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**CFA Practice Question**

A bell-shaped, symmetrical frequency distribution has a mean of 5 and a standard deviation of 2.5. The percentage of observations less than zero is about ______.

A. 2.5%

B. 5%

C. cannot be calculated

**Explanation:**Note that the mean is 2 standard deviations above zero. The fraction of observations in a bell-shaped, symmetrical frequency distribution which lie outside the 2-sigma range around the mean is about 5%. Since the distribution is symmetrical about the mean, the percentage of observations less than zero is approximately equal to 5%/2 = 2.5%.

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**User Contributed Comments**
16

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

shasha |
5% lie outside 2-sigma range is for normal distribution only. not all bell-shaped, symmetrical distribution are normal distribution. i believe we should use Chebyshev's Inequality (1-1/k sqrd) to get 25%, roughly 10% for each tail. |

tony1973 |
Based on the textbook: normal distributions are a family of distributions that have the same general shape. They are symmetric with scores more concentrated in the middle than in the tails. Normal distributions are sometimes described as bell shaped with a single peak at the exact center of the distribution. The tails of the normal curve extends indefinitely in both directions. That is, possible outcomes of a normal distribution lie between - infinite to + infinite. They may differ in how spread out they are. Therefore, the distribution in the q should be treated as a normal distribution. |

labsbamb |
don`t understand this one. anyone has any idea. |

ruckmani |
remember 95% of all observations fall 2 sd of mean. 2 sd of mean will range between 0 to 10 which will comprise of 95% of all values. that leaves only 5% of total observations above 10% and below 0%. Since the distribution is symmetrical about the mean, the percentage of observations less than zero is approximately equal to 5%/2 = 2.5%. |

dealsoutlook |
good explanation ruckmani..i know i would get this wrong on the exam |

Kuki |
but according to Chebyshevs's inequality, the minimum amt of values tht lie within 2sd of the mean are 75%. Therefore, leaving us with 12.5% on each side. Please correct me if i'm wrong! |

Kuki |
aaahh...got it! this one's a normal distribution. Thats why! |

Muskie |
I think you could have a bell shaped distribution that isn't normal, but then I think statistics is one of my biggest weaknesses... |

escempep |
If you are accurate, C is the good answer because z is note 2 but 1,96 and we have note the z-table. |

psos |
it says 'about' i.e an approximation |

MaresaJaden |
Remember: +/- 1 standard deviation = 68%, +/- 2 standard deviations = 95% and +/- 3 standard deviations = 98%. I believe bell shaped and symmetrical says we can use these estimates. |

7Ricky |
The CFA book says chebyshevs inequality sets a lower bound on a std interval not a upper bound. Meaning that at minimum that number is within a certain deviation, but it could go up to a 100%, meaning calculating the % beyond that could not be accurate. But maybe it is different with the Normal Distribution. The range would definitly be between 0-2.5 but I answered the actual # can't be calculated. |

Mikehuynh |
Normal distribution => 2SD = 95% => since it's 2-tailed => less than 0 is 2.5% |

moneyguy |
where does the zero come from???? |

CalebMast |
mean is peak of bell curve, i.e. that about which the curve oscillates. I biffed the 5% trap, but am def glad that I understood what the question was getting at, i.e. that 2.5 was 1 SD, and with 5 as 2 SD's, the question becomes how much of a curve lies outside of 2 SD's? Answer is 5%...just need to divide by 2 to get 1 of the two oscillating (about the mean...which, in this case is 5) outside of 2 SD values. Another questions...how much lies outside of 3 SD's? 1%...3 (really 2.5) SD's comprises 99% of bell curve. 3 SD is 99.7% or so. 68% is 1 SD, as stated in the previous question. |

anatoly87 |
It is not clear that a normal distribution is meant by "bell-shaped". It can also be a t-distribution with any degrees of freedome. As the above question does not indicate that clearly, C should be correct, or am I missing something? |