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To find the mean and standard deviation of the given weight data, we need to follow a series of detailed steps. Let’s break it down one step at a time.
### Step 1: Determine the midpoints (class marks) of each class interval
The midpoint of a class interval is the average of the lower and upper boundaries of the interval.
For the interval [tex]\(30 \leq w < 35\)[/tex]:
[tex]\[ \text{Midpoint} = \frac{30 + 35}{2} = 32.5 \][/tex]
Similarly, we calculate the midpoints for the other intervals:
- [tex]\(35 \leq w < 40\)[/tex]:
[tex]\[ \text{Midpoint} = \frac{35 + 40}{2} = 37.5 \][/tex]
- [tex]\(40 \leq w < 45\)[/tex]:
[tex]\[ \text{Midpoint} = \frac{40 + 45}{2} = 42.5 \][/tex]
- [tex]\(45 \leq w < 50\)[/tex]:
[tex]\[ \text{Midpoint} = \frac{45 + 50}{2} = 47.5 \][/tex]
- [tex]\(50 \leq w < 55\)[/tex]:
[tex]\[ \text{Midpoint} = \frac{50 + 55}{2} = 52.5 \][/tex]
### Step 2: Multiply each midpoint by its corresponding frequency (fx)
Now, we calculate [tex]\( f \times x \)[/tex] for each class interval where [tex]\( f \)[/tex] is the frequency, and [tex]\( x \)[/tex] is the midpoint:
- For [tex]\(30 \leq w < 35\)[/tex]:
[tex]\[ 10 \times 32.5 = 325 \][/tex]
- For [tex]\(35 \leq w < 40\)[/tex]:
[tex]\[ 11 \times 37.5 = 412.5 \][/tex]
- For [tex]\(40 \leq w < 45\)[/tex]:
[tex]\[ 15 \times 42.5 = 637.5 \][/tex]
- For [tex]\(45 \leq w < 50\)[/tex]:
[tex]\[ 7 \times 47.5 = 332.5 \][/tex]
- For [tex]\(50 \leq w < 55\)[/tex]:
[tex]\[ 4 \times 52.5 = 210 \][/tex]
### Step 3: Sum up all [tex]\( f \times x \)[/tex]
[tex]\[ \text{Sum of } f \times x = 325 + 412.5 + 637.5 + 332.5 + 210 = 1917.5 \][/tex]
### Step 4: Sum up all frequencies
[tex]\[ \text{Sum of frequencies } = 10 + 11 + 15 + 7 + 4 = 47 \][/tex]
### Step 5: Calculate the Mean
[tex]\[ \text{Mean} = \frac{\sum (f \times x)}{\sum f} = \frac{1917.5}{47} \approx 40.7979 \][/tex]
### Step 6: Calculate [tex]\( f \times x^2 \)[/tex]
Next, we calculate [tex]\( f \times x^2 \)[/tex] for each class interval:
- For [tex]\(30 \leq w < 35\)[/tex]:
[tex]\[ 10 \times (32.5)^2 = 10 \times 1056.25 = 10562.5 \][/tex]
- For [tex]\(35 \leq w < 40\)[/tex]:
[tex]\[ 11 \times (37.5)^2 = 11 \times 1406.25 = 15468.75 \][/tex]
- For [tex]\(40 \leq w < 45\)[/tex]:
[tex]\[ 15 \times (42.5)^2 = 15 \times 1806.25 = 27093.75 \][/tex]
- For [tex]\(45 \leq w < 50\)[/tex]:
[tex]\[ 7 \times (47.5)^2 = 7 \times 2256.25 = 15793.75 \][/tex]
- For [tex]\(50 \leq w < 55\)[/tex]:
[tex]\[ 4 \times (52.5)^2 = 4 \times 2756.25 = 11025 \][/tex]
### Step 7: Sum up all [tex]\( f \times x^2 \)[/tex]
[tex]\[ \text{Sum of } f \times x^2 = 10562.5 + 15468.75 + 27093.75 + 15793.75 + 11025 = 79943.75 \][/tex]
### Step 8: Calculate the variance
The variance [tex]\( \sigma^2 \)[/tex] is given by the formula:
[tex]\[ \sigma^2 = \frac{\sum (f \times x^2)}{\sum f} - \left(\frac{\sum (f \times x)}{\sum f}\right)^2 \][/tex]
### Step 9: Substitute the values and compute
[tex]\[ \sigma^2 = \frac{79943.75}{47} - (40.7979)^2 \][/tex]
[tex]\[ \sigma^2 \approx 1700.9313191489362 - 1664.4668555907641 = 36.46446355817079 \][/tex]
### Step 10: Calculate the standard deviation
The standard deviation [tex]\( \sigma \)[/tex] is the square root of the variance:
[tex]\[ \sigma = \sqrt{36.46446355817079} \approx 6.0386 \][/tex]
### Summary
- Mean: [tex]\( \approx 40.7979 \)[/tex]
- Standard Deviation: [tex]\( \approx 6.0386 \)[/tex]
### Step 1: Determine the midpoints (class marks) of each class interval
The midpoint of a class interval is the average of the lower and upper boundaries of the interval.
