Solved Mean, Median and Mode (30 pts) Find the Mean. Median ... - Free Printable

Let's solve the problem step by step for both ungrouped and grouped data, finding the Mean, Median, and Mode.

---

✔ Part 1: Ungrouped Data

#### 1. Jhaydan’s test scores in 9 subjects
Data:
89, 86, 88, 89, 90, 90, 87, 88, 90

##### Step 1: Mean
Mean = (Sum of all values) / (Number of values)

$$
\text{Sum} = 89 + 86 + 88 + 89 + 90 + 90 + 87 + 88 + 90 = 797
$$
$$
\text{Number of values} = 9
$$
$$
\text{Mean} = \frac{797}{9} \approx 88.56
$$

✔ Mean ≈ 88.56

---

##### Step 2: Median
Arrange data in ascending order:
86, 87, 88, 88, 89, 89, 90, 90, 90

There are 9 values (odd), so median is the 5th value.

$$
\text{Median} = 89
$$

✔ Median = 89

---

##### Step 3: Mode
Mode = most frequent value(s)

- 86 → 1 time
- 87 → 1 time
- 88 → 2 times
- 89 → 2 times
- 90 → 3 times

So, 90 appears most frequently.

✔ Mode = 90

---

#### 2. Measurements of 6 samples
Data:
112, 113, 117, 115, 110, 109

##### Step 1: Mean
$$
\text{Sum} = 112 + 113 + 117 + 115 + 110 + 109 = 676
$$
$$
\text{Number of values} = 6
$$
$$
\text{Mean} = \frac{676}{6} \approx 112.67
$$

✔ Mean ≈ 112.67

---

##### Step 2: Median
Arrange in ascending order:
109, 110, 112, 113, 115, 117

Even number of values (6), so median is average of 3rd and 4th values:

$$
\text{Median} = \frac{112 + 113}{2} = \frac{225}{2} = 112.5
$$

✔ Median = 112.5

---

##### Step 3: Mode
All values appear only once → No mode

✔ Mode = None (or No mode)

---

✔ Part 2: Grouped Data

Given frequency distribution:

| Class Limit | Frequency |
|-------------|-----------|
| 1–9 | 15 |
| 10–18 | 17 |
| 19–27 | 11 |
| 28–36 | 10 |
| 37–45 | 13 |
| 46–54 | 14 |
| Total | 80 |

We are told:
- Median class is 28–36 (since 40th and 41st values fall here)
- Modal class is 10–18

Let’s compute Mean, Median, and Mode using formulas.

---

#### Step 1: Mean (for grouped data)

Use formula:
$$
\text{Mean} = \frac{\sum f_i x_i}{\sum f_i}
$$
Where $x_i$ is the midpoint of each class.

| Class Limit | Midpoint ($x_i$) | Frequency ($f_i$) | $f_i \times x_i$ |
|-------------|------------------|--------------------|-------------------|
| 1–9 | (1+9)/2 = 5 | 15 | 15 × 5 = 75 |
| 10–18 | (10+18)/2 = 14 | 17 | 17 × 14 = 238 |
| 19–27 | (19+27)/2 = 23 | 11 | 11 × 23 = 253 |
| 28–36 | (28+36)/2 = 32 | 10 | 10 × 32 = 320 |
| 37–45 | (37+45)/2 = 41 | 13 | 13 × 41 = 533 |
| 46–54 | (46+54)/2 = 50 | 14 | 14 × 50 = 700 |
| | | Total = 80 | ∑f_ix_i = 2120|

$$
\text{Mean} = \frac{2120}{80} = 26.5
$$

✔ Mean = 26.5

---

#### Step 2: Median (for grouped data)

We use the formula:
$$
\text{Median} = L + \left( \frac{\frac{N}{2} - CF}{f} \right) \times h
$$
Where:
- $L$ = lower limit of median class
- $N$ = total frequency = 80
- $CF$ = cumulative frequency before median class
- $f$ = frequency of median class
- $h$ = class width

From note: Median class is 28–36

- $L = 28$
- $h = 36 - 28 = 8$
- $f = 10$
- $N/2 = 40$

Now find cumulative frequency up to previous class:

| Class | Frequency | Cumulative Freq |
|-------------|-----------|-----------------|
| 1–9 | 15 | 15 |
| 10–18 | 17 | 15 + 17 = 32 |
| 19–27 | 11 | 32 + 11 = 43 |
| 28–36 | 10 | 43 + 10 = 53 |

So, CF (before 28–36) = 43

Now plug into formula:
$$
\text{Median} = 28 + \left( \frac{40 - 43}{10} \right) \times 8 = 28 + \left( \frac{-3}{10} \right) \times 8
$$
$$
= 28 - 2.4 = 25.6
$$

Wait! This gives 25.6, but the median class was said to be 28–36, which should contain the 40th and 41st values.

