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Question
Find the mean of the following frequency distribution using step- deviation method:
| Class | 84 – 90 | 90 – 96 | 96 – 102 | 102 – 108 | 108 – 114 | 114 – 120 |
| Frequency | 15 | 22 | 20 | 18 | 20 | 25 |
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Solution
1. Identify given values
We determine the class width (h) by subtracting the lower limit from the upper limit of any class interval:
h = 90 – 84 = 6
2. Determine class marks
The class mark (xi) for each interval is the midpoint, calculated using the formula `x_i = ("Lower Limit" + "Upper Limit")/2`.
We choose an assumed mean (A) from near the middle of our xi values. Let A = 99.
3. Compute step-deviations
The step-deviation (ui) for each class interval is determined by:
`u_i = (x_i - A)/h = (x_i - 99)/6`
We tabulate the given data and calculate the summation values:
| Class Interval | Frequency (fi) | Class Mark (xi) | `bb(u_i = (x_i - 99)/6)` | fi × ui |
| 84 – 90 | 15 | 87 | –2 | –30 |
| 90 – 96 | 22 | 93 | –1 | –22 |
| 96 – 102 | 20 | 99 (A) | 0 | 0 |
| 102 – 108 | 18 | 105 | 1 | 18 |
| 108 – 114 | 20 | 111 | 2 | 40 |
| 114 – 120 | 25 | 117 | 3 | 75 |
| Total | Σfi = 120 | Σfiui = 81 |
4. Apply step-deviation formula
The formula to calculate the arithmetic mean using the step-deviation method is:
Mean `(barx) = A + ((sumf_iu_i)/(sumf_i)) xx h`
Substitute the totals from our table into the formula:
`barx = 99 + (81/120) xx 6`
`barx = 99 + 486/120`
`barx = 99 + 4.05`
`barx = 103.05`
The mean of the given frequency distribution using the step-deviation method is exactly 103.05.
Notes
The answer in the textbook is incorrect.
