GARP.FRM.PQ.P1 What percentage of the distribution is not between A and B (garp16-p1-66)

[Natalia13]

Hi,

Can someone please explain the solution to the following problem in the GARP textbook:

Assume that a random variable follows a normal distribution with a mean of 80 and a standard deviation of 24. What percentage of this distribution is not between 32 and 116.

The answer: Prob( mean - 2 St dev<X<mean+1.5 St dev)=(0.5-0.0228)+(0.5-0.0668)=0.9104
1- 0.9104=0.0896

How did they get 0.0228 and 0.0668?
Thanks.

 

[David Harper CFA FRM]

Hi Natalia,

Without an HP20 (see here) or calculator with CDF functionality or Excel, you'd need a lookup table.

2.28% is P(Z <= -2) when Z ~ N(0,1); i.e., standard random normal. CDF refers to asking the question, what is the Pr(random variable will be less than or equal to some value X).
Per the symmetry of the normal (this you must be able to do!), P(Z<- = -2 ) = 1 - P(Z<= +2) because both are capturing symmetrical tails to the left/right of two standard deviations. Below P(Z<= +2) = 97.7%, so P(Z < -2) = 1 - 97.7%.

Similarly P(Z>+1.5) = 1 - P(Z <= 1.5) = 1 - 93.3%. I hope that helps!

 

[Natalia13]

Thank you, David!
I have HP 12C Platinum and was wondering how I should calculate it without any given table.
By the way, after searching about Z tables during the exam, I found your answer from 2012. Thanks again!
http://forum.bionicturtle.com/threads/z-table-during-the-exam.5443/

 

[David Harper CFA FRM]

Hi Natalia,

You can't find normal CDF with HP 12C (nor the TA BA II+); hence the exam must give a lookup or give the N(.) values directly.

Oh, yikes I should have searched first myself, thanks for locating prior answer Thanks ,

 

[king_bren777]

Hi All, Pretty sure the the answer is pretty obvious (Just not to me currently)

Where do the probability percentage's come from in this question? I get the standard deviation's from the mean but not sure how the .0228 and .0668 were come by?

66. Assume that a random variable follows a normal distribution with a mean of 80 and a standard deviation of 24. What percentage of this distribution is not between 32 and 116?
a. 4.56%
b. 8.96%
c. 13.36%
d. 18.15%
Answer: B
Explanation:
Prob(mean - 2*σ < X < mean + 1.5*σ) = (0.5 - 0.0228) + (0.5 - 0.0668) = 0.9104
Prob(mean - 2*σ > X or X > mean + 1.5*σ) = 1 - Prob(mean - 2*σ < X < mean + 1.5*σ) = 0.0896

 

[David Harper CFA FRM]

Hi @king_bren777 I moved your question to a pre-existing thread, see above

 

[king_bren777]

Thanks David. I get it. What threw me is they normally give snippets of the look up. I assume they will give us a look up for the exam if required.

 

[kansal7mba]

Hi David,

Can you please explain this part in detail?

Prob(mean - 2*σ < X < mean + 1.5*σ) = (0.5 - 0.0228) + (0.5 - 0.0668) = 0.9104

I know 80 - 32 = 48 means 2 STD DEV from the mean as the STD DEV is 24.
32 is 2 STD DEV from the mean of 80.

Thanks,
Sachin

 

[David Harper CFA FRM]

Hi @kansal7mba Given µ=80 and σ=24, we want 1 - Pr( 32 < X < 116) = 1 - Pr (-2.0 < Z< +1.5) given (32-80)/24 = -2.0 and (116-80)/24 = +1.5
Personally I think their expression (0.5 - 0.0228) + (0.5 - 0.0668) is confusing and doesn't exactly make sense; maybe I am missing something

Here is how I look at it:

• The question is What percentage of this distribution is not between 32 and 116? which translated into normal Z terms, we can see, is equivalent to What percentage of this distribution is not between Z = -2.0 and Z= +1.5?
  