But cumulative frequency before 28–36 is 43, which means the first 43 values are ≤ 27.

So the 40th and 41st values are in the class 28–36, yes — that's correct.

But let's check cumulative frequencies again:

- After 1–9: 15
- After 10–18: 15 + 17 = 32
- After 19–27: 32 + 11 = 43

So the 40th and 41st values are within 28–36 → correct.

But since the cumulative frequency before 28–36 is 43, and we're looking for the 40th value, it's already in the 28–36 class, because 40 > 32 but < 43?

Wait — this is a contradiction!

Wait: The cumulative frequency after 19–27 is 43, meaning that the 43rd value is the last one in 19–27.

So the 40th value is still in the 19–27 class.

But the problem says: *"The Median Class will be the 40th and 41st values which will be in interval 28–36."*

That contradicts our calculation.

Let’s double-check:

- Class 1–9: 15 values → positions 1 to 15
- Class 10–18: 17 values → positions 16 to 32
- Class 19–27: 11 values → positions 33 to 43
- Class 28–36: 10 values → positions 44 to 53

So:
- 40th value → in class 19–27
- 41st value → also in class 19–27

Therefore, median class should be 19–27, not 28–36.

But the note says: *"The Median Class will be the 40th and 41st values which will be in interval 28–36"*

This is incorrect based on frequencies.

Let’s verify the frequencies:

- 1–9: 15
- 10–18: 17 → cum = 32
- 19–27: 11 → cum = 43
- 28–36: 10 → cum = 53

So the 40th value falls in 19–27 (since 32 < 40 ≤ 43)

Hence, median class is 19–27, not 28–36.

But the problem says "Note: The Median Class will be the 40th and 41st values which will be in interval 28–36"

That seems to be an error.

Alternatively, perhaps the class limits are inclusive and there's confusion.

Wait — maybe the class intervals are exclusive, like 1–9 means 1 to 9.99, etc., but that doesn't change the count.

Let me recheck the cumulative frequency:

- 1–9: 15 → positions 1–15
- 10–18: 17 → positions 16–32
- 19–27: 11 → positions 33–43
- So 40th value is at position 40 → still in 19–27

Thus, median class is 19–27

But the note says it's 28–36 — which is wrong.

However, since the problem explicitly states:
> "Note: The Median Class will be the 40th and 41st values which will be in interval 28-36."

And asks us to use that, perhaps we should follow the instruction, even if it's mathematically incorrect.

But let's see: total frequency is 80 → N/2 = 40

If 40th and 41st values are in 28–36, then cumulative frequency before 28–36 must be less than 40

But from table:
- Before 28–36: 15 + 17 + 11 = 43

Which is greater than 40, so 40th value is before 28–36.

So unless there’s a typo in the frequency table, the note is wrong.

Let’s suppose the frequencies are correct.

Then the correct median class is 19–27, not 28–36.

But the problem says: “The Median Class will be the 40th and 41st values which will be in interval 28-36”

This implies that cumulative frequency before 28–36 is less than 40, but we have 43.

So either:
- There's a typo in the frequencies
- Or the note is wrong

Let’s assume the note is given as a hint, and we are to use 28–36 as the median class, despite the numbers.

But that would mean the cumulative frequency before 28–36 is < 40, but we have 43 → impossible.

Wait — unless the classes are not overlapping and the cumulative frequency is calculated differently.

Wait — let’s check total: 15+17+11+10+13+14 = 80 → correct.

Cumulative:
- 1–9: 15
- 10–18: 15+17=32
- 19–27: 32+11=43 → already 43

So 40th value is in 19–27

Therefore, median class is 19–27

So the note is incorrect.

But the problem says: “Note: The Median Class will be the 40th and 41st values which will be in interval 28-36.”

So either the frequencies are wrong or the note is wrong.

But since we have to follow the instructions, perhaps we should assume the note is correct, and re-evaluate.

Wait — maybe the class limits are inclusive and the intervals are written incorrectly.

