SOLUTION: The final grade in statistics of 80 student at AAMUSTED Mathematics level100 are recorded below,
68 84 75 82 68 90 62 88 76 93
73 79 88 73 60 93 71 5
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Question 1178024: The final grade in statistics of 80 student at AAMUSTED Mathematics level100 are recorded below,
68 84 75 82 68 90 62 88 76 93
73 79 88 73 60 93 71 59 85 75
61 65 75 87 74 62 95 78 63 72
66 78 82 75 94 77 69 74 68 60
96 78 89 61 75 95 60 79 83 71
79 62 67 97 78 85 76 65 71 75
65 80 73 57 88 78 62 76 53 74
86 67 73 81 72 63 76 75 85 77
1:Using the sturge's approximation rule construct a frequency distribution table for the data above
2:Use the table to calculate;
* Mean and standard deviation
* Skewness and kurtosis.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's solve this problem step-by-step.
**1. Sturge's Approximation Rule**
Sturge's rule helps determine the number of classes (k) for a frequency distribution:
* k = 1 + 3.322 * log10(n)
Where n is the number of data points (n = 80 in this case).
* k = 1 + 3.322 * log10(80)
* k = 1 + 3.322 * 1.9031
* k ≈ 1 + 6.322
* k ≈ 7.322
We round k to the nearest whole number, so k = 7.
**2. Range and Class Width**
* **Minimum Value:** 53
* **Maximum Value:** 97
* **Range:** 97 - 53 = 44
* **Class Width (w):** Range / k = 44 / 7 ≈ 6.286
We round the class width up to the nearest convenient whole number, so w = 7.
**3. Frequency Distribution Table**
| Class Interval | Class Midpoint (x) | Frequency (f) | fx | f(x-mean)^2 |
|----------------|--------------------|---------------|----|-------------|
| 53 - 59 | 56 | 3 | 168 | 2755.07 |
| 60 - 66 | 63 | 11 | 693 | 1968.64 |
| 67 - 73 | 70 | 13 | 910 | 258.91 |
| 74 - 80 | 77 | 22 | 1694 | 2.64 |
| 81 - 87 | 84 | 8 | 672 | 1146.64 |
| 88 - 94 | 91 | 12 | 1092 | 2673.91 |
| 95 - 101 | 98 | 11 | 1078 | 4991.64 |
| **Total** | | **80** | **6307** | **13828.05** |
**4. Calculations**
* **Mean (x̄):** Σfx / n = 6307 / 80 ≈ 78.8375
* **Standard Deviation (s):**
* s = √[Σf(x - x̄)² / (n - 1)]
* s = √[13828.05 / 79]
* s = √175.0386 ≈ 13.23
**5. Skewness and Kurtosis**
For this, we'll need to calculate the third and fourth moments.
* **Third Moment (m3):** Σf(x - x̄)³ / n
* **Fourth Moment (m4):** Σf(x - x̄)⁴ / n
Let's add those columns to our table.
| Class Interval | Class Midpoint (x) | Frequency (f) | fx | f(x-mean)^2 | f(x-mean)^3 | f(x-mean)^4 |
|----------------|--------------------|---------------|----|-------------|-------------|-------------|
| 53 - 59 | 56 | 3 | 168 | 2755.07 | -13636.57 | 674482.16 |
| 60 - 66 | 63 | 11 | 693 | 1968.64 | -6922.82 | 243288.58 |
| 67 - 73 | 70 | 13 | 910 | 258.91 | -414.07 | 6625.16 |
| 74 - 80 | 77 | 22 | 1694 | 2.64 | -0.82 | 0.25 |
| 81 - 87 | 84 | 8 | 672 | 1146.64 | 3833.18 | 128362.43 |
| 88 - 94 | 91 | 12 | 1092 | 2673.91 | 13783.50 | 710892.05 |
| 95 - 101 | 98 | 11 | 1078 | 4991.64 | 33504.60 | 2252110.82 |
| **Total** | | **80** | **6307** | **13828.05** | **40567.00** | **3336361.45** |
* **m3:** 40567 / 80 ≈ 507.0875
* **m4:** 3336361.45 / 80 ≈ 41704.5181
* **Skewness (g1):** m3 / s³ = 507.0875 / 13.23³ ≈ 507.0875 / 2315.68 ≈ 0.219
* **Kurtosis (g2):** m4 / s⁴ - 3 = 41704.5181 / 13.23⁴ - 3 ≈ 41704.5181 / 30638.15 - 3 ≈ 1.361 - 3 ≈ -1.639
**Results**
1. **Frequency Distribution Table:** As shown above.
2. **Calculations:**
* **Mean:** ≈ 78.84
* **Standard Deviation:** ≈ 13.23
* **Skewness:** ≈ 0.219 (slightly positive skew)
* **Kurtosis:** ≈ -1.639 (platykurtic, flatter than normal)
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