Discussion
The variance value for an F-distribution with parameters V1 = 5 and V2 = 9 is :
Answer: A
The variance is 1.587.
For an F-distribution with \(\nu_{1}\) and \(\nu_{2}\) degrees of freedom (here 5 and 9; the variance exists only when \(\nu_{2}\) is greater than 4):
\[ \begin{aligned} V &= \frac{N}{D} \\ N &= 2\nu_{2}^{2}(\nu_{1} + \nu_{2} - 2) \\ D &= \nu_{1}(\nu_{2} - 2)^{2}(\nu_{2} - 4) \end{aligned} \]
\[ \begin{aligned} V &= \frac{2(81)(12)}{5(49)(5)} \\ &= \frac{1944}{1225} = 1.587 \end{aligned} \]
The mean of this distribution is \(\nu_{2}/(\nu_{2} - 2) = 9/7 = 1.286\), which none of the options gives; 1.378, 2.578 and 1.498 do not follow from the formula.
No comment is present. Be the first to comment.
Loading…
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development