Question:medium

If \(X \sim N(\mu, \sigma^2)\). The maximum ordinate is at \(X = \mu\) and is given by:

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The peak value of any probability density function is its maximum ordinate.
For a standard normal distribution (\(\mu=0, \sigma=1\)), this peak value simplifies directly to \(\frac{1}{\sqrt{2\pi}} \approx 0.3989\).
  • \(\sigma \sqrt{2\pi}\)
  • \(\frac{1}{\sigma \sqrt{2\pi}}\)
  • \(\sigma \sqrt{2\pi} e\)
  • \(\frac{1}{\sigma \sqrt{2\pi e}}\)
Show Solution

The Correct Option is B

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