For these students, find the a) Lower quartile, b) Upper quartile, c) interquartile range.

Let $x_i$ be the midpoint of each class interval.

Class Interval	Frequencies ($f_i$)	Midpoint ($x_i$)	$f_i x_i$
11-13	3	12	36
13-15	6	14	84
15-17	9	16	144
17-19	13	18	234
19-21	f	20	20f
21-23	5	22	110
23-25	1	24	24
Total	$37+f$		$632+20f$

The formula for the mean of grouped data is $\bar{x} = \frac{\sum f_i x_i}{\sum f_i}$.
Given $\bar{x} = 18$:

$$18 = \frac{632 + 20f}{37 + f}$$

Multiply both sides by $(37+f)$:

$$18(37 + f) = 632 + 20f$$ $$666 + 18f = 632 + 20f$$

Subtract $18f$ from both sides and 632 from both sides:

$$666 - 632 = 20f - 18f$$ $$34 = 2f$$ $$f = \frac{34}{2}$$

The value of $f$ is $\boxed{\text{17}}$.

Now that $f=17$, the total frequency $N = \sum f_i = 37 + 17 = 54$.
The sum $\sum f_i x_i = 632 + 20(17) = 632 + 340 = 972$.
The mean $\bar{x} = \frac{972}{54} = 18$.

To calculate the standard deviation, we need $\sum f_i x_i^2$.

Class Interval	Frequencies ($f_i$)	Midpoint ($x_i$)	$x_i^2$	$f_i x_i^2$
11-13	3	12	144	432
13-15	6	14	196	1176
15-17	9	16	256	2304
17-19	13	18	324	4212
19-21	17	20	400	6800
21-23	5	22	484	2420
23-25	1	24	576	576
Total	$N=54$			17920

The formula for the standard deviation ($s$) of grouped data is:

$$s = \sqrt{\frac{\sum f_i x_i^2}{N} - \bar{x}^2}$$

Substitute the values:

$$s = \sqrt{\frac{17920}{54} - (18)^2}$$ $$s = \sqrt{331.85185... - 324}$$ $$s = \sqrt{7.85185...}$$ $$s \approx 2.8021$$

Rounding to two decimal places, the standard deviation is $\boxed{\text{2.80}}$.

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: QUESTION FIVE (Bottom Table)
The data shows the number of times students open their WhatsApp accounts.

The total frequency $N = \sum f_i = 15+17+14+12+7 = 65$.

Class Interval	Frequency ($f_i$)	Cumulative Frequency ($CF$)
1-10	15	15
11-20	17	32
21-30	14	46
31-40	12	58
41-50	7	65
Total	$N=65$

The formula for the $k$-th quartile ($Q_k$) of grouped data is:

$$Q_k = L + \left( \frac{\frac{kN}{4} - CF}{f} \right) \times w$$

where $L$ is the lower boundary of the quartile class, $CF$ is the cumulative frequency of the class before the quartile class, $f$ is the frequency of the quartile class, and $w$ is the class width. The class width for this data is $10.5 - 0.5 = 10$.

The position of $Q_1$ is $\frac{N}{4} = \frac{65}{4} = 16.25$.
The $Q_1$ class is 11-20 (since its $CF=32$ is the first $CF \ge 16.25$).

• $L = 11 - 0.5 = 10.5$
• $CF = 15$ (cumulative frequency of the class before 11-20)
• $f = 17$ (frequency of the 11-20 class)
• $w = 10$

Substitute these values into the formula:

$$Q_1 = 10.5 + \left( \frac{16.25 - 15}{17} \right) \times 10$$ $$Q_1 = 10.5 + \left( \frac{1.25}{17} \right) \times 10$$ $$Q_1 = 10.5 + \frac{12.5}{17}$$ $$Q_1 = 10.5 + 0.73529...$$

Rounding to two decimal places, the lower quartile ($Q_1$) is $\boxed{\text{11.24}}$.

The position of $Q_3$ is $\frac{3N}{4} = \frac{3 \times 65}{4} = \frac{195}{4} = 48.75$.
The $Q_3$ class is 31-40 (since its $CF=58$ is the first $CF \ge 48.75$).

• $L = 31 - 0.5 = 30.5$
• $CF = 46$ (cumulative frequency of the class before 31-40)
• $f = 12$ (frequency of the 31-40 class)
• $w = 10$

Substitute these values into the formula:

$$Q_3 = 30.5 + \left( \frac{48.75 - 46}{12} \right) \times 10$$ $$Q_3 = 30.5 + \left( \frac{2.75}{12} \right) \times 10$$ $$Q_3 = 30.5 + \frac{27.5}{12}$$ $$Q_3 = 30.5 + 2.2916...$$

Rounding to two decimal places, the upper quartile ($Q_3$) is $\boxed{\text{32.79}}$.

The interquartile range is the difference between the upper and lower quartiles:

$$\text{IQR} = Q_3 - Q_1$$ $$\text{IQR} = 32.79 - 11.24$$ $$\text{IQR} = 21.55$$

The interquartile range is $\boxed{\text{21.55}}$.