Comprehensive statistics worksheet designed to help students practice calculating measures of central tendency and variability using grouped data.

Let’s solve this step by step.

We are given a grouped frequency distribution table with class intervals and frequencies. We need to:

1. Complete the table (calculate cf, x, x², fx, fx²)
2. Find: Mean, Median, Mode, Variance, Standard Deviation

---

Step 1: Understand the columns

- Class Interval: Groups of scores (e.g., 46–50)
- Frequency (f): How many scores fall in each group
- cf: Cumulative frequency — running total of frequencies
- x: Midpoint of each class interval → (lower + upper) / 2
- x²: Square of midpoint
- fx: Frequency × midpoint
- fx²: Frequency × (midpoint squared)

---

Step 2: Fill in the table row by row

#### First row: 46 - 50, f = 5

- x = (46 + 50)/2 = 48
- x² = 48² = 2304
- fx = 5 × 48 = 240
- fx² = 5 × 2304 = 11520
- cf = 5 (first row, so just the frequency)

#### Second row: 41 - 45, f = 12

- x = (41 + 45)/2 = 43
- x² = 43² = 1849
- fx = 12 × 43 = 516
- fx² = 12 × 1849 = 22188
- cf = 5 + 12 = 17

#### Third row: 36 - 40, f = 14

- x = (36 + 40)/2 = 38
- x² = 38² = 1444
- fx = 14 × 38 = 532
- fx² = 14 × 1444 = 20216
- cf = 17 + 14 = 31

#### Fourth row: 31 - 35, f = 4

- x = (31 + 35)/2 = 33
- x² = 33² = 1089
- fx = 4 × 33 = 132
- fx² = 4 × 1089 = 4356
- cf = 31 + 4 = 35

---

Step 3: Totals

- N = total frequency = 5 + 12 + 14 + 4 = 35
- i = class width = 50 - 46 + 1? Wait — actually, class width is difference between lower limits: 46 - 41 = 5 → so i = 5
- Σ(fx) = 240 + 516 + 532 + 132 = let’s add:
- 240 + 516 = 756
- 756 + 532 = 1288
- 1288 + 132 = 1420
- Σ(fx²) = 11520 + 22188 + 20216 + 4356 =
- 11520 + 22188 = 33708
- 33708 + 20216 = 53924
- 53924 + 4356 = 58280

So now our completed table looks like:

| CLASS INTERVAL | Frequency (f) | cf | x | x² | fx | fx² |
|----------------|---------------|-----|-----|-------|------|-------|
| 46 - 50 | 5 | 5 | 48 | 2304 | 240 | 11520 |
| 41 - 45 | 12 | 17 | 43 | 1849 | 516 | 22188 |
| 36 - 40 | 14 | 31 | 38 | 1444 | 532 | 20216 |
| 31 - 35 | 4 | 35 | 33 | 1089 | 132 | 4356 |
| i = 5 | N = 35 | | | | Σ(fx)=1420 | Σfx²=58280 |

---

Step 4: Calculate Mean

Formula:
Mean (x̄) = Σ(fx) / N = 1420 / 35

Calculate:
1420 ÷ 35 = 40.5714... ≈ 40.57 (rounded to two decimals)

---

Step 5: Calculate Median

For grouped data, median formula:

Median = L + [(N/2 - cf_b) / f_m] × i

Where:
- L = lower boundary of median class
- N = total frequency = 35
- cf_b = cumulative frequency before median class
- f_m = frequency of median class
- i = class width = 5

First, find median class: N/2 = 35/2 = 17.5 → look for first class where cf ≥ 17.5

Looking at cf column:
- 46-50: cf=5 → too low
- 41-45: cf=17 → still less than 17.5
- 36-40: cf=31 → this is the first class ≥ 17.5 → so median class is 36-40

But wait — we must be careful. The classes are listed from highest to lowest? Actually, in standard practice, we order classes from lowest to highest for cumulative frequency. But here, the table is written from top (46-50) to bottom (31-35). So our cf is calculated from top down? That would be incorrect for median calculation.

Actually, in statistics, when calculating median for grouped data, we usually arrange classes in ascending order (lowest to highest) and compute cf from bottom up or top down consistently.

