- CFA Exams
- CFA Level I Exam
- Topic 1. Quantitative Methods
- Learning Module 4. Common Probability Distributions
- Subject 7. The Standard Normal Distribution

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**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.

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**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. |