DSE 2009 Question 3: Random variables and standard distributions

$\displaystyle\int_{1.96}^{\infty} e^{-x^{2}/2}\,dx$ is approximately

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∫_(1.96)^∞ e^(-x²/2) dx is approximately (A) 0.025 (B) √(2π) (C) 0.025/√π (D) 0.025√(2π)

Also written as: integral_(1.96)^(inf) e^(-x^2/2) dx is approximately

Random variables and standard distributions has come up 173 times in 7 exams since 2005.

This question is on The normal distribution, part of Random variables and standard distributions in Statistics.

The bell curve: standardising, tail probabilities and approximations.

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Plain-text version

∫_(1.96)^∞ e^(-x²/2) dx is approximately (A) 0.025 (B) √(2π) (C) 0.025/√π (D) 0.025√(2π)

Also written as: integral_(1.96)^(inf) e^(-x^2/2) dx is approximately

Random variables and standard distributions has come up 173 times in 7 exams since 2005.

Question from the DSE 2009 paper. Solutions are TestStop's own. Not affiliated with or endorsed by any examination authority.