 This topic has 3 replies, 2 voices, and was last updated Dec204:11 am by Sophie Macon.

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Up::10
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 â€“ F(2) = 1 â€“ 0.9772 = 0.0228.
(Study Session 3, Module 9.2, LOS 9.k)

Up::4
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 zdistribution applies here, so if random variable is 2 standard deviations above the mean, this means that the zscore is 2.
If you look at the ztable (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 zvalue being looked up or calculated, this means that the probability of P(Z>2) is 10.9772 = 0.0228.
Hope this is clear!

Up::1
Thanks for the feedback. I actually noticed it upon rereading 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.

Up::1
@pcunniff – there are 6 critical zvalues you need to memorize (at outlined here https://300hours.com/thefreecfastudymaterialslist/#probabilitytables).
I think the question is cheeky in the sense that to those who are familiar with the 6 critical zvalues and normal distribution, the answer is clear even without any referral to the ztables. 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.


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