Question 1166221
This is a problem involving calculations with a **grouped frequency distribution**, specifically finding missing frequencies using the median, and then calculating key statistics like the mean, interquartile range (IQR), and standard deviation.

The total number of students is $N = 170$.
The frequency distribution is:

| Variable (Score) | Frequency ($f$) |
| :---: | :---: |
| 0-10 | 10 |
| 10-20 | 20 |
| 20-30 | $f_1$ |
| 30-40 | 40 |
| 40-50 | $f_2$ |
| 50-60 | 25 |
| 60-70 | 15 |

---

## (a) Finding the Missing Frequencies

We have two missing frequencies, $f_1$ and $f_2$. We'll use two pieces of information: the total frequency ($N$) and the median value.

### 1. Using the Total Frequency ($N=170$)

The sum of all frequencies must equal the total number of students:
$$10 + 20 + f_1 + 40 + f_2 + 25 + 15 = 170$$
$$110 + f_1 + f_2 = 170$$
$$f_1 + f_2 = 170 - 110$$
$$f_1 + f_2 = 60 \quad \text{(Equation 1)}$$

### 2. Using the Median (Median = 35)

The median position is at $\frac{N}{2} = \frac{170}{2} = 85^{\text{th}}$ observation.

Since the median is $35$ (which falls in the $30-40$ class interval), the **median class** is **$30-40$**.

The formula for the median ($M$) of a grouped data set is:
$$M = L + \left(\frac{\frac{N}{2} - C}{f_m}\right) \times w$$
Where:
* $L$ is the lower boundary of the median class ($30-40$), so $L = 30$.
* $N$ is the total frequency, $N = 170$.
* $C$ is the cumulative frequency of the class *before* the median class ($20-30$).
    $$C = 10 + 20 + f_1 = 30 + f_1$$
* $f_m$ is the frequency of the median class ($30-40$), so $f_m = 40$.
* $w$ is the class width, $w = 10$.

Substitute the known values into the median formula:
$$35 = 30 + \left(\frac{85 - (30 + f_1)}{40}\right) \times 10$$

Solve for $f_1$:
$$35 - 30 = \frac{55 - f_1}{40} \times 10$$
$$5 = \frac{55 - f_1}{4}$$
$$5 \times 4 = 55 - f_1$$
$$20 = 55 - f_1$$
$$f_1 = 55 - 20$$
$$\mathbf{f_1 = 35}$$

### 3. Finding $f_2$

Substitute $f_1 = 35$ into Equation 1:
$$f_1 + f_2 = 60$$
$$35 + f_2 = 60$$
$$f_2 = 60 - 35$$
$$\mathbf{f_2 = 25}$$

The missing frequencies are $\mathbf{f_1 = 35}$ and $\mathbf{f_2 = 25}$.

---

## (b) Calculating Mean, Interquartile Range, and Standard Deviation

Now we have the complete distribution: $N=170$.

| Score Class | Frequency ($f$) | Midpoint ($x$) | $f \cdot x$ | Cum. Freq. ($C_f$) |
| :---: | :---: | :---: | :---: | :---: |
| 0-10 | 10 | 5 | 50 | 10 |
| 10-20 | 20 | 15 | 300 | 30 |
| 20-30 | **35** ($f_1$) | 25 | 875 | 65 |
| 30-40 | 40 | 35 | 1400 | 105 |
| 40-50 | **25** ($f_2$) | 45 | 1125 | 130 |
| 50-60 | 25 | 55 | 1375 | 155 |
| 60-70 | 15 | 65 | 975 | 170 |
| **Total** | **170** | | $\sum f x = 6100$ | |

### 1. Mean ($\bar{x}$)

The formula for the mean of grouped data is:
$$\bar{x} = \frac{\sum f x}{N}$$
$$\bar{x} = \frac{6100}{170}$$
$$\bar{x} \approx \mathbf{35.88}$$

