Discussion
The function f(x) represents a normal distribution whose standard deviation and mean are 1 and 5, respectively. The value of f(x) at x = 5 is
Answer: D
The value is 0.398.
The normal density is \(f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{-(x-\mu)^{2}/2\sigma^{2}}\). At the mean, \(x = \mu = 5\), the exponent is 0, so with \(\sigma = 1\):
\[ \begin{aligned} f(5) &= \frac{1}{\sqrt{2\pi}} \approx \frac{1}{2.5066} \\ &\approx 0.399 \end{aligned} \]
The density is greatest at the mean, so it cannot be 0.0. 0.159 is \(1/2\pi\), which omits the square root, and 0.282 is \(1/(2\sqrt{\pi})\), the value for \(\sigma = \sqrt{2}\), not 1.
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