Table 2 below records the number of mobile phones sold in a sample of 30 electronics shops:
Table 2
Number of Mobile
Phones Sold
Frequency
1 - 9
5
10 - 18
7
19 - 27
10
28 - 36
3
37 - 45
3
46 - 54
2
The class width of the frequency distribution is
The mean number of mobile phones sold is
-
The standard deviation is —
- How many electronics shops sold less than 19 mobile phones?
A. 12
B. 12.71
C. 12.93
D. 9
E. 22.4
F. 6
G. 8

Let's break down the problem step by step:

### 1. Class Width of the Frequency Distribution
The class width is the difference between the upper and lower boundaries of each class. Let's check the first class:

\[
\text{Class width} = 10 - 1 = 9
\]

Thus, the **class width is 9**.

---

### 2. Mean Number of Mobile Phones Sold
The mean is calculated using the formula:

\[
\text{Mean} = \frac{\sum (f \cdot x)}{\sum f}
\]

Where:
- \( f \) is the frequency for each class.
- \( x \) is the midpoint of each class.

We first calculate the midpoints \( x \) for each class:

| Class        | Frequency \( f \) | Midpoint \( x \) |
|--------------|-------------------|------------------|
| 1 - 9        | 5                 | \( \frac{1 + 9}{2} = 5 \) |
| 10 - 18      | 7                 | \( \frac{10 + 18}{2} = 14 \) |
| 19 - 27      | 10                | \( \frac{19 + 27}{2} = 23 \) |
| 28 - 36      | 3                 | \( \frac{28 + 36}{2} = 32 \) |
| 37 - 45      | 3                 | \( \frac{37 + 45}{2} = 41 \) |
| 46 - 54      | 2                 | \( \frac{46 + 54}{2} = 50 \) |

Now, calculate \( f \cdot x \) for each class:

| Class        | Frequency \( f \) | Midpoint \( x \) | \( f \cdot x \) |
|--------------|-------------------|------------------|-----------------|
| 1 - 9        | 5                 | 5                | 25              |
| 10 - 18      | 7                 | 14               | 98              |
| 19 - 27      | 10                | 23               | 230             |
| 28 - 36      | 3                 | 32               | 96              |
| 37 - 45      | 3                 | 41               | 123             |
| 46 - 54      | 2                 | 50               | 100             |

Total frequency \( \sum f = 5 + 7 + 10 + 3 + 3 + 2 = 30 \)

Total of \( f \cdot x = 25 + 98 + 230 + 96 + 123 + 100 = 672 \)

Now, calculate the mean:

\[
\text{Mean} = \frac{672}{30} = 22.4
\]

Thus, the **mean number of mobile phones sold is 22.4**.

---

### 3. Standard Deviation
The formula for the standard deviation is:

\[
\sigma = \sqrt{\frac{\sum (f \cdot (x - \mu)^2)}{\sum f}}
\]

Where:
- \( \mu \) is the mean, which we calculated as 22.4.
- \( x \) is the midpoint for each class.
- \( f \) is the frequency for each class.

We need to compute \( f \cdot (x - \mu)^2 \) for each class:

| Class        | Frequency \( f \) | Midpoint \( x \) | \( (x - \mu)^2 \) | \( f \cdot (x - \mu)^2 \) |
|--------------|-------------------|------------------|------------------|-------------------------|
| 1 - 9        | 5                 | 5                | \( (5 - 22.4)^2 = 306.25 \)  | \( 5 \cdot 306.25 = 1531.25 \) |
| 10 - 18      | 7                 | 14               | \( (14 - 22.4)^2 = 70.56 \)  | \( 7 \cdot 70.56 = 493.92 \) |
| 19 - 27      | 10                | 23               | \( (23 - 22.4)^2 = 0.36 \)   | \( 10 \cdot 0.36 = 3.6 \) |
| 28 - 36      | 3                 | 32               | \( (32 - 22.4)^2 = 92.16 \)  | \( 3 \cdot 92.16 = 276.48 \) |
| 37 - 45      | 3                 | 41               | \( (41 - 22.4)^2 = 345.96 \) | \( 3 \cdot 345.96 = 1037.88 \) |
| 46 - 54      | 2                 | 50               | \( (50 - 22.4)^2 = 756.16 \) | \( 2 \cdot 756.16 = 1512.32 \) |

Sum of \( f \cdot (x - \mu)^2 = 1531.25 + 493.92 + 3.6 + 276.48 + 1037.88 + 1512.32 = 4855.45 \)

Now, calculate the standard deviation:

\[
\sigma = \sqrt{\frac{4855.45}{30}} = \sqrt{161.85} \approx 12.72
\]

Thus, the **standard deviation is approximately 12.72**.

---

### 4. Number of Electronics Shops That Sold Less Than 19 Mobile Phones
To find how many shops sold fewer than 19 phones, we sum the frequencies of the classes where the number of phones sold is less than 19:

\[
\text{Total} = 5 \ (\text{from } 1-9) + 7 \ (\text{from } 10-18) = 12
\]

Thus, **12 shops** sold fewer than 19 phones.

The correct answer is **A. 12**.

---

### Summary:
- **Class width**: 9
- **Mean**: 22.4
- **Standard deviation**: 12.72
- **Number of shops selling less than 19 phones**: 12

---

Do you have any questions or need further details?

---

Here are five related questions to expand your understanding:
1. How would the mean change if another class was added to the table?
2. What is the importance of the class width in frequency distribution?
3. How does the standard deviation help in interpreting data spread?
4. How would the results change if we had a larger sample size?
5. Can you describe how the variance relates to standard deviation?

**Tip**: When calculating the mean, always double-check your midpoints and frequencies for accuracy!

Table 2 below records the number of mobile phones sold in a sample of 30 electronics shops: Number of Mobile Phones Sold: 1 - 9 (5), 10 - 18 (7), 19 - 27 (10), 28 - 36 (3), 37 - 45 (3), 46 - 54 (2). The class width of the frequency distribution is ___, the mean number of mobile phones sold is ___, the standard deviation is ___. How many electronics shops sold less than 19 mobile phones? A. 12 B. 12.71 C. 12.93 D. 9 E. 22.4 F. 6 G. 8.

This problem involves calculating the class width, mean, and standard deviation for mobile phone sales data across 30 electronics shops. The data covers sales ranging from 1 to 54 phones, and the analysis provides insights into the distribution of sales. Learn to compute key statistical measures, including how many shops sold less than 19 phones. Suitable for high school students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation