Question

The variance of the following continuous frequency distribution is 

Class Interval	\(0\!-\!10\)	\(10\!-\!20\)	\(20\!-\!30\)
Frequency	\(3\)	\(4\)	\(3\)

• A. \(15\)
• B. \(30\)
• C. \(45\)
• D. \(60\)

Hint:

For grouped data, \[ \boxed{ \text{Variance} = \frac{\sum f(x-\bar{x})^2}{\sum f} } \] where \(x\) denotes the class marks.

Answer 1

The Correct Option is D

Step 1: Find the class marks.& nbsp;

The class marks are

\[ 5,\;15,\;25. \]

The corresponding frequencies are

\[ 3,\;4,\;3. \]

Therefore,

\[ N=3+4+3=10. \]

Step 2: Calculate the mean.

\[ \bar{x} = \frac{\sum fx}{N} = \frac{3(5)+4(15)+3(25)}{10} = \frac{150}{10} = 15. \]

Step 3: Compute the variance.

\(x\)	\(f\)	\(x-\bar{x}\)	\(f(x-\bar{x})^2\)
\(5\)	\(3\)	\(-10\)	\(300\)
\(15\)	\(4\)	\(0\)	\(0\)
\(25\)	\(3\)	\(10\)	\(300\)

Thus,

\[ \sum f(x-\bar{x})^2=600. \]

Hence,

\[ \text{Variance} = \frac{600}{10} = 60. \]

Therefore,

\[ \boxed{60}. \]

Hence, the correct option is \[ \boxed{(D)}. \]