But in this table, the classes are listed from high to low: 46-50, then 41-45, etc. And cf is computed as running total from top: 5, 17, 31, 35.

That means the “first” class (46-50) has the highest values, and last class (31-35) has lowest.

So if we want to find the median (middle value), we should consider the data ordered from low to high.

Therefore, to correctly compute median, we need to reorder the classes from lowest to highest:

Reordered table (for median calculation):

| Class Interval | f | cf (from low to high) |
|----------------|----|------------------------|
| 31 - 35 | 4 | 4 |
| 36 - 40 | 14 | 18 |
| 41 - 45 | 12 | 30 |
| 46 - 50 | 5 | 35 |

Now, N/2 = 17.5 → find first class where cf ≥ 17.5 → that’s 36-40 (cf=18)

So:
- L = lower boundary of 36-40 → since classes are inclusive, lower boundary is 35.5? Wait — in grouped data, if classes are 31-35, 36-40, etc., the real lower limit is 35.5? Actually, no — if it's discrete integers, the class 36-40 starts at 36, but for continuity, we use 35.5 as lower boundary? Let me clarify.

In many textbooks, for class intervals like 31-35, 36-40, the true class boundaries are 30.5-35.5, 35.5-40.5, etc., to avoid gaps.

But in this problem, since it's not specified, and the intervals are whole numbers, we can assume the class boundaries are:

- 31-35 → 30.5 to 35.5
- 36-40 → 35.5 to 40.5
- 41-45 → 40.5 to 45.5
- 46-50 → 45.5 to 50.5

So for median class 36-40, L = 35.5

cf_b = cumulative frequency before median class = 4 (from 31-35)

f_m = 14

i = 5

So:

Median = L + [(N/2 - cf_b) / f_m] × i
= 35.5 + [(17.5 - 4) / 14] × 5
= 35.5 + [13.5 / 14] × 5
= 35.5 + (0.9643) × 5
= 35.5 + 4.8214
≈ 40.32

Wait — that seems high. Let me double-check.

N/2 = 17.5

cf_b = 4 (cumulative before 36-40)

So position within class: 17.5 - 4 = 13.5 into the class of 14 items.

So fraction = 13.5 / 14 ≈ 0.9643

Multiply by class width 5: 0.9643 × 5 ≈ 4.8215

Add to lower boundary 35.5: 35.5 + 4.8215 = 40.3215 → ≈ 40.32

But intuitively, since most data is in 36-40 and 41-45, and median is around there, 40.32 seems reasonable.

Alternatively, some sources use the lower limit as 36 without adjustment. Let’s check both.

If we use L = 36 (without boundary adjustment):

Median = 36 + [(17.5 - 4)/14] × 5 = 36 + (13.5/14)*5 = same as above but starting at 36 → 36 + 4.8215 = 40.8215 ≈ 40.82

This is inconsistent. I think the correct approach is to use class boundaries.

Standard method: for class 36-40, if data is continuous, the lower boundary is 35.5.

I’ll go with 35.5.

So Median ≈ 40.32

But let me confirm with another way.

Total N=35, median is 18th value (since (35+1)/2 = 18th? No — for median in grouped data, we use N/2 = 17.5, so between 17th and 18th.

With cf from low to high:

- Up to 31-35: 4 values (positions 1-4)
- Up to 36-40: 18 values (positions 5-18) → so 17th and 18th values are in 36-40 class.

The 17th value is the 13th in the 36-40 class (since 17 - 4 = 13), and 18th is 14th.

Since class width is 5, and 14 values span 5 units, each value spans 5/14 ≈ 0.357 units.

So 13th value in class: start at 35.5 + 13*(5/14) = 35.5 + 65/14 ≈ 35.5 + 4.6429 ≈ 40.1429

14th value: 35.5 + 14*(5/14) = 35.5 + 5 = 40.5

Average of 17th and 18th: (40.1429 + 40.5)/2 ≈ 40.32145 → same as before.

