CFA Practice Question
There are 434 practice questions for this study session.
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
Correct Answer: A
User Contributed Comments 16
|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|