For the interval [tex]\(30 \leq w < 35\)[/tex]:
[tex]\[ \text{Midpoint} = \frac{30 + 35}{2} = 32.5 \][/tex]
Similarly, we calculate the midpoints for the other intervals:
- [tex]\(35 \leq w < 40\)[/tex]:
[tex]\[ \text{Midpoint} = \frac{35 + 40}{2} = 37.5 \][/tex]
- [tex]\(40 \leq w < 45\)[/tex]:
[tex]\[ \text{Midpoint} = \frac{40 + 45}{2} = 42.5 \][/tex]
- [tex]\(45 \leq w < 50\)[/tex]:
[tex]\[ \text{Midpoint} = \frac{45 + 50}{2} = 47.5 \][/tex]
- [tex]\(50 \leq w < 55\)[/tex]:
[tex]\[ \text{Midpoint} = \frac{50 + 55}{2} = 52.5 \][/tex]
### Step 2: Multiply each midpoint by its corresponding frequency (fx)
Now, we calculate [tex]\( f \times x \)[/tex] for each class interval where [tex]\( f \)[/tex] is the frequency, and [tex]\( x \)[/tex] is the midpoint:
- For [tex]\(30 \leq w < 35\)[/tex]:
[tex]\[ 10 \times 32.5 = 325 \][/tex]
- For [tex]\(35 \leq w < 40\)[/tex]:
[tex]\[ 11 \times 37.5 = 412.5 \][/tex]
- For [tex]\(40 \leq w < 45\)[/tex]:
[tex]\[ 15 \times 42.5 = 637.5 \][/tex]
- For [tex]\(45 \leq w < 50\)[/tex]:
[tex]\[ 7 \times 47.5 = 332.5 \][/tex]
- For [tex]\(50 \leq w < 55\)[/tex]:
[tex]\[ 4 \times 52.5 = 210 \][/tex]
### Step 3: Sum up all [tex]\( f \times x \)[/tex]
[tex]\[ \text{Sum of } f \times x = 325 + 412.5 + 637.5 + 332.5 + 210 = 1917.5 \][/tex]
### Step 4: Sum up all frequencies
[tex]\[ \text{Sum of frequencies } = 10 + 11 + 15 + 7 + 4 = 47 \][/tex]
### Step 5: Calculate the Mean
[tex]\[ \text{Mean} = \frac{\sum (f \times x)}{\sum f} = \frac{1917.5}{47} \approx 40.7979 \][/tex]
### Step 6: Calculate [tex]\( f \times x^2 \)[/tex]
Next, we calculate [tex]\( f \times x^2 \)[/tex] for each class interval:
- For [tex]\(30 \leq w < 35\)[/tex]:
[tex]\[ 10 \times (32.5)^2 = 10 \times 1056.25 = 10562.5 \][/tex]
- For [tex]\(35 \leq w < 40\)[/tex]:
[tex]\[ 11 \times (37.5)^2 = 11 \times 1406.25 = 15468.75 \][/tex]
- For [tex]\(40 \leq w < 45\)[/tex]:
[tex]\[ 15 \times (42.5)^2 = 15 \times 1806.25 = 27093.75 \][/tex]
- For [tex]\(45 \leq w < 50\)[/tex]:
[tex]\[ 7 \times (47.5)^2 = 7 \times 2256.25 = 15793.75 \][/tex]
- For [tex]\(50 \leq w < 55\)[/tex]:
[tex]\[ 4 \times (52.5)^2 = 4 \times 2756.25 = 11025 \][/tex]
### Step 7: Sum up all [tex]\( f \times x^2 \)[/tex]
[tex]\[ \text{Sum of } f \times x^2 = 10562.5 + 15468.75 + 27093.75 + 15793.75 + 11025 = 79943.75 \][/tex]
### Step 8: Calculate the variance
The variance [tex]\( \sigma^2 \)[/tex] is given by the formula:
[tex]\[ \sigma^2 = \frac{\sum (f \times x^2)}{\sum f} - \left(\frac{\sum (f \times x)}{\sum f}\right)^2 \][/tex]
### Step 9: Substitute the values and compute
[tex]\[ \sigma^2 = \frac{79943.75}{47} - (40.7979)^2 \][/tex]
[tex]\[ \sigma^2 \approx 1700.9313191489362 - 1664.4668555907641 = 36.46446355817079 \][/tex]
### Step 10: Calculate the standard deviation
The standard deviation [tex]\( \sigma \)[/tex] is the square root of the variance:
[tex]\[ \sigma = \sqrt{36.46446355817079} \approx 6.0386 \][/tex]
### Summary
- Mean: [tex]\( \approx 40.7979 \)[/tex]
- Standard Deviation: [tex]\( \approx 6.0386 \)[/tex]
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