So Median = 40.32

---

Step 6: Calculate Mode

For grouped data, mode formula:

Mode = L + [(f_m - f_{m-1}) / (2f_m - f_{m-1} - f_{m+1})] × i

Where:
- L = lower boundary of modal class
- f_m = frequency of modal class
- f_{m-1} = frequency of class before modal class
- f_{m+1} = frequency of class after modal class
- i = class width

First, find modal class: the class with highest frequency.

Frequencies: 5, 12, 14, 4 → highest is 14 → class 36-40

So:
- L = 35.5 (lower boundary)
- f_m = 14
- f_{m-1} = frequency of class before 36-40 → which is 31-35 → f=4
- f_{m+1} = frequency of class after 36-40 → which is 41-45 → f=12
- i = 5

Plug in:

Mode = 35.5 + [(14 - 4) / (2*14 - 4 - 12)] × 5
= 35.5 + [10 / (28 - 16)] × 5
= 35.5 + [10 / 12] × 5
= 35.5 + (5/6) × 5
= 35.5 + 25/6
= 35.5 + 4.1667
≈ 39.67

---

Step 7: Calculate Variance

For grouped data, variance formula:

Variance (σ²) = [Σ(fx²) - (Σ(fx))² / N] / N

This is population variance. Since the problem doesn't specify sample, and it's "measures of central tendency and variability" for grouped data, likely population variance.

Formula given in image is for mean, but for variance, standard formula is:

σ² = [Σ(fx²) - (Σfx)² / N] / N

We have:
- Σ(fx²) = 58280
- Σ(fx) = 1420
- N = 35

First, (Σfx)² = 1420² = 2,016,400

Then, (Σfx)² / N = 2,016,400 / 35 = 57,611.4286

Then, Σ(fx²) - that = 58280 - 57611.4286 = 668.5714

Then, variance = 668.5714 / 35 ≈ 19.102

So σ² ≈ 19.10

---

Step 8: Standard Deviation

σ = √variance = √19.102 ≈ 4.37

---

Final Values:

- Mean = 1420 / 35 = 40.5714 ≈ 40.57
- Median = 40.32
- Mode = 39.67
- Variance = 19.10
- Standard Deviation = 4.37

Let me verify variance calculation again.

Alternative formula for variance (sometimes used):

σ² = [Σf(x - x̄)²] / N

But we don't have that, so we use the computational formula.

Σ(fx²) = 58280

(Σfx)² / N = (1420)^2 / 35 = 2016400 / 35

Calculate 2016400 ÷ 35:

35 * 57600 = 35*50000=1,750,000; 35*7600=266,000; total 2,016,000 → remainder 400

400 / 35 ≈ 11.4286 → so 57600 + 11.4286 = 57611.4286 → correct.

58280 - 57611.4286 = 668.5714

668.5714 / 35 = 19.10204 → yes.

√19.10204 ≈ ? 4.37^2 = 19.0969, close to 19.102, so 4.37 is fine.

---

Now, fill in the unknown table:

| UNKNOWN | VALUE |
|---------------------|-----------|
| MEAN (x̄) | 40.57 |
| MEDIAN (x̃) | 40.32 |
| MODE (x̂) | 39.67 |
| VARIANCE (σ²) | 19.10 |
| STANDARD DEVIATION (σ) | 4.37 |

Note: In the table, it says "MEDIAN (x̄)" but that's a typo — it should be x̃ or just median. Similarly for mode. But we'll use the labels as given.

Also, in the original table, it has "i=" and "N=" — we filled i=5, N=35.

And Σ(fx)=1420, Σfx²=58280.

---

Final Answer:

MEAN (x̄) = 40.57
MEDIAN (x̃) = 40.32
MODE (x̂) = 39.67
VARIANCE (σ²) = 19.10
STANDARD DEVIATION (σ) = 4.37

But to match the table format exactly, and since the problem might expect specific rounding, I'll use two decimal places as standard.

Also, note: in some curricula, they might calculate median differently, but I believe 40.32 is correct.

Another thing: in the median calculation, if we used the lower limit as 36 instead of 35.5, we'd get 40.82, but I think 35.5 is more accurate for continuous data.

To confirm, let's see the class intervals: 31-35, 36-40 — these are discrete, but in statistics, we treat them as continuous for such calculations, so boundaries are 30.5-35.5, etc.

Yes.

So final values are as above.