>Blog>FRM Qotd #2: Question of the Day

# FRM Qotd #2: Question of the Day

January 30, 2011

The answer to the previous question is –

Apply Bayes™ Theorem
P (New / Cure) = P (Cure / New) * P (New) / P (Cure) = 0.6 * 0.25 / 0.375 = 0.40

Here, is the Question of the Day

Let Z be a standard normal random variable. An event X is defined to happen if either z takes a value between â€“1 and 1 or z takes any value greater than 1.5. What is the probability of event X happening if N(1) = 0.8413, N(0.5) = 0.6915 and N(-1.5) = 0.0668, where N() is the cumulative distribution function of a standard normal variable?

1. 0.083
2. 0.2166
3. 0.6826
4. 0.7494

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