77.0 | 77.0 - 67.0 = 10.0 |
### Step 5: Compute the Product of Frequency and Deviation [tex]\( f_i \cdot d_i \)[/tex]
Next, calculate [tex]\( f_i \cdot d_i \)[/tex] for each class:
| Frequency [tex]\( f_i \)[/tex] | Deviation [tex]\( d_i \)[/tex] | [tex]\( f_i \cdot d_i \)[/tex] |
|---------------------|---------------------|---------------------|
| 8 | -15.0 | 8 \cdot (-15.0) = -120.0 |
| 12 | -10.0 | 12 \cdot (-10.0) = -120.0 |
| 20 | -5.0 | 20 \cdot (-5.0) = -100.0 |
| 25 | 0.0 | 25 \cdot 0.0 = 0.0 |
| 10 | 5.0 | 10 \cdot 5.0 = 50.0 |
| 5 | 10.0 | 5 \cdot 10.0 = 50.0 |
### Step 6: Calculate [tex]\( \sum f_i \)[/tex] and [tex]\( \sum (f_i \cdot d_i) \)[/tex]
Sum of frequencies [tex]\( \sum f_i \)[/tex]:
[tex]\[ \sum f_i = 8 + 12 + 20 + 25 + 10 + 5 = 80 \][/tex]
Sum of products [tex]\( \sum (f_i \cdot d_i) \)[/tex]:
[tex]\[ \sum (f_i \cdot d_i) = -120 + -120 + -100 + 0 + 50 + 50 = -240 \][/tex]
### Step 7: Compute the Mean [tex]\( \bar{x} \)[/tex]
Finally, we use the Assumed Mean Formula to calculate the mean [tex]\( \bar{x} \)[/tex]:
[tex]\[ \bar{x} = A + \frac{\sum (f_i \cdot d_i)}{\sum f_i} \][/tex]
[tex]\[ \bar{x} = 67.0 + \frac{-240}{80} \][/tex]
[tex]\[ \bar{x} = 67.0 - 3 \][/tex]
[tex]\[ \bar{x} = 64.0 \][/tex]
### Conclusion:
The mean mass of the 80 items, using the Assumed Mean Method, is [tex]\( 64.0 \, \text{kg} \)[/tex].
### Step 1: Define the Given Data
Suppose we are given the following frequency distribution of masses:
| Mass Range (x) | Midpoint [tex]\( m_i \)[/tex] | Frequency [tex]\( f_i \)[/tex] |
|----------------|-----------------|--------------------|
| 50 - 54 | 52.0 | 8 |
| 55 - 59 | 57.0 | 12 |
| 60 - 64 | 62.0 | 20 |
| 65 - 69 | 67.0 | 25 |
| 70 - 74 | 72.0 | 10 |
| 75 - 79 | 77.0 | 5 |
### Step 2: Calculate the Midpoints [tex]\( m_i \)[/tex]
The midpoints have already been calculated and provided in the table above.
### Step 3: Choose an Assumed Mean (A)
Let's choose [tex]\( A \)[/tex] as the midpoint of the class with the highest frequency. Here, the class with the highest frequency is 65 - 69, and its midpoint [tex]\( m_i \)[/tex] is:
[tex]\[ A = 67.0 \][/tex]
### Step 4: Calculate the Deviations [tex]\( d_i = m_i - A \)[/tex]
Now, calculate [tex]\( d_i \)[/tex] for each class:
| Midpoint [tex]\( m_i \)[/tex] | Deviation [tex]\( d_i = m_i - 67.0 \)[/tex] |
|--------------------|----------------------------------|
| 52.0 | 52.0 - 67.0 = -15.0 |
| 57.0 | 57.0 - 67.0 = -10.0 |
| 62.0 | 62.0 - 67.0 = -5.0 |
| 67.0 | 67.0 - 0.0 |
| 72.0 | 72.0 - 67.0 = 5.0 |
| 77.0 | 77.0 - 67.0 = 10.0 |
### Step 5: Compute the Product of Frequency and Deviation [tex]\( f_i \cdot d_i \)[/tex]
Next, calculate [tex]\( f_i \cdot d_i \)[/tex] for each class:
| Frequency [tex]\( f_i \)[/tex] | Deviation [tex]\( d_i \)[/tex] | [tex]\( f_i \cdot d_i \)[/tex] |
|---------------------|---------------------|---------------------|
| 8 | -15.0 | 8 \cdot (-15.0) = -120.0 |
| 12 | -10.0 | 12 \cdot (-10.0) = -120.0 |
| 20 | -5.0 | 20 \cdot (-5.0) = -100.0 |
| 25 | 0.0 | 25 \cdot 0.0 = 0.0 |
| 10 | 5.0 | 10 \cdot 5.0 = 50.0 |
| 5 | 10.0 | 5 \cdot 10.0 = 50.0 |
### Step 6: Calculate [tex]\( \sum f_i \)[/tex] and [tex]\( \sum (f_i \cdot d_i) \)[/tex]
Sum of frequencies [tex]\( \sum f_i \)[/tex]:
[tex]\[ \sum f_i = 8 + 12 + 20 + 25 + 10 + 5 = 80 \][/tex]
Sum of products [tex]\( \sum (f_i \cdot d_i) \)[/tex]:
[tex]\[ \sum (f_i \cdot d_i) = -120 + -120 + -100 + 0 + 50 + 50 = -240 \][/tex]
### Step 7: Compute the Mean [tex]\( \bar{x} \)[/tex]
Finally, we use the Assumed Mean Formula to calculate the mean [tex]\( \bar{x} \)[/tex]:
[tex]\[ \bar{x} = A + \frac{\sum (f_i \cdot d_i)}{\sum f_i} \][/tex]
[tex]\[ \bar{x} = 67.0 + \frac{-240}{80} \][/tex]
[tex]\[ \bar{x} = 67.0 - 3 \][/tex]
[tex]\[ \bar{x} = 64.0 \][/tex]
### Conclusion:
The mean mass of the 80 items, using the Assumed Mean Method, is [tex]\( 64.0 \, \text{kg} \)[/tex].
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