Alternatively, perhaps the frequency for 19–27 is not 11, but something else.

But the table says:
- 1–9: 15
- 10–18: 17
- 19–27: 11
- 28–36: 10
- 37–45: 13
- 46–54: 14
- Total: 80

15+17=32; 32+11=43 → yes.

So unless the class 19–27 has fewer values, it's impossible for 40th value to be in 28–36.

Perhaps the note is misstated.

But the problem says: “The Median Class will be the 40th and 41st values which will be in interval 28-36”

So perhaps we should trust the note, and recompute frequencies?

Alternatively, maybe the classes are not continuous? But they seem continuous.

Another possibility: the class limits are inclusive, but the next class starts at 28, so 19–27 includes 19 to 27, then 28–36 includes 28 to 36.

So no overlap.

So cumulative before 28–36 is 15+17+11 = 43

So 40th value is in 19–27

So median class is 19–27

But the note says 28–36 — contradiction.

Let’s proceed with correct calculation, and mention the discrepancy.

But since the problem says “Note: ...”, and assigns points, we may need to follow it.

Wait — perhaps the total is not 80?

15+17=32; 32+11=43; 43+10=53; 53+13=66; 66+14=80 → yes.

So I think the note is wrong.

But to comply, let’s assume the note is correct, and see what happens.

Suppose the median class is 28–36, then cumulative frequency before it is less than 40

But we have 43 → impossible.

Unless the frequency of 19–27 is not 11, but lower.

But it’s given as 11.

So unless the problem meant that the 40th value is in 28–36, but that would require cumulative before it to be <40.

So if cumulative before 28–36 is 39, then 40th is in 28–36.

But currently it’s 43.

So unless the frequency of 19–27 is only 1, but it’s 11.

So I think the note is erroneous.

But let’s continue with correct values.

---

#### ✔ Correct Median Calculation

Median class: 19–27

- $L = 19$
- $h = 27 - 19 = 8$
- $f = 11$
- $CF = 32$ (cumulative before 19–27)
- $N/2 = 40$

$$
\text{Median} = 19 + \left( \frac{40 - 32}{11} \right) \times 8 = 19 + \left( \frac{8}{11} \right) \times 8
$$
$$
= 19 + \frac{64}{11} = 19 + 5.818 = 24.818
$$

✔ Median ≈ 24.82

But the note says it should be in 28–36 — so this is inconsistent.

But we’ll go with correct math.

---

#### Step 3: Mode (for grouped data)

Formula:
$$
\text{Mode} = L + \left( \frac{f_m - f_1}{2f_m - f_1 - f_2} \right) \times h
$$
Where:
- $L$ = lower limit of modal class
- $f_m$ = frequency of modal class
- $f_1$ = frequency of class before
- $f_2$ = frequency of class after
- $h$ = class width

Given: Modal class is 10–18 (from note)

- $L = 10$
- $f_m = 17$
- $f_1 = 15$ (class 1–9)
- $f_2 = 11$ (class 19–27)
- $h = 18 - 10 = 8$

$$
\text{Mode} = 10 + \left( \frac{17 - 15}{2(17) - 15 - 11} \right) \times 8
= 10 + \left( \frac{2}{34 - 26} \right) \times 8
= 10 + \left( \frac{2}{8} \right) \times 8
= 10 + 2 = 12
$$

✔ Mode = 12

---

✔ Final Answers

---

#### Ungrouped Data

1. Jhaydan’s test scores:

- Mean = 88.56
- Median = 89
- Mode = 90

2. Measurements of 6 samples:

- Mean = 112.67
- Median = 112.5
- Mode = None

---

#### Grouped Data

- Mean = 26.5
- Median = 24.82 (using correct class 19–27)
- Mode = 12

⚠️ Note: The provided note says the median class is 28–36, but based on cumulative frequencies, the median class is 19–27. Therefore, the median calculated using 28–36 would be incorrect. However, if you are required to follow the note strictly, please inform your instructor about the inconsistency.

---

✔ Summary Table

| Data Set | Mean | Median | Mode |
|-----------------------------|------------|------------|----------|
| Jhaydan's scores | 88.56 | 89 | 90 |
| 6 measurements | 112.67 | 112.5 | None |
| Grouped data | 26.5 | 24.82 | 12 |

Let me know if you'd like to recalculate assuming the note is correct (e.g., if frequencies were different).