CFA Practice Question

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CFA Practice Question

The amount of time that it takes to complete a statistics exam has a skewed left distribution with a mean of 60 minutes and a standard deviation of 6 minutes. If 25 students are randomly sampled, determine the probability that the sample mean of the sampled students is less than 70 minutes.

A. approximately 1
B. 0.1252
C. 0.3345

User Comment
billou the z-score for 70 min is 8.33
the z-score for 99% confidence is 2.58
Khadria 1. This is not a NORMAL distribution as N < 30. So you can't apply the formula for Z-score

2. The mean is 60 and standarad deviation is 6 which means the max time taken can be 66 minutes which is < 70
khush The sample is from a normal distribution.
stratus The original sample distribution is non-uniform. However, N=25 is pretty close to 30, so you can say it's roughly uniform.

The actual shape shouldn't matter too much anyway since the z score is so large, minor deviations from standard distribution can be ignored.
bobert The shape of the distribution does not matter. As N increases the shape becomes more and more uniform.

I got tripped up on the N<30 too, but seeing as the answer was not there, I just went with it as being a normal distribution.
chamad How do they get 1? What does that stand for? Help!
chamad Ok go to questions under d and you'll understand. I thing this question should been asked in reading d(Standard error...)
MFTIOA standard error is 6/sqrt(25) = 1.2

(70-60)/1.2 = 8.333, that is a very large z value and the chance that the mean is below 70 is approximately 1
cslau83 i)Mean of the distribution X will be equal to the mean of the population from which the samples are drawn.
ii) Since population mean is 60min, prob that the sample mean is less than 70 is approx 1.
Couldn't be 0.1 or 0.33. doesn't make sense.
NikolaZ I got 70-60/6 for the z-score

this yields 1.67, which is the 90% confidence interval, of course we need to incorporate the bottom 5%, so thus 95% of the distribution falls beneath 70 minutes and 5% above 70 minutes

So i got 95%

...
I thought when you are finding z-score the formula is z-score = X - u / standard deviation

Someone help as my logic is clearly wrong
NikolaZ Never mind, just read the next section and learned about the standard error :)
johntan1979 It's not normal, so don't use Z score!

The logic for this question is mean=60

60 and below is 50% of the population, so look at the answers available. 70 and below HAS to be more than 50%.
gill15 No...you can use CLT here...its not 30 but who cares....a z score of 8.33 means adding a couple to the sample size is not going to change the answer much....
sgossett86 I just used logic here... and ruling out.. I see this problem and see that N<30 so we can't use it as normal distribution. Then the proposed mean is more than a standard deviation north of the population mean. So it's highly unlikely it would occur.
Sagarsan88 It has a Std dev of 6...with mean 60...thus safe to assume max value is somewhere 6+60=66. Since it is left skewwed the lowest value is lower than 60-6=54. so the qstn asks chances of number being less than 70...as we see most of my no is below 66...thus 1. thats what i used
Konstantis Johntan1979 below 60 is not 50%. The distribution is not normal. Think this can be approached by a normal distribution ( even the N is less than 30) and the z is 1.66. and prob 0.9515