- CFA Exams
- CFA Level I Exam
- Study Session 3. Quantitative Methods (2)
- Reading 9. Common Probability Distributions
- Subject 8. The Standard Normal Distribution

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**CFA Practice Question**

For x, a random variable from a normal distribution with mean 40 and standard deviation 5, P(28 < x < 38) = ______ (to nearest 0.1%).

B. 34.5%

C. 33.6%

A. 35.3%

B. 34.5%

C. 33.6%

Correct Answer: C

P(28 < x < 38) = 0.3446 - 0.0082 = 0.3364 or 33.6% (to the nearest 0.1%).

The z-score for 38 is (38 - 40)/5 = -0.4. The table value for -0.4, P(x < 38), is 0.3446. The z-score for 28 is -1.2 and the table value is 0.0082.

P(28 < x < 38) = 0.3446 - 0.0082 = 0.3364 or 33.6% (to the nearest 0.1%).

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**User Contributed Comments**
18

User |
Comment |
---|---|

woojacky |
z table? |

ENDER |
the z-table for 28 is -2.4 (28-40)/5=-2.4 |

doze |
Do I trully need to memorize all the Z-table....? Miserable brazilian life. |

BonnieG |
yeah, z-score is -2.4? this is wrong! also, how can we possibly memorize a z-table, that should be provided. |

tenny45 |
the cfa website (faq) says that we'd need to "estimate" it based on whatever is given (maybe a portion of the table would be given?). But No complete tables will be provided. |

kaliokale |
if CFA wants us to "estimate"... the answers sholdn't be that close to each other. |

bobert |
Yeah, without the table this is impossible. |

epizi |
if it was so easy to memorize this table they wouldn't had provided it in the curriculum notes, at the Appendix B exactly |

Becker |
So P(x>28) is -1.2 or 2.4 ? Why ? |

StanleyMo |
strange,i always get -2.4.. |

Murrayman |
You should get -2.4. Also, the z score mentioned above, 0.0082, is the z-score for -2.4, not -1.2. |

freda |
Why not A? ie why not 0.3446+0.0082? |

anastasiya |
I agree with Murrayman. freda, here is the explanation you need: P(-2.4<z<-0.4)=P(z<-0.4) - P(z<-2.4)= 0.3446 - 0.0082=0.3364 |

bundy |
You are looking for the area between 28 and 38 The corresponding Z score for 28 is -2.4 and prob that observation lies below that is .0082. The corresponding Z score for 38 is -.4 and the prob that observation lies below that is .3446. SO the area between the two is the prob of the observation falling btw them hence .3446 - .0082 = 33.6% |

bkballa |
you guys worry too much. There's no way you would be asked to do this without a table. Also keep in mind that the whole quant section has about 24 ques so there might only be 1 or 2 on this particular topic. |

johntan1979 |
Agree with bkballa. |

PSVC |
Z- table extracts are provided for questions to which they will be applied. |

ashish100 |
WE ARE SCREWED!!!!! Everybody panic so I can score above the 60th percentile. :D jk |