SOLUTION: Mr Cheelo�s perfomance of his 170 Imionology class of students is given in an incomplete distribution below. Variable 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Frequence

Question 1166221: Mr Cheelo�s perfomance of his 170 Imionology class of students is given in an incomplete distribution below.
Variable 0-10 10-20 20-30 30-40 40-50 50-60 60-70
Frequence 10 20 f1 40 f2 25 15
(a) If the median is 35. Find the missing frequence.
(b) Calculate mean, Interquartile range and standard deviation. Hence explain using the mean and standard devi-ation, the meaning of these values in relation to Mr cheelo results.
Answer by CPhill(2189) (Show Source): You can put this solution on YOUR website!
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.
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