CFA Practice Question

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CFA Practice Question

The random variable X denotes the time taken for a computer link to be made between the terminal in an executive's office and the computer at a remote factory site. It is known that X has a normal distribution with a mean of 15 seconds and a standard deviation of 3 seconds. P(X > 20) has value (choose the closest option) ______.

A. 0.048
B. 0.952
C. 1.67
Correct Answer: A

To compute P(X > 20), we must first find the z-score of 20. We compute this z-score as z-score = (20 - μ)/σ = (20 - 15)/3 = 1.67.

Then we need to find the area to the right of this z-score. From the z-score table, we find that the area to the left of 1.67 is 0.9525. To find the area to the right of 1.67, we simply subtract 0.9525 from 1, which yields 0.0475, which, after rounding, is 0.048.

User Contributed Comments 6

User Comment
tanyak Since we don't have z-tables on the exam, you know 1.65 is for 0.95, so 1-0.95 = 0.05, so 0.048 is the closest
octavianus 1.65 for a Z-score is equivalent to 1.65 standard deviations.

1.65 standard deviations is equal to a 90% confidence interval.

5% left tail + 5% right tail = 10% error

P(Z>20) is in the right tail, so the answer is approximately 5% without using Z-tables, which is close to answer A, 0.048
MICHAEL1982 It is totally unlikely to have a prob of 95,25% as the mean is 15 seconds. The most obvious answer without any calculation is A.
tschorsch C is impossible in this universe

P(X > mean+4) must be less than .5, thus B is impossible

so it must be A
sgossett86 obvious
ashish100 You don't NEED to use the z table for this.
We are looking for the area on the right side of the mean which will obviously be less than .50.

.048 is less than .50 and .952 is greater than .50. So it can't be .952 or 1.167 (which is just the z score)
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