- 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
The distribution of the annual incomes of a group of middle management employees approximated a normal distribution with a mean of $37,200 and a standard deviation of $800. About 68 percent of the incomes lie between what two incomes?
A. $34,800 and $39,600
B. $36,400 and $38,000
C. $35,600 and $38,800
Explanation: 0.68/2 = 0.34. The z-value for 0.34 is 1. x = u ±- z*sigma. So, x = 37200 ± 1*800. So, x is between 36,400 and 38,000.
User Contributed Comments 7
| User | Comment |
|---|---|
| tanyak | Why are we dividing .68/2? |
| SuperKnight | Because 68% of values fall within 1 standard deviation.. Therefore there is 34% on each side of the mean. |
| SCBAnalyst | how do we find the z value? |
| Jmcb | An easier way is just to know that 68% of the values will be within one std of the mean. Since the std is 800, just subtract it and add it to the mean salary. |
| thekobe | rule of thumb 68% = 1z 90% = 1.645z 95% = 1.96z |
| ioanaN | actually, the table gives z-value for .34 = 0.955 z=1 is for alpha .1587 with is a 68.26% confidence interval. I'm confused from what table they took the number from |
| kingirm | same here iona. but i guess the answer should be like 1 standard deviation to the left and right of the mean. so its 37,2 plus/minus 800 |