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**CFA Practice Question**

The mean amount of gasoline and services charged by Key Refining Company credit customers is $70 per month. The distribution of amounts spent is approximately normal with a standard deviation of $10. What is the probability of selecting a credit card customer at random and finding the customer charged between $70 and $83?

A. 0.1962

B. 0.4032

C. 0.3413

**Explanation:**z = (x-u)/sigma. z1 = 70 - 70/10 = 0 and z2 = 83 - 70/10 = 1.3. For z = 1.3, the area under the curve is 0.4032.

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**User Contributed Comments**
8

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

ahan |
Since we can't use z-table in the exam, I don't think this type of question will be on the exam. |

Flyfree |
we will be provided w/ a set a z-table including the numbers we need. |

lwang014 |
r u sure? |

gweiden |
This problem can be solved by approximation. You have to draw out the bell curve and plot the area. You know that 70 (the mean) to 80 equals one standard deviation. This represents 34% of the area. 83 is just above this so you can eliminate answers A. C would be nearly the entire half of the curve so that can be eliminated. This leaves B as your answer. |

jackwez |
i did qweiden's method as well... if you are in a rush just draw a picture. |

panvino |
Definitely agree, I always draw out a quick diagram for these and probability type questions. 99% success rate! |

Nitishm |
How do they get 0.4032 by looking at z table. I looked at 1.3 in z table and get 0.9032? |

davcer |
nitishm you have to eliminate 50% below the mean as the question is asking for probability to the right, 0.9032-0.50 =0.4032 |