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# normally distributed random variable

Hello All. Can someone help me figure out this problem? I can't seem to understand the explanation. Thanks!

The probability that a normally distributed random variable will be more than two standard deviations above its mean is:

A)

0.9772.

B)

0.0228.

C)

0.4772.

Explanation

1 – F(2) = 1 – 0.9772 = 0.0228.

(Study Session 3, Module 9.2, LOS 9.k)

• Hi @pcunniff , this is a standard normal distribution, i.e. it is a normal distribution with a mean of 0 and a standard deviation of 1.

The z-distribution applies here, so if random variable is 2 standard deviations above the mean, this means that the z-score is 2.

If you look at the z-table (you can download probability tables you need for CFA exams for free here), where z value is 2, you'll get 0.9772 (where P(Z<2)).

If you recall that the table gives probabilities of the form: P(Z < z), i.e, probabilities that are less than the z-value being looked up or calculated, this means that the probability of P(Z>2) is 1-0.9772 = 0.0228.

Hope this is clear!

• Thanks for the feedback. I actually noticed it upon re-reading the LOS. Will we have to memorize the Z table stats or will they provide those tables? Didn't look like there was an option for this particular problem.

• @pcunniff - there are 6 critical z-values you need to memorize (at outlined here https://www.300hours.com/articles/the-free-cfa-study-material-list#probabilitytables).

I think the question is cheeky in the sense that to those who are familiar with the 6 critical z-values and normal distribution, the answer is clear even without any referral to the z-tables. This is because C isn't an option, 2 std deviation above the mean wouldn't be around 50% of the distribution.

And if you know that P(Z<1.645) = 95%, and P(Z<2.33)= 99%, you know that P(Z<2) is between 95% and 99% (i.e. 97% here). But P(Z>2) has to a tiny number. So the answer is B.