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:
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!
@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.
- You must be logged in to reply to this topic.