Given the frequency distribution of ages, calculate the 1st quartile, 2nd quartile, 3rd quartile, 4th decile, and 5th decile.

To compute the quartiles ($Q_1, Q_2, Q_3$) and deciles ($D_4, D_5$), we first need to determine the class boundaries and cumulative frequencies.

The given data is:

Age (Class Interval)	Frequency (f)
0-9	19
10-19	47
20-29	105
30-39	93
40-49	16

The class width ($h$) is $10 - 0 = 10$ (or $19.5 - 9.5 = 10$).
The total frequency ($N$) is $19 + 47 + 105 + 93 + 16 = 280$.

First, we construct the cumulative frequency table:

Class Interval	Class Boundaries	Frequency (f)	Cumulative Frequency (cf)
0-9	-0.5 - 9.5	19	19
10-19	9.5 - 19.5	47	$19 + 47 = 66$
20-29	19.5 - 29.5	105	$66 + 105 = 171$
30-39	29.5 - 39.5	93	$171 + 93 = 264$
40-49	39.5 - 49.5	16	$264 + 16 = 280$

The general formula for quartiles ($Q_j$) and deciles ($D_j$) for grouped data is:

$$P_k = L + \left(\frac{\frac{kN}{X} - cf_b}{f_p}\right)h$$

where:

• $P_k$ is the $k$-th percentile (or quartile/decile).
• $L$ is the lower class boundary of the class containing $P_k$.
• $N$ is the total frequency.
• $X$ is 4 for quartiles, 10 for deciles.
• $cf_b$ is the cumulative frequency of the class before the class containing $P_k$.
• $f_p$ is the frequency of the class containing $P_k$.
• $h$ is the class width.

Step 1: Calculate $Q_1$ (First Quartile)
The position of $Q_1$ is $\frac{1N}{4} = \frac{1 \times 280}{4} = 70$.
The 70th value falls in the 20-29 class (since $cf=66$ for 10-19 and $cf=171$ for 20-29).
For $Q_1$: $L = 19.5$, $cf_b = 66$, $f_Q = 105$, $h = 10$.

$$Q_1 = 19.5 + \left(\frac{70 - 66}{105}\right)10$$ $$Q_1 = 19.5 + \left(\frac{4}{105}\right)10$$ $$Q_1 = 19.5 + \frac{40}{105}$$ $$Q_1 = 19.5 + 0.38095$$ $$\boxed{Q_1 = 19.881}$$

Step 2: Calculate $Q_2$ (Second Quartile / Median)
The position of $Q_2$ is $\frac{2N}{4} = \frac{280}{2} = 140$.
The 140th value falls in the 20-29 class.
For $Q_2$: $L = 19.5$, $cf_b = 66$, $f_Q = 105$, $h = 10$.

$$Q_2 = 19.5 + \left(\frac{140 - 66}{105}\right)10$$ $$Q_2 = 19.5 + \left(\frac{74}{105}\right)10$$ $$Q_2 = 19.5 + \frac{740}{105}$$ $$Q_2 = 19.5 + 7.04762$$ $$\boxed{Q_2 = 26.548}$$

Step 3: Calculate $Q_3$ (Third Quartile)
The position of $Q_3$ is $\frac{3N}{4} = \frac{3 \times 280}{4} = 210$.
The 210th value falls in the 30-39 class (since $cf=171$ for 20-29 and $cf=264$ for 30-39).
For $Q_3$: $L = 29.5$, $cf_b = 171$, $f_Q = 93$, $h = 10$.

$$Q_3 = 29.5 + \left(\frac{210 - 171}{93}\right)10$$ $$Q_3 = 29.5 + \left(\frac{39}{93}\right)10$$ $$Q_3 = 29.5 + \frac{390}{93}$$ $$Q_3 = 29.5 + 4.19355$$ $$\boxed{Q_3 = 33.694}$$

Step 4: Calculate $D_4$ (Fourth Decile)
The position of $D_4$ is $\frac{4N}{10} = \frac{4 \times 280}{10} = 112$.
The 112th value falls in the 20-29 class.
For $D_4$: $L = 19.5$, $cf_b = 66$, $f_D = 105$, $h = 10$.

$$D_4 = 19.5 + \left(\frac{112 - 66}{105}\right)10$$ $$D_4 = 19.5 + \left(\frac{46}{105}\right)10$$ $$D_4 = 19.5 + \frac{460}{105}$$ $$D_4 = 19.5 + 4.38095$$ $$\boxed{D_4 = 23.881}$$

Step 5: Calculate $D_5$ (Fifth Decile)
The position of $D_5$ is $\frac{5N}{10} = \frac{5 \times 280}{10} = 140$.
This is the same position as $Q_2$. The 140th value falls in the 20-29 class.
For $D_5$: $L = 19.5$, $cf_b = 66$, $f_D = 105$, $h = 10$.

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