### 2. Interquartile Range (IQR)

$IQR = Q_3 - Q_1$

#### i. First Quartile ($Q_1$)
$Q_1$ position is at $\frac{N}{4} = \frac{170}{4} = 42.5^{\text{th}}$ observation.
The $42.5^{\text{th}}$ observation falls in the **$20-30$ class** ($C_f$ goes from 30 to 65).
* $L = 20$
* $C = 30$
* $f_q = 35$
* $w = 10$

$$Q_1 = L + \left(\frac{\frac{N}{4} - C}{f_q}\right) \times w = 20 + \left(\frac{42.5 - 30}{35}\right) \times 10$$
$$Q_1 = 20 + \left(\frac{12.5}{35}\right) \times 10 \approx 20 + 3.5714$$
$$\mathbf{Q_1 \approx 23.57}$$

#### ii. Third Quartile ($Q_3$)
$Q_3$ position is at $\frac{3N}{4} = \frac{3 \times 170}{4} = 127.5^{\text{th}}$ observation.
The $127.5^{\text{th}}$ observation falls in the **$40-50$ class** ($C_f$ goes from 105 to 130).
* $L = 40$
* $C = 105$
* $f_q = 25$
* $w = 10$

$$Q_3 = L + \left(\frac{\frac{3N}{4} - C}{f_q}\right) \times w = 40 + \left(\frac{127.5 - 105}{25}\right) \times 10$$
$$Q_3 = 40 + \left(\frac{22.5}{25}\right) \times 10 = 40 + 0.9 \times 10 = 40 + 9$$
$$\mathbf{Q_3 = 49}$$

#### iii. IQR
$$IQR = Q_3 - Q_1 = 49 - 23.57 \approx \mathbf{25.43}$$

### 3. Standard Deviation ($\sigma$)

We need to calculate $\sum f x^2$:

| Class | $f$ | $x$ | $f \cdot x$ | $x^2$ | $f \cdot x^2$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 0-10 | 10 | 5 | 50 | 25 | 250 |
| 10-20 | 20 | 15 | 300 | 225 | 4500 |
| 20-30 | 35 | 25 | 875 | 625 | 21875 |
| 30-40 | 40 | 35 | 1400 | 1225 | 49000 |
| 40-50 | 25 | 45 | 1125 | 2025 | 50625 |
| 50-60 | 25 | 55 | 1375 | 3025 | 75625 |
| 60-70 | 15 | 65 | 975 | 4225 | 63375 |
| **Total** | **170** | | $\sum f x = 6100$ | | $\sum f x^2 = 265250$ |

The formula for the standard deviation ($\sigma$) is:
$$\sigma = \sqrt{\frac{\sum f x^2}{N} - \bar{x}^2}$$
$$\sigma = \sqrt{\frac{265250}{170} - (35.88235...)^2}$$
$$\sigma = \sqrt{1560.2941 - 1287.545}$$
$$\sigma = \sqrt{272.7491}$$
$$\sigma \approx \mathbf{16.51}$$

### Summary of Results
* **Mean ($\bar{x}$):** $35.88$
* **Interquartile Range (IQR):** $25.43$
* **Standard Deviation ($\sigma$):** $16.51$

---

## Explanation of Mean and Standard Deviation

### Meaning of the Mean ($\bar{x} \approx 35.88$)

The **mean** of $35.88$ represents the **average score** obtained by the students in Mr. Cheelo's Imionology class. This is the central tendency point—if the scores were redistributed equally among all 170 students, each student would have approximately $35.88$ points. This score falls within the middle range of the possible scores (0 to 70).

### Meaning of the Standard Deviation ($\sigma \approx 16.51$)

The **standard deviation** of $16.51$ is a measure of the **dispersion or variability** of the scores around the mean.
* **Large Standard Deviation (like 16.51, which is $\approx 46\%$ of the mean):** Indicates that the scores are **widely spread out** from the average score. A significant number of students scored either much higher (e.g., above $35.88 + 16.51 = 52.39$) or much lower (e.g., below $35.88 - 16.51 = 19.37$).
* **Conclusion for Mr. Cheelo's Class:** The high standard deviation suggests the class performance was **not homogeneous**. There was a large difference in performance between the highest and lowest-scoring students, indicating a wide range of academic ability or preparation in the class.