Howdy, Stranger!

It looks like you're new here. If you want to get involved, click one of these buttons!

CFA Events Calendar

View full calendar

CFA Events Calendar

View full calendar

normally distributed random variable

MilwaukeePosts: 30 Associate

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)

• Posts: 1,999 Sr Partner

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!

• MilwaukeePosts: 30 Associate

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.

• Posts: 1,999 Sr Partner

@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.