- 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
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%).
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 |