- CFA Exams
- CFA Level I Exam
- Topic 1. Quantitative Methods
- Learning Module 4. Common Probability Distributions
- Subject 7. The Standard Normal Distribution
CFA Practice Question
As part of a promotion for a new type of cracker, free trial samples are offered to shoppers in a local supermarket. The probability that a shopper will buy a packet of crackers after testing the free sample is 0.20. Different shoppers can be regarded as independent trials. Let X be the number among the next 100 shoppers who buy a packet of the crackers after tasting a free sample. The probability that fewer than 30 buy a packet after tasting a free sample is approximately ______.
B. 0.9938
C. None of the above
A. 0.20
B. 0.9938
C. None of the above
Correct Answer: B
X, the number among the next 100 shoppers who buy a packet of the crackers after tasting a free sample, has approximately an N(20, 4) distribution. You need to evaluate P(X <= 30), which equals 0.9938.
User Contributed Comments 21
User | Comment |
---|---|
humphrey | do I have a Z table to look up during exam? |
murli | From where you got 4? |
katybo | looks like binomial to me. |
LondonBoy | mmm... also not sure where 4 came from? |
nsmwaura | Yikes... this scares me |
lawrence | 4 is the standard deviation. [100 x 0.2 x (1-0.2)]^(1/2) = 4. N(m, std) is the convention for standard normal distribution with m as the mean adn std as the standard deviation. |
danlan | Binomial will be approximated by normal distribution. |
cwrolfe | Yes you do need a z-table...or you can extrapolate... 1st calculate mean = p(x) times n = .2 x 100 = 20 2nd calculate std. deviation (see "lawrence" above) = 4 3rd convert 30 to a z-value above mean = (30-20)/4 = 2.5 std deviations 4th, look up this z-value on a table or know that 99% of values of x will fall within +/- 2.58 std deviations of the mean (it will be slightly less than 99% because of our 2.5 z-value, but close enough) 5th take half of this amount as our probability above the mean of 20, and you get 49.5% 6th add that to the 50% probability below the mean and you get 99.5%, plus we know its a little lower than this because of our assumption is step 4. |
khush | Yes, you get a book of tables in the exam |
bobert | Crwolfe, I think you added an extra step. Do steps 1, 2, 3. For 4, 2.5 on the Z-table = .4938 Then recognize that the Z table gives values from 0 to 4, not negatives. Since this is a positive Z-score simply add the 50% from the below 0, or the negative Z-score side which is completely encompassed. p(x)= x<30. That gets you the exact answer. Your way also works though as none of the other answers were even close. |
Tomcat82 | mean of binomial random variable=np=100*0.2=20. variance of binomial random variable = np(1-P), where n=100, p=.2. var=100*0.2*0.8=16. Now you want the standard deviation, which is (16)^(1/2)=4. Now the z-score=(x-M)sd, which is (30-20)/4=2.5. Look up the probablility for the z-value on the z-table, which is equal to 0.9938 |
StanleyMo | This question include binomial theorem and standard normal distribution. nice question. |
Yurik74 | Using binomianl got pretty much the same result 0.9948. |
JKiro | good question... thanks Tomcat82 |
bantoo | great explanation tomcat! |
gulfa99 | Tomcat82...i am lost..100*0.2*0.8 where did 0.8 come from? |
gulfa99 | oops i got it..thanks |
Kaloyan | Tomcat82, excellent explanation! |
sgossett86 | Gulfa .8 = 1-P from multiple variable bernoulli formula to get variance. sigma^2=np(1-p) |
Rachelle3 | I do it like Tomcat82 but I find this power ^ 0.5 thing weird just do sq root of 16 = 4 any time you see power ^ half or 0.5 or 1/2 like I have seen on this site it is BETTER to sq root same thing OK? Do not make it longer than it needs to be I am sure you have this button on your calculators!!! SQ ROOT PLZ |
ashish100 | Rachelle3 most people use that sq root function i'm sure. its just easier to type ^.5 thatn whatever the sign for sq root is on here. see, i dont even know what the sign for sq root is on here. |