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**CFA Practice Question**

The sum of the squares of 1,200 observations equals 9,830. The sum of the observations equals 1,510. The population standard deviation of the observations equals ______.

A. 5.31

B. 4.22

C. 2.57

**Explanation:**For N observations, it is easy to show that: population variance*N = (sum of squares) - N*(mean

^{2})

The mean equals 1,510/1,200 = 1.258. Hence, population variance = (9,830 - 1,200*1.258

^{2})/1,200= 6.608. The standard deviation then equals sqrt(6.608) = 2.57.

Note: You should be careful to note the difference between population variance and sample variance. The formula for sample variance is: sample variance*(N-1) = (sum of squares) - N*(mean

^{2}).

You can expect an exam question that asks for population variance, with the choices given containing both the population and the sample variances.

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**User Contributed Comments**
17

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

Rguerra |
Where did you get that equation from? pop var*N = (sum of squares) - N*(mean2)? Shouldn´t: pop var*N = Sum(X - mean)squared? In this case: pop var*N = (sum of squares) - 2*X*mean) + mean(square). I don´t see how you get to the first equation. |

PedroEdmundo |
What about 2*mean*SumX? |

bhawa |
agreed with Rguerra and Pedro, can someone please explain from where the equation came from in the answer given?? |

SunilRaja |
This is how it shud be done. variance = (Sum(X-Mean X)squared)/N Now expand (A-B)squared = A squared + B squared - 2AB So variance = {Sum(Xsquared)+Sum(Mean Xsquared) - 2 Sum(X*Mean X)}/N Now Sum(Mean Xsquared) = Sum(Mean Xsquared * 1) = Mean Xsquared * Sum(1) = Mean Xsquared * N since Mean Xsquared is not affected by Sum and is constant and Sum(1) = 1+1+.....N times = N Similarly Sum(X*Mean X) = Mean X * Sum(X) Mean X = Sum(X)/N = 1510/1200 = 1.26 We are given Sum(X) and Sum(Xsquared) (MeanX)squared = 1.26*1.26 = 1.5876 variance={9830 + 1.5876*1200 - 2*1.26*1510}/1200 = 7929.92/1200 = 6.608 standard deviation = square root(variance) = 2.57 |

antony |
Thanks for the explanation. That helped. |

surob |
Wow, hope this is not going to be a question in the real test. |

Xocrevilo |
SHORT SOLUTION: Given that the sample is large relative to the population (i.e. (1200/1510) = 79.5%), I assumed that the sum of the squares of of mean variances for the sample is approximately equal to that of the population. Therefore, population variance = sum of squares / n population variance = 9830 / 1510 = 6.51 square root = 2.55 Though this is not the exact answer, it is different enough to A and B, and close enough to C, to confirm that C is correct. |

grezavi |
Xocrevilo is right, we are not expected to the Algebra, the logic fits for large sample theory |

apiccion |
Here's how I did it. Var(x) = sum(x-Xbar)^2/n <= sum(X)^2/n = 9830/1200 = 8.19 Therefore, std(x) <= sqrt(8.19) = 2.86 Looking at the answers, there is only one number which is less than 2.86. Pick that number. |

moritz |
Great solution, apiccion! Thx! |

7Ricky |
Variance = mean of squares - square of means = E[x^2]-E[x]^2 =(9830/1200) - (1510/1200)^2 =6.608 sqr rt=2.57 Not sure if the equation is in CFA text but should be in any intro stat book |

Ifi2703 |
Very useful eplanation by 7Ricky - will help for time management in exam knowing such tips...thanks! |

bidisha |
Thanks ricky |

Procbaby1 |
7Ricky that's really helpful! |

birdperson |
thanks ricky. |

Rsanches |
7Rick, awesome explanation. |

ashish100 |
7Ricky! Where are you been in the last 197 hours? (time i spent on AN) Thank you! |