- CFA Exams
- CFA Level I Exam
- Topic 1. Quantitative Methods
- Learning Module 5. Sampling and Estimation
- Subject 3. Standard Error of the Sample Mean

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**CFA Practice Question**

The SAT scores of entering freshman at University X have a N(1200, 90) distribution. The SAT scores of entering freshman at University Y have a N(1215, 110) distribution. A random sample of 100 freshmen is taken from University X, and x-bar, the sample mean of the 100 scores from University X, is computed. The probability that x-bar is greater than 1215 is ______.

B. 0.5000

C. 0.4325

A. 0.0475

B. 0.5000

C. 0.4325

Correct Answer: A

x-bar has a N(1200, 90/100

^{1/2}) distribution. You now need to evaluate P(x-bar > 1215). The result is 0.0475.###
**User Contributed Comments**
11

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

vincenthuang |
(1200-1215)/9=1.67--->z check z table z(1.67)=.9525 P(x)=1-.9525=.0475 |

sivenkova |
Why 1200-1215 and why divided by 9, please? |

capform |
90 divided by square root of N (100) |

surob |
Look at it this way: (90^2/100)^(1/2) |

JimM |
Without a z-table, 1.67 z-score (as vincenthuang showed) is real close to a 90% confidence interval (1.645). 5% lie above that interval, so answer A is correct. |

boegs |
If a normal distribution is completely described by its mean and variance i.e. N(mean, variance), why do we not take the square-root of the variance (90) in the calculation of the standard error? |

sgossett86 |
Yeah seriously! Shouldn't it be calculated 90^.5/100^.5 = .949 = sigma and the z calculation should be 1215-1200.. but that isn't important. i guess whatever. i get it in this context. |

srdgreen |
Can someone please explain the N(1200,90) format? |

raghu2gd |
Any normal distribution is denoted by N( Mean, variance). That means 1200 is mean and 90 is variance. |

mzaheedihm |
Please check the solution of the next question. It is actually N (Mean, Standard Deviation) |

chrismoore |
Concur with mzaheedihm - can we please verify the proper interpretation of N(1200,90)? |