Calculate quartile deviation: Marks 50 60 70 80 90 : : : : : : No. of students 2 4 8 4 1

Calculation of Quartile Deviation

Step 1: Find the total number of students

The given frequency distribution is:

Total number of students ($N$) = $2+4+8+4+1=19$.

Step 2: Find the position of the first quartile ($Q_1$) and third quartile ($Q_3$)

The formula for quartile position is: $$Q_n=\frac{n(N+1)}{4}$$ where $n=1$ for $Q_1$ and $n=3$ for $Q_3$.

• First quartile ($Q_1$): $$Q_1=\frac{1(19+1)}{4}=\frac{20}{4}=5$$ The 5th value falls in the 70 marks class (since cumulative frequency up to 60 is $2+4=6$, which exceeds 5).

• Third quartile ($Q_3$): $$Q_3=\frac{3(19+1)}{4}=\frac{60}{4}=15$$ The 15th value falls in the 80 marks class (since cumulative frequency up to 70 is $2+4+8=14$, and the next class is 80).

Step 3: Find the exact values of $Q_1$ and $Q_3$ using interpolation

The formula for interpolation is: $$Q_n=L+\left(\frac{\frac{n(N+1)}{4}-C}{f}\right)\times i$$ where:

• $L$ = lower boundary of the quartile class,

• $C$ = cumulative frequency before the quartile class,

• $f$ = frequency of the quartile class,

• $i$ = class width.

• For $Q_1$ (70 marks class):

  • $L=60$ (lower boundary of 70 marks class),
  • $C=6$ (cumulative frequency up to 60),
  • $f=8$ (frequency of 70 marks class),
  • $i=10$ (class width).

  $$Q_1=60+\left(\frac{5-6}{8}\right)\times 10=60-\frac{10}{8}=60-1.25=58.75$$

• For $Q_3$ (80 marks class):

  • $L=70$ (lower boundary of 80 marks class),
  • $C=14$ (cumulative frequency up to 70),
  • $f=4$ (frequency of 80 marks class),
  • $i=10$ (class width).

  $$Q_3=70+\left(\frac{15-14}{4}\right)\times 10=70+\frac{10}{4}=70+2.5=72.5$$

Step 4: Calculate the quartile deviation

The formula for quartile deviation ($Q.D.$) is: $$Q.D.=\frac{Q_3-Q_1}{2}$$

$$Q.D.=\frac{72.5-58.75}{2}=\frac{13.75}{2}=6.875$$

Final Answer: The quartile deviation is 6.875.