• This is asking for the area in both tails. We can either get them directly with: Pr[Z < -2.0] + Pr[Z > 1.5] = Pr[Z < -2.0] + (1 - Pr[Z < 1.5]) = 2.28% + ( 1 - 93.319%) = 8.96%; or
• We can solve for the inside and invert: inside is Pr[Z < +1.5] - Pr[Z < -2.0] so the tails are given by 1 - (Pr[Z < +1.5] - Pr[Z < -2.0]) = 1 - (93.32% - 2.28%) = 1 - 91.04% = 8.96%. I suspect the given answer is not exactly logical, or perhaps presumes a Z truncated Z lookup table that somehow requires the operation. Not sure, but answer looks correct.

 

[RaDi7]

Hi David,

could you please explain the calculations in question 66 of the practice exam?



We got 2*σ since 80 - 32 = 48 = 2*24 and 1.5*σ since 116 - 80 = 36 = 1.5*24. But I don't get the rest of the calculation.

Thank you very much und kind regards!

 

[ShaktiRathore]

Hi,
Pr(X>116)+Pr(X<32) ..(116=mean+1.5*σ=80+36,32=mean-2*σ=80-48)
=Pr(z>(116-80)/24)+Pr(z<(32-80)/24) (convert to standard Normal)
=Pr(z>36/24)+Pr(z<-48/24)
=Pr(z>1.5)+Pr(z<-2)
=0.0668+0.0228
=0.0896
thanks

 

[Nicole Seaman]

Hi David,

could you please explain the calculations in question 66 of the practice exam?

View attachment 927

We got 2*σ since 80 - 32 = 48 = 2*24 and 1.5*σ since 116 - 80 = 36 = 1.5*24. But I don't get the rest of the calculation.

Thank you very much und kind regards!

Hello@RaDi7

Please note that I moved your question here, where this question has been discussed already

Thank you,

Nicole

 

[David Harper CFA FRM]

(garp16-p1-66)Thank you @Nicole Seaman
And please notice @RaDi7 in the original thread above, we do a couple of things to make it easier to find these Q&A

1. In addition to the green thread prefix (GARP.FRM.PQ.P1), the title ends with "(garp16-p1-66)"
2. The question is also tagged with garp16-p1-66. Tags can be searched here at https://forum.bionicturtle.com/tags/ and there is a natural hierarchy to these tag labels: you can just type "garp" and they start to appear. Thanks,

 

[RaDi7]

Hi,
Pr(X>116)+Pr(X<32) ..(116=mean+1.5*σ=80+36,32=mean-2*σ=80-48)
=Pr(z>(116-80)/24)+Pr(z<(32-80)/24) (convert to standard Normal)
=Pr(z>36/24)+Pr(z<-48/24)
=Pr(z>1.5)+Pr(z<-2)
=0.0668+0.0228
=0.0896
thanks

Thank you ShaktiRathore!

 

[RaDi7]

(garp16-p1-66)Thank you @Nicole Seaman
And please notice @RaDi7 in the original thread above, we do a couple of things to make it easier to find these Q&A

1. In addition to the green thread prefix (GARP.FRM.PQ.P1), the title ends with "(garp16-p1-66)"
2. The question is also tagged with garp16-p1-66. Tags can be searched here at https://forum.bionicturtle.com/tags/ and there is a natural hierarchy to these tag labels: you can just type "garp" and they start to appear. Thanks,

Hi Nicole, hi David,

thank you for the moving my question to the right thread. Sorry, I couldn't find this thread. :-( Searched for GARP Practice exam 2016 question 66.
Best regards!

 

[Nicole Seaman]

Hi Nicole, hi David,

thank you for the moving my question to the right thread. Sorry, I couldn't find this thread. :-( Searched for GARP Practice exam 2016 question 66.
Best regards!

@RaDi7

You're welcome! I find that searching the first sentence of the actual question brings up the thread easily, but as David mentioned above, our system in the forum for organizing the GARP questions has made these questions easier to find.

Nicole